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Is This Correct?

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  Is This Correct? hdblenner@gmail.com - 2015-08-25 03:48 -0700
    Re: Is This Correct? Poutnik <Poutnik4NNTP@gmail.com> - 2015-08-25 13:52 +0200
      Re: Is This Correct? hdblenner@gmail.com - 2015-08-26 07:59 -0700
        Re: Is This Correct? Poutnik <Poutnik4NNTP@gmail.com> - 2015-08-26 17:23 +0200
          Re: Is This Correct? R Kym Horsell <kym@kymhorsell.com> - 2015-08-26 15:45 +0000
            Re: Is This Correct? Sam Wormley <swormley1@gmail.com> - 2015-08-26 11:23 -0500
        Re: Is This Correct? Odd Bodkin <bodkinodd@gmail.com> - 2015-08-26 11:28 -0500
          Re: Is This Correct? Poutnik <poutnik4nntp@gmail.com> - 2015-08-26 20:27 +0200
        Re: Is This Correct? "nuny@bid.nes" <Alien8752@gmail.com> - 2015-08-26 10:20 -0700

#516722 — Is This Correct?

"Again, I don't want to try to snow the panel with a lot of equations. I think it is important here to point out that there is a significant difference between kinetic energy and momentum. As you see on top of this exhibit F-303, the energy, one-half mass times the velocity squared, is an expression of what shall we say, the destructive capability of the projectile, and as we all know from our familiarity with Einstein, that energy is conserved. Also momentum is conserved. But in this case, the conservation of momentum is slightly different from the conservation of energy. "
"Conservation of momentum is a vector quantity, that is, it has direction. If a projectile were moving along and then struck another object, then both of those objects would move off with exactly the same momentum that the first object had coming in. In other words, the linear momentum, the product of the mass and velocity, is conserved and the direction is conserved.
Let's apply both of these to a hypothetical bullet that is striking a head and losing some velocity."
"Now, the next line labeled momentum lost, all I have done is taken the product of the mass-this is 162 grains divided by 7,000 - which gives us the mass of the bullet in pounds. Multiply that mass of bullet in pounds times 800 - feet per second, the velocity lost, and we have a quantity, an unusual quantity, 18.4 pound feet per second of momentum which has been deposited by the bullet."

I find multiple errors. 

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#516729

On 08/25/2015 12:48 PM, hdblenner@gmail.com wrote:
> 
> "Again, I don't want to try to snow the panel with a lot of equations. I think it is important here to point out that there is a significant difference between kinetic energy and momentum. As you see on top of this exhibit F-303, the energy, one-half mass times the velocity squared, is an expression of what shall we say, the destructive capability of the projectile, and as we all know from our familiarity with Einstein, that energy is conserved. Also momentum is conserved. But in this case, the conservation of momentum is slightly different from the conservation of energy. "
> "Conservation of momentum is a vector quantity, that is, it has direction. If a projectile were moving along and then struck another object, then both of those objects would move off with exactly the same momentum that the first object had coming in. In other words, the linear momentum, the product of the mass and velocity, is conserved and the direction is conserved.
> Let's apply both of these to a hypothetical bullet that is striking a head and losing some velocity."
> "Now, the next line labeled momentum lost, all I have done is taken the product of the mass-this is 162 grains divided by 7,000 - which gives us the mass of the bullet in pounds. Multiply that mass of bullet in pounds times 800 - feet per second, the velocity lost, and we have a quantity, an unusual quantity, 18.4 pound feet per second of momentum which has been deposited by the bullet."
> 
> I find multiple errors. 
> 
why not use few formulas, instead of msny words, and SI units, that I
hoped are used in imperial regions at least withing scientific community ?

(162/7000 ) pounds  * 800 feet/s = linear momentum cca 18.5 pound x feet/s

Where did you find multiple errors ?
If the bullet is trapped by e.g. a block
used to determine the bullet speed,
the block gains initial momentum
18.5 pound x feet/s

-- 
Poutnik ( the Czech word for a wanderer )

Knowledge makes a great man humble, but a small man arrogant.

