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Shooting to the Moon

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  Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-15 05:20 -0700
    Re: Shooting to the Moon Wally W. <ww84wa@aim.com> - 2018-06-15 08:55 -0400
      Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-15 08:34 -0700
        Re: Shooting to the Moon Sergio <invalid@invalid.com> - 2018-06-15 13:42 -0500
          Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-15 14:26 -0700
          Re: Shooting to the Moon benj <benj@nobody.net> - 2018-06-15 19:17 -0400
            Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-15 18:21 -0700
              Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 05:55 -0700
              Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-17 17:04 -0700
                Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 18:13 -0700
                  Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-17 18:44 -0700
                    Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 19:07 -0700
                      Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-18 13:35 -0700
                        Re: Shooting to the Moon Odd Bodkin <bodkinodd@gmail.com> - 2018-06-18 21:02 +0000
                          Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-19 10:39 -0700
                            Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-19 15:35 -0700
                              Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-19 21:09 -0700
                              Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-20 10:49 -0700
                                Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-20 16:18 -0700
                                  Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-21 15:05 -0700
                                    Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-21 16:57 -0700
                        Re: Shooting to the Moon Sergio <invalid@invalid.com> - 2018-06-18 16:18 -0500
                          Re: Shooting to the Moon Steve Carroll <frelwizzen@gmail.com> - 2018-06-19 10:40 -0700
                          Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-19 15:20 -0700
                        Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-18 22:40 -0700
        Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-15 14:29 -0700
          Re: Shooting to the Moon Wally W. <ww84wa@aim.com> - 2018-06-16 11:10 -0400
            Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 15:13 -0700
            Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 19:04 -0700
              Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 20:05 -0700
                Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 21:18 -0700
                Re: Shooting to the Moon Wally W. <ww84wa@aim.com> - 2018-06-17 05:35 -0400
                  Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 03:08 -0700
                    Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 06:00 -0700
                      Re: Shooting to the Moon Wally W. <ww84wa@aim.com> - 2018-06-17 13:28 -0400
                        Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 16:15 -0700
      Re: Shooting to the Moon benj <benj@nobody.net> - 2018-06-15 11:48 -0400
    Re: Shooting to the Moon numbernumber1964@gmail.com - 2018-06-18 15:29 -0700
    Re: Shooting to the Moon numbernumber1964@gmail.com - 2018-06-19 12:21 -0700
      Re: Shooting to the Moon Odd Bodkin <bodkinodd@gmail.com> - 2018-06-19 21:47 +0000
    Re: Shooting to the Moon numbernumber1964@gmail.com - 2018-06-19 12:29 -0700
    Re: Shooting to the Moon furthermore7990662@gmail.com - 2018-06-19 16:03 -0700
    Re: Shooting to the Moon furthermore7990662@gmail.com - 2018-06-19 16:03 -0700
    Re: Shooting to the Moon carleto4157990662@gmail.com - 2018-06-19 16:05 -0700
    Re: Shooting to the Moon carleto4157990662@gmail.com - 2018-06-19 16:06 -0700

Page 2 of 3 — ← Prev page 1 [2] 3  Next page →

#691496

On Friday, June 22, 2018 at 8:05:20 AM UTC+10, Edward Prochak wrote:
> On Wednesday, June 20, 2018 at 7:18:25 PM UTC-4, Arindam Banerjee wrote:
> > On Thursday, June 21, 2018 at 3:49:35 AM UTC+10, Edward Prochak wrote:
> > > 
> > > Agree, this is not nuclear level EMP, but a 25 million amp
> > > circuit over 2 KM. (How big is that Magnetic field?)
> > > So I am still curious for an answer to my question:
> > 
> > Whatever big that is, it will be contained locally,
> > affecting mostly the armature. Biot-Savart law will
> > show pretty clearly how much the magnitude of the
> > magnetic field will be at any distance. It is a
> > inverse distance law relation.
> > https://en.wikipedia.org/wiki/Magnetic_field
> >  tells us:
> > 
> > B=muI/(2*pi*r)
> > B=4piE-7*25E6/(2*pi*r)
> > B=5/r where r is in meters.
> > 
> > Don't think there is much to worry, what.
> > 
> >  I have no doubt there will be thus a fairly strong
> > magnetic field generated for 0.036seconds, so proper
> > precautions need to be taken for those wearing say
> > pacemakers or other electronic implants within say 20 meters.
> 
> Really that short a distance? I trust you Arindam.
> But I may check it offline anyway 8^)
> 
> Ed

When you do generate a large current like 25 million A, it is not that it will be 25 million in zero seconds. There is an inductance involved in the circuit, and that will slow down the rate of rise of the current. It is the rate of rise of the current, which relates to the rate of rise of the magnetic field. It is the rate of the rise of the magnetic field, which creates the induced voltage that will burn your electronics - with the voltage spike that will cause dielectric breeakdown; meaning sparking; meaning destruction of circuits where there are very small distances between the conductors as electronics circuits must be. In a nuclear explosion, the rate of change of the magnetic field is extremely rapid, much more than this 25 millionamps created over say 10-50 milliseconds. There we are looking at much vaster energies meaning much higher changes of magnetic field over much shorter times.

No doubt all conducting stuff worn by personnel should be kept away, and the electronics equipement kept in Faraday cages when firing is on. 

Not that I am particularly keen about making a 1000m gun firing Mach160 or more. I am interested in turning that gun into a motor by arresting the much heavier low speed bullet, and making a two-stroke engine, as I have explained in my video films. Still, the gun if made of gold or silver will be very impressive to look at, and all the gold and silver will be very safe and make money as a tourist attraction. 

So will Trump be the man with the golden gun? Will all the gold in Fort Knox be melted to make this gun? It cannot be safer, than to melt it all in one huge piece everyone will see and long for, touch, but regretfully, cannot take, not a milligram! 

Cheers,
Arindam Banerjee

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#691111

On 6/18/2018 3:35 PM, Edward Prochak wrote:
> On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
>> On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
>>> On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
>>>> On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
>>>>> On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
>>>>>> Did 3200 amps with own two brown hands
>>>>>
>>>>> you are off by 3 orders of magnitudes!
>>>>
>>>> No.
>>>>
>>>> I shorted with a 16.5 V 250F capacitor bank with a low milliohm
>>>> circuit resistance to get 3200 amps on the average.  The 3200 A
>>>> was calculated from the drop in the voltage across the capacitor
>>>> bank. Say when it dropped from 16.5 V to 7.5 V then the charge
>>>> discharged is 250 (16.5 - 7.5) = 2250 coulombs.
>>>>
>>>> Agreed?
>>>
>>> Yes, but
>>> That is still 3 orders of magnitude less than the
>>> 25,000,000 Amps needed for the 1km rail gun.
>>> I don't know what materials will carry that high a current.
>>>
>>> ED
>>
>> In my PhD work, my guide asked me to do a worksheet showing
>> how much a conductor of a given dimension will heat up over
>> time with current.
>>
>> That I did, and with that help I designed my 3200A gun. It does not become hot by the way, thanks to the dimensioning.
>>
>> Extending that to my motor (the IFE) I worked out that with a 500mm*500mm copper cross section, the heat increase would be manageable for 3 million amperes.
>>
>> I can use the same worksheet to work out the dimensions for 25million amps.
>>
>> My guess is that it would be around 2000mm*1000mm.
>>
>> That is a lot of copper, for 2000m rails, but it will carry 25 million amps. Note, this are rough figures. We also need to know that the whole barrel will be vacuum sealed and as cold as possible. Colder the temperature, better the conductivity, lower the contact resistance, etc. I am not talking about superconductivity, just higher conductivity with lower temperature.
>>
>> I am giving over-the-top back of the envelope estimates. Naturally, when we get down to the calculas and do the fine slicing, the actual figures for a given design will emerge. Reducing the contact resistance is the key. 
>>
>> Another point - we need existing defunct mineshafts going down 1-2Km for this gun. No point creating a 1000 m vertical structure, what! Natural side-support, in a hole!
>>
>> I think we can with a limited budget make a gun that can launch small satellites into space. It can compete with rocket launches. 
>>
>> Cheers,
>> Arindam Banerjee
> 
> Earlier you Arindam) posted:
> "25 million amps is nothing but 25000 volts divided by 1 milliohm. 
> No problem at all, both in theory and practice."
> 
> the total resistance needs to be 1 milliohm (or less)
> 
> So what is the resistance?
> to obtain a voltage drop of 25000V over 2 kilometers,
> R=rl/A
> l - length 2 km (=2000 m)
> A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
> r - rho resistivity for copper is 0.000 000 017 ohm-meter
> 
> R= 0.000 017 ohm = 0.017 milliohm.
> 
> I'm impressed. about a factor of 5 margin of safety.
> 
> Now 2 Km by 2m^2 is a volume of 4000 m^3
> Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
> total weight 35,840,000 kg
> Now I'm impressed again. Cost is a little over $25 million.
> 
> Of course it will be the assembly costs that swamp the material costs.
> You'd likely want the rails to be continuous. so you'd want to caste the rails in place maybe?Or can it be multiple parallel sheets of copper?
> (how big is a roll of copper 1km long?)
> 
> Physically doable, though challenging. that the navy gave up on their
> railgun, suggests there are other challenges we haven't identified.


Interesting on why the navy gave up, it was trying to get 2K amps
reliablely from rail to projectile, 0r 32 megajoles, but there always
was ablation where the current melted the rails and Sabot, even with
conductive lubes.  transfering large current on moving/sliding contact

nice pix here, all the flame is metal burning from rails and sabot.
http://www.thefiscaltimes.com/2016/06/04/Navy-Abandoning-Railgun

> 
> Nothing better be around the launch tip, that is going to be
> one hell of a sonic boom!
> 
> Ed
> Meaning the resistance needs to be 
> 

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#691225

On Monday, June 18, 2018 at 2:18:30 PM UTC-7, Sergio wrote:
> On 6/18/2018 3:35 PM, Edward Prochak wrote:
> > On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
> >> On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
> >>> On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> >>>> On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> >>>>> On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> >>>>>> Did 3200 amps with own two brown hands
> >>>>>
> >>>>> you are off by 3 orders of magnitudes!
> >>>>
> >>>> No.
> >>>>
> >>>> I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> >>>> circuit resistance to get 3200 amps on the average.  The 3200 A
> >>>> was calculated from the drop in the voltage across the capacitor
> >>>> bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> >>>> discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> >>>>
> >>>> Agreed?
> >>>
> >>> Yes, but
> >>> That is still 3 orders of magnitude less than the
> >>> 25,000,000 Amps needed for the 1km rail gun.
> >>> I don't know what materials will carry that high a current.
> >>>
> >>> ED
> >>
> >> In my PhD work, my guide asked me to do a worksheet showing
> >> how much a conductor of a given dimension will heat up over
> >> time with current.
> >>
> >> That I did, and with that help I designed my 3200A gun. It does not become hot by the way, thanks to the dimensioning.
> >>
> >> Extending that to my motor (the IFE) I worked out that with a 500mm*500mm copper cross section, the heat increase would be manageable for 3 million amperes.
> >>
> >> I can use the same worksheet to work out the dimensions for 25million amps.
> >>
> >> My guess is that it would be around 2000mm*1000mm.
> >>
> >> That is a lot of copper, for 2000m rails, but it will carry 25 million amps. Note, this are rough figures. We also need to know that the whole barrel will be vacuum sealed and as cold as possible. Colder the temperature, better the conductivity, lower the contact resistance, etc. I am not talking about superconductivity, just higher conductivity with lower temperature.
> >>
> >> I am giving over-the-top back of the envelope estimates. Naturally, when we get down to the calculas and do the fine slicing, the actual figures for a given design will emerge. Reducing the contact resistance is the key. 
> >>
> >> Another point - we need existing defunct mineshafts going down 1-2Km for this gun. No point creating a 1000 m vertical structure, what! Natural side-support, in a hole!
> >>
> >> I think we can with a limited budget make a gun that can launch small satellites into space. It can compete with rocket launches. 
> >>
> >> Cheers,
> >> Arindam Banerjee
> > 
> > Earlier you Arindam) posted:
> > "25 million amps is nothing but 25000 volts divided by 1 milliohm. 
> > No problem at all, both in theory and practice."
> > 
> > the total resistance needs to be 1 milliohm (or less)
> > 
> > So what is the resistance?
> > to obtain a voltage drop of 25000V over 2 kilometers,
> > R=rl/A
> > l - length 2 km (=2000 m)
> > A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
> > r - rho resistivity for copper is 0.000 000 017 ohm-meter
> > 
> > R= 0.000 017 ohm = 0.017 milliohm.
> > 
> > I'm impressed. about a factor of 5 margin of safety.
> > 
> > Now 2 Km by 2m^2 is a volume of 4000 m^3
> > Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
> > total weight 35,840,000 kg
> > Now I'm impressed again. Cost is a little over $25 million.
> > 
> > Of course it will be the assembly costs that swamp the material costs.
> > You'd likely want the rails to be continuous. so you'd want to caste the rails in place maybe?Or can it be multiple parallel sheets of copper?
> > (how big is a roll of copper 1km long?)
> > 
> > Physically doable, though challenging. that the navy gave up on their
> > railgun, suggests there are other challenges we haven't identified.
> 
> 
> Interesting on why the navy gave up, it was trying to get 2K amps
> reliablely from rail to projectile, 0r 32 megajoles, but there always
> was ablation where the current melted the rails and Sabot, even with
> conductive lubes.  transfering large current on moving/sliding contact
> 
> nice pix here, all the flame is metal burning from rails and sabot.
> http://www.thefiscaltimes.com/2016/06/04/Navy-Abandoning-Railgun
> 
> > 
> > Nothing better be around the launch tip, that is going to be
> > one hell of a sonic boom!
> > 
> > Ed
> > Meaning the resistance needs to be 
> >



But Sandman feels the need to insult the cult-like herd of convenient friends. 

I have worked in the VAR field for more than 25 years. I've had a contract with top OEMs making deals with the aeronautics industry and the worlds largest companies. Never, not one single time, were we on any occasion instructed to not load any other OS. We sold AIXL, Ultix, Red Hat, among others as well as MS OSs very successfully. Sandman is far too unintelligent to write a quality script. The only scripting he's able to do is simple stuff I did years ago. 

--
Best CMS Solution of 2017!
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Jonas Eklundh

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#691286

On Tuesday, 19 June 2018 07:18:30 UTC+10, Sergio  wrote:
> On 6/18/2018 3:35 PM, Edward Prochak wrote:
> > On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
> >> On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
> >>> On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> >>>> On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> >>>>> On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> >>>>>> Did 3200 amps with own two brown hands
> >>>>>
> >>>>> you are off by 3 orders of magnitudes!
> >>>>
> >>>> No.
> >>>>
> >>>> I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> >>>> circuit resistance to get 3200 amps on the average.  The 3200 A
> >>>> was calculated from the drop in the voltage across the capacitor
> >>>> bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> >>>> discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> >>>>
> >>>> Agreed?
> >>>
> >>> Yes, but
> >>> That is still 3 orders of magnitude less than the
> >>> 25,000,000 Amps needed for the 1km rail gun.
> >>> I don't know what materials will carry that high a current.
> >>>
> >>> ED
> >>
> >> In my PhD work, my guide asked me to do a worksheet showing
> >> how much a conductor of a given dimension will heat up over
> >> time with current.
> >>
> >> That I did, and with that help I designed my 3200A gun. It does not become hot by the way, thanks to the dimensioning.
> >>
> >> Extending that to my motor (the IFE) I worked out that with a 500mm*500mm copper cross section, the heat increase would be manageable for 3 million amperes.
> >>
> >> I can use the same worksheet to work out the dimensions for 25million amps.
> >>
> >> My guess is that it would be around 2000mm*1000mm.
> >>
> >> That is a lot of copper, for 2000m rails, but it will carry 25 million amps. Note, this are rough figures. We also need to know that the whole barrel will be vacuum sealed and as cold as possible. Colder the temperature, better the conductivity, lower the contact resistance, etc. I am not talking about superconductivity, just higher conductivity with lower temperature.
> >>
> >> I am giving over-the-top back of the envelope estimates. Naturally, when we get down to the calculas and do the fine slicing, the actual figures for a given design will emerge. Reducing the contact resistance is the key. 
> >>
> >> Another point - we need existing defunct mineshafts going down 1-2Km for this gun. No point creating a 1000 m vertical structure, what! Natural side-support, in a hole!
> >>
> >> I think we can with a limited budget make a gun that can launch small satellites into space. It can compete with rocket launches. 
> >>
> >> Cheers,
> >> Arindam Banerjee
> > 
> > Earlier you Arindam) posted:
> > "25 million amps is nothing but 25000 volts divided by 1 milliohm. 
> > No problem at all, both in theory and practice."
> > 
> > the total resistance needs to be 1 milliohm (or less)
> > 
> > So what is the resistance?
> > to obtain a voltage drop of 25000V over 2 kilometers,
> > R=rl/A
> > l - length 2 km (=2000 m)
> > A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
> > r - rho resistivity for copper is 0.000 000 017 ohm-meter
> > 
> > R= 0.000 017 ohm = 0.017 milliohm.
> > 
> > I'm impressed. about a factor of 5 margin of safety.
> > 
> > Now 2 Km by 2m^2 is a volume of 4000 m^3
> > Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
> > total weight 35,840,000 kg
> > Now I'm impressed again. Cost is a little over $25 million.
> > 
> > Of course it will be the assembly costs that swamp the material costs.
> > You'd likely want the rails to be continuous. so you'd want to caste the rails in place maybe?Or can it be multiple parallel sheets of copper?
> > (how big is a roll of copper 1km long?)
> > 
> > Physically doable, though challenging. that the navy gave up on their
> > railgun, suggests there are other challenges we haven't identified.
> 
> 
> Interesting on why the navy gave up, it was trying to get 2K amps


Oh no, it was over 2 million amps they were looking for. Even I got more than 2000 amps for all my firings for the rail gun I made. At a conference in China I met a professor who was working on rail guns. He said the power needs were enormous and that matches with the literature.

> reliablely from rail to projectile, 0r 32 megajoles, but there always
> was ablation where the current melted the rails and Sabot, even with
> conductive lubes.  transfering large current on moving/sliding contact
> 
> nice pix here, all the flame is metal burning from rails and sabot.

Yes if the current is too high the thin rails will melt, but the sabot is supposed to melt. What goes through is the plastic bullet in the sabot. If that contains HE then the impact at Mach 7 if horrific. They say that the KE is such that iron will start to burn from the heat of impact. The smoke from the rail gun is the smoke from the burning of the conductor around the bullet. 