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#517006

On Tuesday, August 25, 2015 at 7:52:48 AM UTC-4, Poutnik wrote:
> On 08/25/2015 12:48 PM, hdblenner@gmail.com wrote:
> > 
> > "Again, I don't want to try to snow the panel with a lot of equations. I think it is important here to point out that there is a significant difference between kinetic energy and momentum. As you see on top of this exhibit F-303, the energy, one-half mass times the velocity squared, is an expression of what shall we say, the destructive capability of the projectile, and as we all know from our familiarity with Einstein, that energy is conserved. Also momentum is conserved. But in this case, the conservation of momentum is slightly different from the conservation of energy. "
> > "Conservation of momentum is a vector quantity, that is, it has direction. If a projectile were moving along and then struck another object, then both of those objects would move off with exactly the same momentum that the first object had coming in. In other words, the linear momentum, the product of the mass and velocity, is conserved and the direction is conserved.
> > Let's apply both of these to a hypothetical bullet that is striking a head and losing some velocity."
> > "Now, the next line labeled momentum lost, all I have done is taken the product of the mass-this is 162 grains divided by 7,000 - which gives us the mass of the bullet in pounds. Multiply that mass of bullet in pounds times 800 - feet per second, the velocity lost, and we have a quantity, an unusual quantity, 18.4 pound feet per second of momentum which has been deposited by the bullet."
> > 
> > I find multiple errors. 
> > 
> why not use few formulas, instead of msny words, and SI units, that I
> hoped are used in imperial regions at least withing scientific community ?
> 
> (162/7000 ) pounds  * 800 feet/s = linear momentum cca 18.5 pound x feet/s
> 
> Where did you find multiple errors ?
> If the bullet is trapped by e.g. a block
> used to determine the bullet speed,
> the block gains initial momentum
> 18.5 pound x feet/s
> 
> -- 
> Poutnik ( the Czech word for a wanderer )
> 
> Knowledge makes a great man humble, but a small man arrogant.

The weight, w, of an object equals its mass, m, multiplied by the acceleration of gravity, g. This result follows from the law of motion, f = m a, where the acceleration, a, is the acceleration of gravity, g, and the force, f, equals the weight. 

So a proper calculation of momentum would have divided the weight by the acceleration of gravity and multiplied this quotient by the speed. In this case the dimensions of the momentum become [ pound / ( foot per second squared) ] foot per second = pound second, which is a proper unit for the action of momentum.  

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#517011

On 08/26/2015 04:59 PM, hdblenner@gmail.com wrote:
> On Tuesday, August 25, 2015 at 7:52:48 AM UTC-4, Poutnik wrote:
>>>
>> why not use few formulas, instead of msny words, and SI units, that I
>> hoped are used in imperial regions at least withing scientific community ?
>>
>> (162/7000 ) pounds  * 800 feet/s = linear momentum cca 18.5 pound x feet/s
>>
>> Where did you find multiple errors ?
>> If the bullet is trapped by e.g. a block
>> used to determine the bullet speed,
>> the block gains initial momentum
>> 18.5 pound x feet/s
>>

> The weight, w, of an object equals its mass, m,
> multiplied by the acceleration of gravity, g.
> This result follows from the law of motion, f = m a,
> where the acceleration, a, is the acceleration of gravity, g,
> and the force, f, equals the weight.

Sure, this is obvious.

> 
> So a proper calculation of momentum would have divided the weight
> by the acceleration of gravity and multiplied this quotient by the speed.

Why, if mass is already known ?
BTW, mass of objects of vey big or small mass
is not detemined by weighing.

> In this case the dimensions of the momentum

Dimension of momentum is still kg m/s, resp. pound . foot/s

> become [ pound / ( foot per second squared) ] foot per second = pound second,
> which is a proper unit for the action of momentum.
> 
Weight as force and mass have different dimension,
so pound cannot be used for both.

Weight of mass of 1 pound on the Moon definitely is not 1 pound.
Guess why.

-- 
Poutnik ( the Czech word for a wanderer )

Knowledge makes a great man humble, but a small man arrogant.