> http://www.thefiscaltimes.com/2016/06/04/Navy-Abandoning-Railgun
> 
> > 
> > Nothing better be around the launch tip, that is going to be
> > one hell of a sonic boom!
> > 
> > Ed
> > Meaning the resistance needs to be 
> >

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#691174

On Tuesday, June 19, 2018 at 6:35:07 AM UTC+10, Edward Prochak wrote:
> On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
> > On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
> > > On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> > > > On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> > > > > On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> > > > > > Did 3200 amps with own two brown hands
> > > > > 
> > > > > you are off by 3 orders of magnitudes!
> > > > 
> > > > No.
> > > > 
> > > > I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> > > > circuit resistance to get 3200 amps on the average.  The 3200 A
> > > > was calculated from the drop in the voltage across the capacitor
> > > > bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> > > > discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> > > > 
> > > > Agreed?
> > > 
> > > Yes, but
> > > That is still 3 orders of magnitude less than the
> > > 25,000,000 Amps needed for the 1km rail gun.
> > > I don't know what materials will carry that high a current.
> > > 
> > > ED
> > 
> > In my PhD work, my guide asked me to do a worksheet showing
> > how much a conductor of a given dimension will heat up over
> > time with current.
> > 
> > That I did, and with that help I designed my 3200A gun. It does not become hot by the way, thanks to the dimensioning.
> > 
> > Extending that to my motor (the IFE) I worked out that with a 500mm*500mm copper cross section, the heat increase would be manageable for 3 million amperes.
> > 
> > I can use the same worksheet to work out the dimensions for 25million amps.
> > 
> > My guess is that it would be around 2000mm*1000mm.
> > 
> > That is a lot of copper, for 2000m rails, but it will carry 25 million amps. Note, this are rough figures. We also need to know that the whole barrel will be vacuum sealed and as cold as possible. Colder the temperature, better the conductivity, lower the contact resistance, etc. I am not talking about superconductivity, just higher conductivity with lower temperature.
> > 
> > I am giving over-the-top back of the envelope estimates. Naturally, when we get down to the calculas and do the fine slicing, the actual figures for a given design will emerge. Reducing the contact resistance is the key. 
> > 
> > Another point - we need existing defunct mineshafts going down 1-2Km for this gun. No point creating a 1000 m vertical structure, what! Natural side-support, in a hole!
> > 
> > I think we can with a limited budget make a gun that can launch small satellites into space. It can compete with rocket launches. 
> > 
> > Cheers,
> > Arindam Banerjee
> 
> Earlier you Arindam) posted:
> "25 million amps is nothing but 25000 volts divided by 1 milliohm. 
> No problem at all, both in theory and practice."
> 
> the total resistance needs to be 1 milliohm (or less)
> 
> So what is the resistance?
> to obtain a voltage drop of 25000V over 2 kilometers,
> R=rl/A
> l - length 2 km (=2000 m)
> A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
> r - rho resistivity for copper is 0.000 000 017 ohm-meter
> 
> R= 0.000 017 ohm = 0.017 milliohm.
> 
> I'm impressed. about a factor of 5 margin of safety.

Well, it will be even less with cooling to as low as possible.
I was talking about room temperatures.
As you understand, the need is to get 25million amps with as low as voltage as possible, otherwise the capacitance and power will be huge.
So thanks for your effort, but as you see if we bring it down to say 10 microamps we need not have 25000V for our charge, but only 250V which is a lot more manageable.
> 
> Now 2 Km by 2m^2 is a volume of 4000 m^3
> Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
> total weight 35,840,000 kg
> Now I'm impressed again. Cost is a little over $25 million.

Look at it this way. We do need that much copper. Now if it is silver it will work better. If all the gold and silver were used in that gun, then it would be very safe. So heavy, no one could take it. Maybe it will be billions if it is gold, but so what. Better to use the gold as rails for a gun than keep it in pieces wherever. It will be a tourist site... Anyway, whether gold or copper or silver that much mass will remain put. Unlike the rocket parts which are all wasted.
> 
> Of course it will be the assembly costs that swamp the material costs.
> You'd likely want the rails to be continuous. so you'd want to caste the rails in place maybe?Or can it be multiple parallel sheets of copper?
> (how big is a roll of copper 1km long?)

I would use a mine shaft. An already dug 1Km hole. Or using a steep cliff, dig down below. Unused mine shafts will be ideal.
Make some formers of steel or so, do some fine machining there.
They would be filled withe the copper or gold upto 10m per go, say.
Then lift up the former, slide it up with varying temperature for more or less expansion wrt copper.
And repeat the process say 100+ times, keeping it straight.
A high Farad 250V capacitor is certainly not too difficult to make, we will need plenty of coppper sheets and high epsilon dielectric.


Naturally heating or IIR losses will go down immensely with low R.

> 
> Physically doable, though challenging. that the navy gave up on their
> railgun, suggests there are other challenges we haven't identified.

They had to make the barrel small, following traditional thinking about guns. As as result they had a high resistance armature, needing extremely high voltages so extremely high power requirements.
> 
> Nothing better be around the launch tip, that is going to be
> one hell of a sonic boom!

Won't last long.

Cheers,
Arindam Banerjee
> 
> Ed
> Meaning the resistance needs to be

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#690764

Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.

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#690848

On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:

>Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.

If my math is right, a rail gun at 100% efficiency shooting your 400
kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
seconds) would need an average power input of 12,000 terawatts.

Compare with:
https://en.wikipedia.org/wiki/World_energy_consumption
By the end of 2014, the total installed electricity generating
capacity worldwide was nearly 6.142 TW (million MW) which only
includes generation connected to local electricity grids.

So the average power input would be approximately 2000 times the
existing worldwide generating capacity.

But constant acceleration would require crazier amounts of power at
the end of the run:
https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration

So you need a *big* constant-power source, with accelerations during
the run exceeding the 150,000-g needed for constant acceleration.

The conclusion in the OP seems dubious: "What will happen if some
lifeform is put in it? I don't think it will get harmed".

Feel free to correct my math if I got it wrong.

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#690906

Thanks. We will do the maths a bit later when  I will be on  my computer 

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#690936

On Sunday, 17 June 2018 01:10:52 UTC+10, Wally W.  wrote:
> On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:
> 
> >Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.
> 
> If my math is right, a rail gun at 100% efficiency shooting your 400
> kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
> seconds) would need an average power input of 12,000 terawatts.

First let us check your maths.
For an exit velocity of 55,000 m/s over constant accn over 1000m, using the formula v=u+at and keeping u or initial velocity as 0, we have at = 55000.
Then v^2=u^2 + 2as, where for s=1000m and again u=0 we have as = 55000^2/2
or a = 55000*55/2
In which case t = 55000/a or t=55000/(55000*55/2)=2/55 seconds = 0.036 and so here we agree.

To get a mass of 400 Kg to accelerate to 55,000 m/s will give it a kinetic energy KE of 0.5mVV = 0.5*400*55*55*10^6 joules
So much energy has to be provided by the power source.
KE = 60.5*10^10 Joules which is very substantial indeed.
And the average power will be energy/time = 60.5*10^10/.036 = 60.5*10^11/.36 
= 181*10^11 watts = 18*10^12 watts.
Your figure of 12000 terawatts or 12*10^15 watts is off from my calculation by a factor of 727.

So I wonder how you got that figure.

> Compare with:
> https://en.wikipedia.org/wiki/World_energy_consumption
> By the end of 2014, the total installed electricity generating
> capacity worldwide was nearly 6.142 TW (million MW) which only
> includes generation connected to local electricity grids.

Well yes, but for our gun we need say a few days for a low power plant to charge up the capacitors that will fire at once to provide the necessary power. That power up can be done using say a 200MW power plant.

> 
> So the average power input would be approximately 2000 times the
> existing worldwide generating capacity.

The energy has to be stored in the capacitors. It should be say 10 times that of the KE required at launch so that should be 6*10^12 joules.

A 200MW plant or 2*10^8 plant should charge up the capacitors in 6*10^12/2*10^8 seconds or 30,000 seconds or 8 hours. 

> But constant acceleration would require crazier amounts of power at
> the end of the run:
> https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration
> 
> So you need a *big* constant-power source, with accelerations during
> the run exceeding the 150,000-g needed for constant acceleration.
> 
> The conclusion in the OP seems dubious: "What will happen if some
> lifeform is put in it? I don't think it will get harmed".

We can only try. No one seems harmed in going up the fast lifts that are run by linear motors as opposed to the pull-up lifts in Einstein's time, the sorts of lift that boosted his theory of General Relativity (please go through Einstein's own explanations).

There is a vast difference between a pull-up old life and a new linear motor lift using internal force continuously accelerating the lift cage as the armature of the linear motor, much like a rail gun bullet.

> Feel free to correct my math if I got it wrong.

Did that, just above. 
Cheers,
Arindam Banerjee

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#690938

On Sunday, 17 June 2018 12:04:45 UTC+10, Arindam Banerjee  wrote:
> On Sunday, 17 June 2018 01:10:52 UTC+10, Wally W.  wrote:
> > On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:
> > 
> > >Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.
> > 
> > If my math is right, a rail gun at 100% efficiency shooting your 400
> > kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
> > seconds) would need an average power input of 12,000 terawatts.
> 
> First let us check your maths.
> For an exit velocity of 55,000 m/s over constant accn over 1000m, using the formula v=u+at and keeping u or initial velocity as 0, we have at = 55000.
> Then v^2=u^2 + 2as, where for s=1000m and again u=0 we have as = 55000^2/2
> or a = 55000*55/2
> In which case t = 55000/a or t=55000/(55000*55/2)=2/55 seconds = 0.036 and so here we agree.
> 
> To get a mass of 400 Kg to accelerate to 55,000 m/s will give it a kinetic energy KE of 0.5mVV = 0.5*400*55*55*10^6 joules
> So much energy has to be provided by the power source.
> KE = 60.5*10^10 Joules which is very substantial indeed.
> And the average power will be energy/time = 60.5*10^10/.036 = 60.5*10^11/.36 
> = 181*10^11 watts = 18*10^12 watts.
> Your figure of 12000 terawatts or 12*10^15 watts is off from my calculation by a factor of 727.
> 
> So I wonder how you got that figure.
> 
> > Compare with:
> > https://en.wikipedia.org/wiki/World_energy_consumption
> > By the end of 2014, the total installed electricity generating
> > capacity worldwide was nearly 6.142 TW (million MW) which only
> > includes generation connected to local electricity grids.
> 
> Well yes, but for our gun we need say a few days for a low power plant to charge up the capacitors that will fire at once to provide the necessary power. That power up can be done using say a 200MW power plant.
> 
> > 
> > So the average power input would be approximately 2000 times the
> > existing worldwide generating capacity.
> 
> The energy has to be stored in the capacitors. It should be say 10 times that of the KE required at launch so that should be 6*10^12 joules.
> 
> A 200MW plant or 2*10^8 plant should charge up the capacitors in 6*10^12/2*10^8 seconds or 30,000 seconds or 8 hours. 
> 
> > But constant acceleration would require crazier amounts of power at
> > the end of the run:
> > https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration
> > 
> > So you need a *big* constant-power source, with accelerations during
> > the run exceeding the 150,000-g needed for constant acceleration.
> > 
> > The conclusion in the OP seems dubious: "What will happen if some
> > lifeform is put in it? I don't think it will get harmed".
> 
> We can only try. No one seems harmed in going up the fast lifts that are run by linear motors as opposed to the pull-up lifts in Einstein's time, the sorts of lift that boosted his theory of General Relativity (please go through Einstein's own explanations).
> 
> There is a vast difference between a pull-up old life and a new linear motor lift using internal force continuously accelerating the lift cage as the armature of the linear motor, much like a rail gun bullet.
> 
> > Feel free to correct my math if I got it wrong.
> 
> Did that, just above. 
> Cheers,
> Arindam Banerjee

So if we all agree that my maths is correct then let us work out costs.
One key cost is the energy, in this case 8 hours of 200MW.
Or 1600MW hours, or 1600000 KWhr or units.
One unit here is some 25cents say, haggled down.
So the launch cost for energy alone is 400,000 dollars. 
Multiply that by ten for profit, interest on infrastructure, etc. and we have the cost of launching a 400Kg mass to the Moon at $4 million.

Now, if instead of the Moon we try shooting up satellites, than the launch cost will be much less. By a factor of at least 20 by my reckoning, considering the much lower energy and much smaller gun.

That is, only $200,000 to launch a satellite in space using my new design rail gun with broad beam, low voltage and low resistance.

Is this an attractive proposition? Are their bold venture capitalists licking their lips/

Cheers,
Arindam Banerjee

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#690941

On Sunday, 17 June 2018 13:05:17 UTC+10, Arindam Banerjee  wrote:
> On Sunday, 17 June 2018 12:04:45 UTC+10, Arindam Banerjee  wrote:
> > On Sunday, 17 June 2018 01:10:52 UTC+10, Wally W.  wrote:
> > > On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:
> > > 
> > > >Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.
> > > 
> > > If my math is right, a rail gun at 100% efficiency shooting your 400
> > > kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
> > > seconds) would need an average power input of 12,000 terawatts.
> > 
> > First let us check your maths.
> > For an exit velocity of 55,000 m/s over constant accn over 1000m, using the formula v=u+at and keeping u or initial velocity as 0, we have at = 55000.
> > Then v^2=u^2 + 2as, where for s=1000m and again u=0 we have as = 55000^2/2
> > or a = 55000*55/2
> > In which case t = 55000/a or t=55000/(55000*55/2)=2/55 seconds = 0.036 and so here we agree.
> > 
> > To get a mass of 400 Kg to accelerate to 55,000 m/s will give it a kinetic energy KE of 0.5mVV = 0.5*400*55*55*10^6 joules
> > So much energy has to be provided by the power source.
> > KE = 60.5*10^10 Joules which is very substantial indeed.
> > And the average power will be energy/time = 60.5*10^10/.036 = 60.5*10^11/.36 
> > = 181*10^11 watts = 18*10^12 watts.
> > Your figure of 12000 terawatts or 12*10^15 watts is off from my calculation by a factor of 727.
> > 
> > So I wonder how you got that figure.
> > 
> > > Compare with:
> > > https://en.wikipedia.org/wiki/World_energy_consumption
> > > By the end of 2014, the total installed electricity generating
> > > capacity worldwide was nearly 6.142 TW (million MW) which only
> > > includes generation connected to local electricity grids.
> > 
> > Well yes, but for our gun we need say a few days for a low power plant to charge up the capacitors that will fire at once to provide the necessary power. That power up can be done using say a 200MW power plant.
> > 
> > > 
> > > So the average power input would be approximately 2000 times the
> > > existing worldwide generating capacity.
> > 
> > The energy has to be stored in the capacitors. It should be say 10 times that of the KE required at launch so that should be 6*10^12 joules.
> > 
> > A 200MW plant or 2*10^8 plant should charge up the capacitors in 6*10^12/2*10^8 seconds or 30,000 seconds or 8 hours. 
> > 
> > > But constant acceleration would require crazier amounts of power at
> > > the end of the run:
> > > https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration
> > > 
> > > So you need a *big* constant-power source, with accelerations during
> > > the run exceeding the 150,000-g needed for constant acceleration.
> > > 
> > > The conclusion in the OP seems dubious: "What will happen if some
> > > lifeform is put in it? I don't think it will get harmed".
> > 
> > We can only try. No one seems harmed in going up the fast lifts that are run by linear motors as opposed to the pull-up lifts in Einstein's time, the sorts of lift that boosted his theory of General Relativity (please go through Einstein's own explanations).
> > 
> > There is a vast difference between a pull-up old life and a new linear motor lift using internal force continuously accelerating the lift cage as the armature of the linear motor, much like a rail gun bullet.
> > 
> > > Feel free to correct my math if I got it wrong.
> > 
> > Did that, just above. 
> > Cheers,
> > Arindam Banerjee
> 
> So if we all agree that my maths is correct then let us work out costs.
> One key cost is the energy, in this case 8 hours of 200MW.
> Or 1600MW hours, or 1600000 KWhr or units.
> One unit here is some 25cents say, haggled down.
> So the launch cost for energy alone is 400,000 dollars. 
> Multiply that by ten for profit, interest on infrastructure, etc. and we have the cost of launching a 400Kg mass to the Moon at $4 million.
> 
> Now, if instead of the Moon we try shooting up satellites, than the launch cost will be much less. By a factor of at least 20 by my reckoning, considering the much lower energy and much smaller gun.
> 
> That is, only $200,000 to launch a satellite in space using my new design rail gun with broad beam, low voltage and low resistance.
> 
> Is this an attractive proposition? Are their bold venture capitalists licking their lips/
> 
> Cheers,
> Arindam Banerjee

So, have all the greedy venture capitalists turned into hermits or what?

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#690958

On Sat, 16 Jun 2018 20:05:13 -0700 (PDT), Arindam Banerjee wrote:

>On Sunday, 17 June 2018 12:04:45 UTC+10, Arindam Banerjee  wrote:
>> On Sunday, 17 June 2018 01:10:52 UTC+10, Wally W.  wrote:
>> > On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:
>> > 
>> > >Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.
>> > 
>> > If my math is right, a rail gun at 100% efficiency shooting your 400
>> > kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
>> > seconds) would need an average power input of 12,000 terawatts.
>> 
>> First let us check your maths.
>> For an exit velocity of 55,000 m/s over constant accn over 1000m, using the formula v=u+at and keeping u or initial velocity as 0, we have at = 55000.
>> Then v^2=u^2 + 2as, where for s=1000m and again u=0 we have as = 55000^2/2
>> or a = 55000*55/2
>> In which case t = 55000/a or t=55000/(55000*55/2)=2/55 seconds = 0.036 and so here we agree.
>> 
>> To get a mass of 400 Kg to accelerate to 55,000 m/s will give it a kinetic energy KE of 0.5mVV = 0.5*400*55*55*10^6 joules
>> So much energy has to be provided by the power source.
>> KE = 60.5*10^10 Joules which is very substantial indeed.
>> And the average power will be energy/time = 60.5*10^10/.036 = 60.5*10^11/.36 
>> = 181*10^11 watts = 18*10^12 watts.
>> Your figure of 12000 terawatts or 12*10^15 watts is off from my calculation by a factor of 727.
>> 
>> So I wonder how you got that figure.
>> 
>> > Compare with:
>> > https://en.wikipedia.org/wiki/World_energy_consumption
>> > By the end of 2014, the total installed electricity generating
>> > capacity worldwide was nearly 6.142 TW (million MW) which only
>> > includes generation connected to local electricity grids.
>> 
>> Well yes, but for our gun we need say a few days for a low power plant to charge up the capacitors that will fire at once to provide the necessary power. That power up can be done using say a 200MW power plant.
>> 
>> > 
>> > So the average power input would be approximately 2000 times the
>> > existing worldwide generating capacity.
>> 
>> The energy has to be stored in the capacitors. It should be say 10 times that of the KE required at launch so that should be 6*10^12 joules.
>> 
>> A 200MW plant or 2*10^8 plant should charge up the capacitors in 6*10^12/2*10^8 seconds or 30,000 seconds or 8 hours. 
>> 
>> > But constant acceleration would require crazier amounts of power at
>> > the end of the run:
>> > https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration
>> > 
>> > So you need a *big* constant-power source, with accelerations during
>> > the run exceeding the 150,000-g needed for constant acceleration.
>> > 
>> > The conclusion in the OP seems dubious: "What will happen if some
>> > lifeform is put in it? I don't think it will get harmed".
>> 
>> We can only try. No one seems harmed in going up the fast lifts that are run by linear motors as opposed to the pull-up lifts in Einstein's time, the sorts of lift that boosted his theory of General Relativity (please go through Einstein's own explanations).
>> 
>> There is a vast difference between a pull-up old life and a new linear motor lift using internal force continuously accelerating the lift cage as the armature of the linear motor, much like a rail gun bullet.
>> 
>> > Feel free to correct my math if I got it wrong.
>> 
>> Did that, just above. 
>> Cheers,
>> Arindam Banerjee
>
>So if we all agree that my maths is correct then let us work out costs.

Agreed: 6.05e11 watts at 100% efficiency. There was an error in my
power calc.

>One key cost is the energy, in this case 8 hours of 200MW.
>Or 1600MW hours, or 1600000 KWhr or units.
>One unit here is some 25cents say, haggled down.
>So the launch cost for energy alone is 400,000 dollars. 
>Multiply that by ten for profit, interest on infrastructure, etc. and we have the cost of launching a 400Kg mass to the Moon at $4 million.
>
>Now, if instead of the Moon we try shooting up satellites, than the launch cost will be much less. By a factor of at least 20 by my reckoning, considering the much lower energy and much smaller gun.
>
>That is, only $200,000 to launch a satellite in space using my new design rail gun with broad beam, low voltage and low resistance.
>
>Is this an attractive proposition? Are their bold venture capitalists licking their lips/
>
>Cheers,
>Arindam Banerjee

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#690959

On Sunday, 17 June 2018 19:35:23 UTC+10, Wally W.  wrote:
> On Sat, 16 Jun 2018 20:05:13 -0700 (PDT), Arindam Banerjee wrote:
> 
> >On Sunday, 17 June 2018 12:04:45 UTC+10, Arindam Banerjee  wrote:
> >> On Sunday, 17 June 2018 01:10:52 UTC+10, Wally W.  wrote:
> >> > On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:
> >> > 
> >> > >Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.
> >> > 
> >> > If my math is right, a rail gun at 100% efficiency shooting your 400
> >> > kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
> >> > seconds) would need an average power input of 12,000 terawatts.
> >> 
> >> First let us check your maths.
> >> For an exit velocity of 55,000 m/s over constant accn over 1000m, using the formula v=u+at and keeping u or initial velocity as 0, we have at = 55000.
> >> Then v^2=u^2 + 2as, where for s=1000m and again u=0 we have as = 55000^2/2
> >> or a = 55000*55/2
> >> In which case t = 55000/a or t=55000/(55000*55/2)=2/55 seconds = 0.036 and so here we agree.
> >> 
> >> To get a mass of 400 Kg to accelerate to 55,000 m/s will give it a kinetic energy KE of 0.5mVV = 0.5*400*55*55*10^6 joules
> >> So much energy has to be provided by the power source.
> >> KE = 60.5*10^10 Joules which is very substantial indeed.
> >> And the average power will be energy/time = 60.5*10^10/.036 = 60.5*10^11/.36 
> >> = 181*10^11 watts = 18*10^12 watts.
> >> Your figure of 12000 terawatts or 12*10^15 watts is off from my calculation by a factor of 727.
> >> 
> >> So I wonder how you got that figure.
> >> 
> >> > Compare with:
> >> > https://en.wikipedia.org/wiki/World_energy_consumption
> >> > By the end of 2014, the total installed electricity generating
> >> > capacity worldwide was nearly 6.142 TW (million MW) which only
> >> > includes generation connected to local electricity grids.
> >> 
> >> Well yes, but for our gun we need say a few days for a low power plant to charge up the capacitors that will fire at once to provide the necessary power. That power up can be done using say a 200MW power plant.
> >> 
> >> > 
> >> > So the average power input would be approximately 2000 times the
> >> > existing worldwide generating capacity.
> >> 
> >> The energy has to be stored in the capacitors. It should be say 10 times that of the KE required at launch so that should be 6*10^12 joules.
> >> 
> >> A 200MW plant or 2*10^8 plant should charge up the capacitors in 6*10^12/2*10^8 seconds or 30,000 seconds or 8 hours. 
> >> 
> >> > But constant acceleration would require crazier amounts of power at
> >> > the end of the run:
> >> > https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration
> >> > 
> >> > So you need a *big* constant-power source, with accelerations during
> >> > the run exceeding the 150,000-g needed for constant acceleration.
> >> > 
> >> > The conclusion in the OP seems dubious: "What will happen if some
> >> > lifeform is put in it? I don't think it will get harmed".
> >> 
> >> We can only try. No one seems harmed in going up the fast lifts that are run by linear motors as opposed to the pull-up lifts in Einstein's time, the sorts of lift that boosted his theory of General Relativity (please go through Einstein's own explanations).
> >> 
> >> There is a vast difference between a pull-up old life and a new linear motor lift using internal force continuously accelerating the lift cage as the armature of the linear motor, much like a rail gun bullet.
> >> 
> >> > Feel free to correct my math if I got it wrong.
> >> 
> >> Did that, just above. 
> >> Cheers,
> >> Arindam Banerjee
> >
> >So if we all agree that my maths is correct then let us work out costs.
> 
> Agreed: 6.05e11 watts at 100% efficiency. There was an error in my
> power calc.

Thanks. Well, it doesn't look too difficult after all, what. 
> 
> >One key cost is the energy, in this case 8 hours of 200MW.
> >Or 1600MW hours, or 1600000 KWhr or units.
> >One unit here is some 25cents say, haggled down.
> >So the launch cost for energy alone is 400,000 dollars. 
> >Multiply that by ten for profit, interest on infrastructure, etc. and we have the cost of launching a 400Kg mass to the Moon at $4 million.
> >
> >Now, if instead of the Moon we try shooting up satellites, than the launch cost will be much less. By a factor of at least 20 by my reckoning, considering the much lower energy and much smaller gun.
> >
> >That is, only $200,000 to launch a satellite in space using my new design rail gun with broad beam, low voltage and low resistance.
> >
> >Is this an attractive proposition? Are their bold venture capitalists licking their lips/
> >
> >Cheers,
> >Arindam Banerjee

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#690960

On Sunday, 17 June 2018 20:08:51 UTC+10, Arindam Banerjee  wrote:
> On Sunday, 17 June 2018 19:35:23 UTC+10, Wally W.  wrote:
> > On Sat, 16 Jun 2018 20:05:13 -0700 (PDT), Arindam Banerjee wrote:
> > 
> > >On Sunday, 17 June 2018 12:04:45 UTC+10, Arindam Banerjee  wrote:
> > >> On Sunday, 17 June 2018 01:10:52 UTC+10, Wally W.  wrote:
> > >> > On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:
> > >> > 
> > >> > >Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.
> > >> > 
> > >> > If my math is right, a rail gun at 100% efficiency shooting your 400
> > >> > kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
> > >> > seconds) would need an average power input of 12,000 terawatts.
> > >> 
> > >> First let us check your maths.
> > >> For an exit velocity of 55,000 m/s over constant accn over 1000m, using the formula v=u+at and keeping u or initial velocity as 0, we have at = 55000.
> > >> Then v^2=u^2 + 2as, where for s=1000m and again u=0 we have as = 55000^2/2
> > >> or a = 55000*55/2
> > >> In which case t = 55000/a or t=55000/(55000*55/2)=2/55 seconds = 0.036 and so here we agree.
> > >> 
> > >> To get a mass of 400 Kg to accelerate to 55,000 m/s will give it a kinetic energy KE of 0.5mVV = 0.5*400*55*55*10^6 joules
> > >> So much energy has to be provided by the power source.
> > >> KE = 60.5*10^10 Joules which is very substantial indeed.
> > >> And the average power will be energy/time = 60.5*10^10/.036 = 60.5*10^11/.36 
> > >> = 181*10^11 watts = 18*10^12 watts.
> > >> Your figure of 12000 terawatts or 12*10^15 watts is off from my calculation by a factor of 727.
> > >> 
> > >> So I wonder how you got that figure.
> > >> 
> > >> > Compare with:
> > >> > https://en.wikipedia.org/wiki/World_energy_consumption
> > >> > By the end of 2014, the total installed electricity generating
> > >> > capacity worldwide was nearly 6.142 TW (million MW) which only
> > >> > includes generation connected to local electricity grids.
> > >> 
> > >> Well yes, but for our gun we need say a few days for a low power plant to charge up the capacitors that will fire at once to provide the necessary power. That power up can be done using say a 200MW power plant.
> > >> 
> > >> > 
> > >> > So the average power input would be approximately 2000 times the
> > >> > existing worldwide generating capacity.
> > >> 
> > >> The energy has to be stored in the capacitors. It should be say 10 times that of the KE required at launch so that should be 6*10^12 joules.
> > >> 
> > >> A 200MW plant or 2*10^8 plant should charge up the capacitors in 6*10^12/2*10^8 seconds or 30,000 seconds or 8 hours. 
> > >> 
> > >> > But constant acceleration would require crazier amounts of power at
> > >> > the end of the run:
> > >> > https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration
> > >> > 
> > >> > So you need a *big* constant-power source, with accelerations during
> > >> > the run exceeding the 150,000-g needed for constant acceleration.
> > >> > 
> > >> > The conclusion in the OP seems dubious: "What will happen if some
> > >> > lifeform is put in it? I don't think it will get harmed".
> > >> 
> > >> We can only try. No one seems harmed in going up the fast lifts that are run by linear motors as opposed to the pull-up lifts in Einstein's time, the sorts of lift that boosted his theory of General Relativity (please go through Einstein's own explanations).
> > >> 
> > >> There is a vast difference between a pull-up old life and a new linear motor lift using internal force continuously accelerating the lift cage as the armature of the linear motor, much like a rail gun bullet.
> > >> 
> > >> > Feel free to correct my math if I got it wrong.
> > >> 
> > >> Did that, just above. 
> > >> Cheers,
> > >> Arindam Banerjee
> > >
> > >So if we all agree that my maths is correct then let us work out costs.
> > 
> > Agreed: 6.05e11 watts at 100% efficiency. There was an error in my
> > power calc.
> 
> Thanks. Well, it doesn't look too difficult after all, what. 

It is easy to calculate the capacitance required, for say 10% efficiency of the rail gun which means it has to store 6e12 joules.

At 25000V, that will mean 0.5CV*V=6e12 so with V=25000V we have
0.5C*625e6=6e12
C=12e6/625
C=19200Farads holding 25000V 
Now making this size of capacitor is indeed a technical challenge. It should be very large, and thus be in proportion to the size of the 1000 m long barrel!

> > 
> > >One key cost is the energy, in this case 8 hours of 200MW.
> > >Or 1600MW hours, or 1600000 KWhr or units.
> > >One unit here is some 25cents say, haggled down.
> > >So the launch cost for energy alone is 400,000 dollars. 
> > >Multiply that by ten for profit, interest on infrastructure, etc. and we have the cost of launching a 400Kg mass to the Moon at $4 million.
> > >
> > >Now, if instead of the Moon we try shooting up satellites, than the launch cost will be much less. By a factor of at least 20 by my reckoning, considering the much lower energy and much smaller gun.
> > >
> > >That is, only $200,000 to launch a satellite in space using my new design rail gun with broad beam, low voltage and low resistance.
> > >
> > >Is this an attractive proposition? Are their bold venture capitalists licking their lips/
> > >
> > >Cheers,
> > >Arindam Banerjee

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#690972

On Sun, 17 Jun 2018 06:00:22 -0700 (PDT), Arindam Banerjee wrote:

>On Sunday, 17 June 2018 20:08:51 UTC+10, Arindam Banerjee  wrote:
>> On Sunday, 17 June 2018 19:35:23 UTC+10, Wally W.  wrote:
>> > On Sat, 16 Jun 2018 20:05:13 -0700 (PDT), Arindam Banerjee wrote:
>> > 
>> > >On Sunday, 17 June 2018 12:04:45 UTC+10, Arindam Banerjee  wrote:
>> > >> On Sunday, 17 June 2018 01:10:52 UTC+10, Wally W.  wrote:
>> > >> > On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:
>> > >> > 
>> > >> > >Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.
>> > >> > 
>> > >> > If my math is right, a rail gun at 100% efficiency shooting your 400
>> > >> > kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
>> > >> > seconds) would need an average power input of 12,000 terawatts.
>> > >> 
>> > >> First let us check your maths.
>> > >> For an exit velocity of 55,000 m/s over constant accn over 1000m, using the formula v=u+at and keeping u or initial velocity as 0, we have at = 55000.
>> > >> Then v^2=u^2 + 2as, where for s=1000m and again u=0 we have as = 55000^2/2
>> > >> or a = 55000*55/2
>> > >> In which case t = 55000/a or t=55000/(55000*55/2)=2/55 seconds = 0.036 and so here we agree.
>> > >> 
>> > >> To get a mass of 400 Kg to accelerate to 55,000 m/s will give it a kinetic energy KE of 0.5mVV = 0.5*400*55*55*10^6 joules
>> > >> So much energy has to be provided by the power source.
>> > >> KE = 60.5*10^10 Joules which is very substantial indeed.
>> > >> And the average power will be energy/time = 60.5*10^10/.036 = 60.5*10^11/.36 
>> > >> = 181*10^11 watts = 18*10^12 watts.
>> > >> Your figure of 12000 terawatts or 12*10^15 watts is off from my calculation by a factor of 727.
>> > >> 
>> > >> So I wonder how you got that figure.
>> > >> 
>> > >> > Compare with:
>> > >> > https://en.wikipedia.org/wiki/World_energy_consumption
>> > >> > By the end of 2014, the total installed electricity generating
>> > >> > capacity worldwide was nearly 6.142 TW (million MW) which only
>> > >> > includes generation connected to local electricity grids.
>> > >> 
>> > >> Well yes, but for our gun we need say a few days for a low power plant to charge up the capacitors that will fire at once to provide the necessary power. That power up can be done using say a 200MW power plant.
>> > >> 
>> > >> > 
>> > >> > So the average power input would be approximately 2000 times the
>> > >> > existing worldwide generating capacity.
>> > >> 
>> > >> The energy has to be stored in the capacitors. It should be say 10 times that of the KE required at launch so that should be 6*10^12 joules.
>> > >> 
>> > >> A 200MW plant or 2*10^8 plant should charge up the capacitors in 6*10^12/2*10^8 seconds or 30,000 seconds or 8 hours. 
>> > >> 
>> > >> > But constant acceleration would require crazier amounts of power at
>> > >> > the end of the run:
>> > >> > https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration
>> > >> > 
>> > >> > So you need a *big* constant-power source, with accelerations during
>> > >> > the run exceeding the 150,000-g needed for constant acceleration.
>> > >> > 
>> > >> > The conclusion in the OP seems dubious: "What will happen if some
>> > >> > lifeform is put in it? I don't think it will get harmed".
>> > >> 
>> > >> We can only try. No one seems harmed in going up the fast lifts that are run by linear motors as opposed to the pull-up lifts in Einstein's time, the sorts of lift that boosted his theory of General Relativity (please go through Einstein's own explanations).
>> > >> 
>> > >> There is a vast difference between a pull-up old life and a new linear motor lift using internal force continuously accelerating the lift cage as the armature of the linear motor, much like a rail gun bullet.
>> > >> 
>> > >> > Feel free to correct my math if I got it wrong.
>> > >> 
>> > >> Did that, just above. 
>> > >> Cheers,
>> > >> Arindam Banerjee
>> > >
>> > >So if we all agree that my maths is correct then let us work out costs.
>> > 
>> > Agreed: 6.05e11 watts at 100% efficiency. There was an error in my
>> > power calc.
>> 
>> Thanks. Well, it doesn't look too difficult after all, what. 
>
>It is easy to calculate the capacitance required, for say 10% efficiency of the rail gun which means it has to store 6e12 joules.
>
>At 25000V, that will mean 0.5CV*V=6e12 so with V=25000V we have
>0.5C*625e6=6e12
>C=12e6/625
>C=19200Farads holding 25000V 
>Now making this size of capacitor is indeed a technical challenge. It should be very large, and thus be in proportion to the size of the 1000 m long barrel!
>
>> > 
>> > >One key cost is the energy, in this case 8 hours of 200MW.
>> > >Or 1600MW hours, or 1600000 KWhr or units.
>> > >One unit here is some 25cents say, haggled down.
>> > >So the launch cost for energy alone is 400,000 dollars. 
>> > >Multiply that by ten for profit, interest on infrastructure, etc. and we have the cost of launching a 400Kg mass to the Moon at $4 million.
>> > >
>> > >Now, if instead of the Moon we try shooting up satellites, than the launch cost will be much less. By a factor of at least 20 by my reckoning, considering the much lower energy and much smaller gun.
>> > >
>> > >That is, only $200,000 to launch a satellite in space using my new design rail gun with broad beam, low voltage and low resistance.
>> > >
>> > >Is this an attractive proposition? Are their bold venture capitalists licking their lips/
>> > >
>> > >Cheers,
>> > >Arindam Banerjee

This was interesting to look at, but it seems to have some issues.

We agree on the the value of the acceleration:

a = 55000*55/2 = 1,512,500 m/s^2

Let's give it some context:

1,512,500 m/s^2 / (9.8 m/s^2) = 154,000-g

Can any payload, living or not, withstand 154,000-g?

This is with a power source that can produce constant acceleration,
which would have instantaneous loads higher than 6.02e11 watts.

BTW: What is the physical size of a 19200-Farad capacitor bank?

Then, with the 1 km rail gun aimed vertically, the launched mass (even
if accelerated in an evaculated tube) will encounter atmosphere at at
altitude of 1000 meters, where the speed of sound is:

https://www.engineeringtoolbox.com/elevation-speed-sound-air-d_1534.html
336.4 m/sec

55,000 / 336.4 = Mach 160, as previously noted by Edward Prochak.

Using numbers from the link above, the relative air density at at
altitude of 1000 meters is:

(89.84 / (6.9 + 273)) / (101.35 / (15 + 273)) = 0.91

The drag on the launched mass would be huge at Mach 160 through an
atmosphere that has 91% of sea-level density.

Imagine the abuse experienced by the payload:
1. Enduring 150,000-g while accelerating from zero to 55,000 m/s.
2. Zero acceleration when leaving the evacuated tube.
3. Deceleration due to encountering atmosphere (with 90% of sea level
density) at Mach 160.

Scaling up a rail gun to serve as a space-vehicle launcher seems
impractical.

Building a prototype would be premature until we have a way for the
payload survive 150,000-g.

Also needed for the 150,000-g version: a 1-km-long, large-diameter,
evaculated tube (with a quick-opening door at the exit). Lower
g-forces would need a longer tube (and/or lower exit velocity).

Of possible interest:

<https://www.quora.com/How-does-the-US-Navy-rail-gun-work-That-must-need-an-insane-instantaneous-current-How-do-they-get-that-How-do-they-keep-it-from-melting>

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#690998

On Monday, 18 June 2018 03:28:46 UTC+10, Wally W.  wrote:
> On Sun, 17 Jun 2018 06:00:22 -0700 (PDT), Arindam Banerjee wrote:
> 
> >On Sunday, 17 June 2018 20:08:51 UTC+10, Arindam Banerjee  wrote:
> >> On Sunday, 17 June 2018 19:35:23 UTC+10, Wally W.  wrote:
> >> > On Sat, 16 Jun 2018 20:05:13 -0700 (PDT), Arindam Banerjee wrote:
> >> > 
> >> > >On Sunday, 17 June 2018 12:04:45 UTC+10, Arindam Banerjee  wrote:
> >> > >> On Sunday, 17 June 2018 01:10:52 UTC+10, Wally W.  wrote:
> >> > >> > On Fri, 15 Jun 2018 14:29:49 -0700 (PDT), Arindam Banerjee wrote:
> >> > >> > 
> >> > >> > >Air resistance will be overcome with vacuum over the 1 Km. Different design using a 200 MW powerplant.
> >> > >> > 
> >> > >> > If my math is right, a rail gun at 100% efficiency shooting your 400
> >> > >> > kg mass with constant acceleration (to achieve 55,000 m/s in 0.036
> >> > >> > seconds) would need an average power input of 12,000 terawatts.
> >> > >> 
> >> > >> First let us check your maths.
> >> > >> For an exit velocity of 55,000 m/s over constant accn over 1000m, using the formula v=u+at and keeping u or initial velocity as 0, we have at = 55000.
> >> > >> Then v^2=u^2 + 2as, where for s=1000m and again u=0 we have as = 55000^2/2
> >> > >> or a = 55000*55/2
> >> > >> In which case t = 55000/a or t=55000/(55000*55/2)=2/55 seconds = 0.036 and so here we agree.
> >> > >> 
> >> > >> To get a mass of 400 Kg to accelerate to 55,000 m/s will give it a kinetic energy KE of 0.5mVV = 0.5*400*55*55*10^6 joules
> >> > >> So much energy has to be provided by the power source.
> >> > >> KE = 60.5*10^10 Joules which is very substantial indeed.
> >> > >> And the average power will be energy/time = 60.5*10^10/.036 = 60.5*10^11/.36 
> >> > >> = 181*10^11 watts = 18*10^12 watts.
> >> > >> Your figure of 12000 terawatts or 12*10^15 watts is off from my calculation by a factor of 727.
> >> > >> 
> >> > >> So I wonder how you got that figure.
> >> > >> 
> >> > >> > Compare with:
> >> > >> > https://en.wikipedia.org/wiki/World_energy_consumption
> >> > >> > By the end of 2014, the total installed electricity generating
> >> > >> > capacity worldwide was nearly 6.142 TW (million MW) which only
> >> > >> > includes generation connected to local electricity grids.
> >> > >> 
> >> > >> Well yes, but for our gun we need say a few days for a low power plant to charge up the capacitors that will fire at once to provide the necessary power. That power up can be done using say a 200MW power plant.
> >> > >> 
> >> > >> > 
> >> > >> > So the average power input would be approximately 2000 times the
> >> > >> > existing worldwide generating capacity.
> >> > >> 
> >> > >> The energy has to be stored in the capacitors. It should be say 10 times that of the KE required at launch so that should be 6*10^12 joules.
> >> > >> 
> >> > >> A 200MW plant or 2*10^8 plant should charge up the capacitors in 6*10^12/2*10^8 seconds or 30,000 seconds or 8 hours. 
> >> > >> 
> >> > >> > But constant acceleration would require crazier amounts of power at
> >> > >> > the end of the run:
> >> > >> > https://www.quora.com/Does-constant-value-of-mechanical-power-mean-constant-acceleration
> >> > >> > 
> >> > >> > So you need a *big* constant-power source, with accelerations during
> >> > >> > the run exceeding the 150,000-g needed for constant acceleration.
> >> > >> > 
> >> > >> > The conclusion in the OP seems dubious: "What will happen if some
> >> > >> > lifeform is put in it? I don't think it will get harmed".
> >> > >> 
> >> > >> We can only try. No one seems harmed in going up the fast lifts that are run by linear motors as opposed to the pull-up lifts in Einstein's time, the sorts of lift that boosted his theory of General Relativity (please go through Einstein's own explanations).
> >> > >> 
> >> > >> There is a vast difference between a pull-up old life and a new linear motor lift using internal force continuously accelerating the lift cage as the armature of the linear motor, much like a rail gun bullet.
> >> > >> 
> >> > >> > Feel free to correct my math if I got it wrong.
> >> > >> 
> >> > >> Did that, just above. 
> >> > >> Cheers,
> >> > >> Arindam Banerjee
> >> > >
> >> > >So if we all agree that my maths is correct then let us work out costs.
> >> > 
> >> > Agreed: 6.05e11 watts at 100% efficiency. There was an error in my
> >> > power calc.
> >> 
> >> Thanks. Well, it doesn't look too difficult after all, what. 
> >
> >It is easy to calculate the capacitance required, for say 10% efficiency of the rail gun which means it has to store 6e12 joules.
> >
> >At 25000V, that will mean 0.5CV*V=6e12 so with V=25000V we have
> >0.5C*625e6=6e12
> >C=12e6/625
> >C=19200Farads holding 25000V 
> >Now making this size of capacitor is indeed a technical challenge. It should be very large, and thus be in proportion to the size of the 1000 m long barrel!
> >
> >> > 
> >> > >One key cost is the energy, in this case 8 hours of 200MW.
> >> > >Or 1600MW hours, or 1600000 KWhr or units.
> >> > >One unit here is some 25cents say, haggled down.
> >> > >So the launch cost for energy alone is 400,000 dollars. 
> >> > >Multiply that by ten for profit, interest on infrastructure, etc. and we have the cost of launching a 400Kg mass to the Moon at $4 million.
> >> > >
> >> > >Now, if instead of the Moon we try shooting up satellites, than the launch cost will be much less. By a factor of at least 20 by my reckoning, considering the much lower energy and much smaller gun.
> >> > >
> >> > >That is, only $200,000 to launch a satellite in space using my new design rail gun with broad beam, low voltage and low resistance.
> >> > >
> >> > >Is this an attractive proposition? Are their bold venture capitalists licking their lips/
> >> > >
> >> > >Cheers,
> >> > >Arindam Banerjee
> 
> This was interesting to look at, but it seems to have some issues.
> 
> We agree on the the value of the acceleration:
> 
> a = 55000*55/2 = 1,512,500 m/s^2

yes. It is caused by the so-called Lorenz force, which should be called the Thomson-Heaviside force as the latter were the finders and theoreticians for same. Call it THE force.

> Let's give it some context:
> 
> 1,512,500 m/s^2 / (9.8 m/s^2) = 154,000-g

True.

> Can any payload, living or not, withstand 154,000-g?

Don't see why not. THE force operates by straining the electrons in the circuit continuously. It is not the same thing as a rope pulling up the payload, more like a rocket thrusting up mass. There is fantastic stress in the former, but in the latter there is less pressure overall upon the container, being more evenly distributed. Things have to be properly strapped otherwise thunk they will go and break. When nicely packed up, so that they are no spaces between the walls of the container and the masses that are not supposed to break, there should be no issue. Yes one cannot jump up as the floor will rush up with the jump and that will give a sensation of getting stuck. But so long as one is sticking well, there should be no survival issues. And don't forget this pressure will last for only 2/55 seconds, a blink of the eye. Not that anyone should go to Space or to the Moon in such a style. For that, we need to make Internal Force Engines and put them on planes so that travel will be done in great comfort. My video films show how this can be done.

For a non-life cargo, properly strapped down, there should be no issue. This is thus an excellent platform for shooting satellites for the Earth of Moon, or sending cargo there. In fact, rail guns were supposed to send stuff into outer space as an idea back in the 80s as part of Reagan's Star Wars program. But those designs involved high resistances needing very very high voltages, and that was too expensive. In my design, of low resistance armatures, those problems are avoided. With increasingly low resistance, the cost will be much cheaper.
> 
> This is with a power source that can produce constant acceleration,
> which would have instantaneous loads higher than 6.02e11 watts.

Don't confuse watts with joules. The energy required is as I found out. It all has to be discharged in 2/55 seconds and that will be the wattage over that period.
 
> BTW: What is the physical size of a 19200-Farad capacitor bank?

These days they have 3000F 2.7V capacitors. I used 6 1500F 2.7V capacitors in my experiments. It is not just the capacitance but the voltage as well. A 19200F 27000V capacitor has to be custom made. It should be the size of a large building, with many layers of plates. On the other hand, if the resistance is reduced from 1000 microohms to say 10 microohms, then the whole thing will become much smaller and cheaper. It is all a matter of design. 
> 
> Then, with the 1 km rail gun aimed vertically, the launched mass (even
> if accelerated in an evaculated tube) will encounter atmosphere at at
> altitude of 1000 meters, where the speed of sound is:
> 
> https://www.engineeringtoolbox.com/elevation-speed-sound-air-d_1534.html
> 336.4 m/sec
> 
> 55,000 / 336.4 = Mach 160, as previously noted by Edward Prochak.

Yes. Not too difficult a calculation, but nice.
 
> Using numbers from the link above, the relative air density at at
> altitude of 1000 meters is:
> 
> (89.84 / (6.9 + 273)) / (101.35 / (15 + 273)) = 0.91
> 
> The drag on the launched mass would be huge at Mach 160 through an
> atmosphere that has 91% of sea-level density.

That will depend upon how much the projectile is streamlined, designed, oriented, etc. I don't think anyone has fired anything at Mach160. To reduce drag the tip of the projectile could emit some air or gas at Mach 160 and that will simply blow the air in front of it away, and thus drag will be reduced. The projectile will have an air cushion around it, travelling at its own speed. Let us not forget, that with this speed the projectile will be out of the atmosphere in less than half a second.

No doubt there will be technical issues of the kind you have raised, but I don't see how they are very serious. Ideas such as I just thought up, and wrote above, will be generated to get around the problems as they come. Genius works like that, see.
> 
> Imagine the abuse experienced by the payload:
> 1. Enduring 150,000-g while accelerating from zero to 55,000 m/s.

Not a problem if properly strapped. There is no strain upon any body in the container. The strain is in the large current carrying conductor. That MAY get strained with the electrons forced by the huge magnetic field to push in the direction of motion. It will be a new effect if observed.

> 2. Zero acceleration when leaving the evacuated tube.

True. No problem there. Once there is zero acceleration things need not be tightly bound up.

> 3. Deceleration due to encountering atmosphere (with 90% of sea level
> density) at Mach 160.

Not much, with technical innovations as I have mentioned earlier. In any case, this rail gun will work much better when fired from an orbiting space station. The great cold in space will make the design a lot smaller and cheaper.
 
> Scaling up a rail gun to serve as a space-vehicle launcher seems
> impractical.

It is practical with my new rail gun design involving a large bore, and low contact resistance. To repeat, it will work best from an orbiting station in space. It could be used to send cargo to the Moon, or to break up any asteroids threatening Earth, etc. On the other hand construction on Earth will be easier. 

Sending cargo to orbit Earth will be much cheaper and simpler with this new design rail gun. The idea of sending cargo to the Moon was inspired by a thread in alt.usage.english about Jules Verne's novel about sending people to go around the Moon using a very large gun.
> 
> Building a prototype would be premature until we have a way for the
> payload survive 150,000-g.

It can survive any g if tightly packed up. Just see how things crash in car accidents. If you are properly strapped you survive quite easily. Yes if you are not strapped up, then yes the side of the container will go right thunk against you and that will be not very nice.

Of course, before making a 55000m/s gun, they can start with much less speeds. They can put objects inside the shells of existing rail guns that go at Mach 7. See what happens to a cockroach if it is tightly strapped inside such a shell. My bet is that is will survive and not simply because it is a cockroach (great survivor!).
> 
> Also needed for the 150,000-g version: a 1-km-long, large-diameter,
> evaculated tube (with a quick-opening door at the exit). Lower
> g-forces would need a longer tube (and/or lower exit velocity).

Um, a plastic shell for the vacuum could be easily pierced.
> 
> Of possible interest:
> 
> <https://www.quora.com/How-does-the-US-Navy-rail-gun-work-That-must-need-an-insane-instantaneous-current-How-do-they-get-that-How-do-they-keep-it-from-melting>

Mine is a totally different design. I did PhD research in RMIT University, Melbourne for 4 years on the subject of rail gun recoil, so I am well aware of contemporary rail gun workings. I made a new design to find whether they have a recoil, or not. I have made my findings know to the world through detailed video films showing the gun, its design and workings, experiments with it, and the projections relating to travel with the new engine that could be formed with extending that design.

Cheers,
Arindam Banerjee

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#690718

On 6/15/2018 8:55 AM, Wally W. wrote:
> On Fri, 15 Jun 2018 05:20:36 -0700 (PDT), Arindam Banerjee wrote:
> 
>> To send a 400Kg object to the Moon, you need a 1Km long rail gun (of my new, low voltage design) using 25 million average amperes of current on the rails.
>>
>> Quite doable.
>>
>> The exit velocity should be about 55,000m/s which would mean  a lot of acceleration. What will happen if some lifeform is put in it? I don't think it will get harmed, going by the experience of the high rise lifts that use internal force linear motors for great acceleration, as opposed to the kind of pull-up external force stuff that Einstein bunglingly used to justify the notions of GR.
>>
>> would ,
>> Arindam Banerjee
> 
> Do you have values for:
> 1. The Mach number at your exit velocity?
> 2. The acceleration, assuming constant acceleration ... over a
> distance of 1 km ... from zero initial velocity?
> 
> 
Sorry Wally, but Banjo is an "idea man". He doesn't do math.

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#691122

It is impossible to travel to the moon using rocket engines since the effiency of a rocket is .01 % which is a joke but scientist cannot be anything real since to produce a discovery in physical reality requires experiments and guess what experiments require. Yes you got it measurements but relativity manipulates time and space and we get stuff like LIGO 10^-18 m armature contraction and the CERn accelerator with quadapoles and dipoles..together. My words proves that everything and anything is now possible. What do you think is not that great. Give us Free. Give us Free. Go out an take off all your clothes and say to everyone you meet Give us free and shake it.

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#691234

During the Apollo 11 lander descent from the moon's orbit to the surface of the moon, the lander achieves a velocity of 2 km/s then the lander's rocket engine is ignited and produces a thrust of 10,000 lb that reduces the lander's downward velocity. Near the moon's surface the Lander's thrust is reduced to 3,000 lb. At the moon's surface, the lander's rocket engine's thrust would result in the formation of an enormous amount of rocket smoke yet the lunar descent film does not depict rocket smoke. The lack of the moon's atmosphere is used to justify the non-existence of the rocket smoke but the production of the rocket smoke is caused by the combustion of the Aerozine rocket fuel (kerosene) and the oxidizer (liquid oxygen) which would result in the formation of an enormous amount of rocket smoke which is not depicted in the lander descent film. The decent film of the lunar lander shows the lander propagating in a horizontal direction that would require a thrust in the horizontal direction and a constant center of mass of the lander yet the lander would experience an enormous change in the center of mass because of the fuel being consumed during the landing which would shift the center of mass and render it nearly impossible for the horizontal motion of the lander during the descent depicted in the lander descent video. The Lunar Landing Research Vehicle (LLRV) that propagates in the horizontal direction is used to justify the lander propagating in the horizontal direction but the LLRV main engine is a jet that would not function on the moon. The descent film depicts the Apollo astronauts having a conversation that is being recorded in the descent film but the rocket engine would produce an enormous roar which would prevent the audio feed of the Apollo astronauts have their conversation recorded in the Apollo 11 landing film. The landing and ascent of the Apollo 11 astronauts would be a life and death experience yet the Apollo descent and ascent lander films do not depict the intensity that would be expected in such a dangerous endeavor. The photographic images of the Apollo 11 lander does not depict a blast zone beneath the exhaust nozzle of the lander caused by the 3,000 lb rocket thrust during the final descent. The argument that the 3,000 lb thrust is not significant enough to produce a blast zone beneath the lander is used to justify the non-existence of the blast zone. Using an analogy, a Lear jet engine is rated at 3,500 lb thrust. The 3,000 lb rocket thrust would result in a blast zone beneath the Apollo 11 lunar lander yet the Apollo 11 photographs (fig 22) show absolutely no disturbance of the fine particle matter in the area beneath the lunar lander. In the Apollo 11 photograph, the shadows of the lunar objects are pointing in different directions which suggest that the Apollo 11 photographs were fabricated. The variation in the contour of the lunar surface is used to explain the multiple directions of the lunar shadows but in an Apollo 11 photograph (fig 23), a rock and the lander are separated by the distance of 300 meter and are forming shadows with an angle of 45o on a near level lunar surface which negates the contour surface argument. In another argument, the earth's intensity is used to represent a second light source that forms the multiple directional  lunar shadows but if the light intensities of the Sun and the Earth formed the lunar shadows then each lunar object would form two separate shadows yet each of the lunar objects is forming a single shadow. No photographs were taken together of the stars of the stellar universe and an Apollo 11 astronaut on the surface of the moon. 

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#691277

<numbernumber1964@gmail.com> wrote:
> During the Apollo 11 lander descent from the moon's orbit to the surface
> of the moon, the lander achieves a velocity of 2 km/s then the lander's
> rocket engine is ignited and produces a thrust of 10,000 lb that reduces
> the lander's downward velocity. Near the moon's surface the Lander's
> thrust is reduced to 3,000 lb. At the moon's surface, the lander's rocket
> engine's thrust would result in the formation of an enormous amount of
> rocket smoke yet the lunar descent film does not depict rocket smoke. The
> lack of the moon's atmosphere is used to justify the non-existence of the
> rocket smoke but the production of the rocket smoke is caused by the
> combustion of the Aerozine rocket fuel (kerosene) and the oxidizer (liquid oxygen) 

Why do you continue to shovel crap like this out? The “facts” you cite are
simply incorrect? 

What do you gain by lying?

> which would result in the formation of an enormous amount of rocket smoke
> which is not depicted in the lander descent film. The decent film of the
> lunar lander shows the lander propagating in a horizontal direction that
> would require a thrust in the horizontal direction and a constant center
> of mass of the lander yet the lander would experience an enormous change
> in the center of mass because of the fuel being consumed during the
> landing which would shift the center of mass and render it nearly
> impossible for the horizontal motion of the lander during the descent
> depicted in the lander descent video. The Lunar Landing Research Vehicle
> (LLRV) that propagates in the horizontal direction is used to justify the
> lander propagating in the horizontal direction but the LLRV main engine
> is a jet that would not function on the moon. The descent film depicts
> the Apollo astronauts having a conversation that is being recorded in the
> descent film but the rocket engine would produce an enormous roar which
> would prevent the audio feed of the Apollo astronauts have their
> conversation recorded in the Apollo 11 landing film. The landing and
> ascent of the Apollo 11 astronauts would be a life and death experience
> yet the Apollo descent and ascent lander films do not depict the
> intensity that would be expected in such a dangerous endeavor. The
> photographic images of the Apollo 11 lander does not depict a blast zone
> beneath the exhaust nozzle of the lander caused by the 3,000 lb rocket
> thrust during the final descent. The argument that the 3,000 lb thrust is
> not significant enough to produce a blast zone beneath the lander is used
> to justify the non-existence of the blast zone. Using an analogy, a Lear
> jet engine is rated at 3,500 lb thrust. The 3,000 lb rocket thrust would
> result in a blast zone beneath the Apollo 11 lunar lander yet the Apollo
> 11 photographs (fig 22) show absolutely no disturbance of the fine
> particle matter in the area beneath the lunar lander. In the Apollo 11
> photograph, the shadows of the lunar objects are pointing in different
> directions which suggest that the Apollo 11 photographs were fabricated.
> The variation in the contour of the lunar surface is used to explain the
> multiple directions of the lunar shadows but in an Apollo 11 photograph
> (fig 23), a rock and the lander are separated by the distance of 300
> meter and are forming shadows with an angle of 45o on a near level lunar
> surface which negates the contour surface argument. In another argument,
> the earth's intensity is used to represent a second light source that
> forms the multiple directional  lunar shadows but if the light
> intensities of the Sun and the Earth formed the lunar shadows then each
> lunar object would form two separate shadows yet each of the lunar
> objects is forming a single shadow. No photographs were taken together of
> the stars of the stellar universe and an Apollo 11 astronaut on the surface of the moon. 
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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