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#517017

Poutnik <Poutnik4NNTP@gmail.com> wrote:
> On 08/26/2015 04:59 PM, hdblenner@gmail.com wrote:
>> On Tuesday, August 25, 2015 at 7:52:48 AM UTC-4, Poutnik wrote:
>>> why not use few formulas, instead of msny words, and SI units, that I
>>> hoped are used in imperial regions at least withing scientific community ?
>>> (162/7000 ) pounds  * 800 feet/s = linear momentum cca 18.5 pound x feet/s
>>> Where did you find multiple errors ?
>>> If the bullet is trapped by e.g. a block
>>> used to determine the bullet speed,
>>> the block gains initial momentum
>>> 18.5 pound x feet/s
>> The weight, w, of an object equals its mass, m,
>> multiplied by the acceleration of gravity, g.
>> This result follows from the law of motion, f = m a,
>> where the acceleration, a, is the acceleration of gravity, g,
>> and the force, f, equals the weight.
> Sure, this is obvious.
>> So a proper calculation of momentum would have divided the weight
>> by the acceleration of gravity and multiplied this quotient by the speed.
> Why, if mass is already known ?
> BTW, mass of objects of vey big or small mass
> is not detemined by weighing.
>> In this case the dimensions of the momentum
> Dimension of momentum is still kg m/s, resp. pound . foot/s
>> become [ pound / ( foot per second squared) ] foot per second = pound second,
>> which is a proper unit for the action of momentum.
> Weight as force and mass have different dimension,
> so pound cannot be used for both.
> Weight of mass of 1 pound on the Moon definitely is not 1 pound.
> Guess why.



Sounds like yet another case of confusion caused by using
the same word "pound" for "pound mass" as well as "pound force".

In SI, of course, N is for force and kg is for mass.

While there is that 9.8 factor converting weight in N to mass in kg there 
is no such factor converting 1 lb force to 1 lb mass because the 32 ft/s2
is "built in" to the defn of lb force.

But  there will be no convincing anyone of this if some kind of bar bet
is involved.

-- 
-- <http://www.forbes.com/sites/adamtanner/2013/06/17/the-web-cookie-is-dying
-heres-the-creepier-technology-that-comes-next/>
[Fingerprinting] allows a web site to look at the characteristics of a
computer such as what plugins and software you have installed, the
size of the screen, the time zone, fonts and other features of any
particular machine. These form a unique signature just like random
skin patterns on a finger. The Electronic Frontier Foundation has
found that 94% of browsers that use Flash or Java -- which enable key
features in Internet browsing -- had unique identities.
Fingerprinting may prove a more robust tracking technology than
cookies because the user's identity endures even if they erase their
cookies. Making changes to your software and settings only makes you
more identifiable, not less.

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#517023

On 8/26/15 10:45 AM, R Kym Horsell wrote:

>
> Sounds like yet another case of confusion caused by using
> the same word "pound" for "pound mass" as well as "pound force".
>
> In SI, of course, N is for force and kg is for mass.
>
> While there is that 9.8 factor converting weight in N to mass in kg there
> is no such factor converting 1 lb force to 1 lb mass because the 32 ft/s2
> is "built in" to the defn of lb force.
>
> But  there will be no convincing anyone of this if some kind of bar bet
> is involved.
>

   Bar bets are similar to USENET arguments. In the case of the latter,
   the participants must be willing to learn something.

-- 

sci.physics is an unmoderated newsgroup dedicated
to the discussion of physics, news from the physics
community, and physics-related social issues.

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#517025

On 8/26/2015 9:59 AM, hdblenner@gmail.com wrote:
> The weight, w, of an object equals its mass, m, multiplied by the acceleration of
> gravity, g. This result follows from the law of motion, f = m a, where the acceleration,
> a, is the acceleration of gravity, g, and the force, f, equals the weight.
>
> So a proper calculation of momentum would have divided the weight by the acceleration of
> gravity and multiplied this quotient by the speed. In this case the dimensions of the
> momentum become [ pound / ( foot per second squared) ] foot per second = pound second,
> which is a proper unit for the action of momentum.

Sticking to pound as a unit of force, I don't see a problem with this.
Have to be careful about the casual use of pound as a unit of mass.

There is an imperial unit for mass called a slug, which is indeed a 
pound second squared per foot. And so a slug foot per second is 
equivalent to a pound second.


-- 
Odd Bodkin --- maker of fine toys, tools, tables

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#517066

Dne 26/08/2015 v 18:28 Odd Bodkin napsal(a):

> 
> Sticking to pound as a unit of force, I don't see a problem with this.
> Have to be careful about the casual use of pound as a unit of mass.
> 
> There is an imperial unit for mass called a slug, which is indeed a 
> pound second squared per foot. And so a slug foot per second is 
> equivalent to a pound second.
> 

Those damned imperial units.                 :-)
If it comes to calculations.. oh my....

-- 
Poutnik ( the Czech word for a wanderer )

Knowledge makes great men humble,
but small men arrogant.

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#517046

On Wednesday, August 26, 2015 at 7:59:28 AM UTC-7, hdbl...@gmail.com wrote:
> On Tuesday, August 25, 2015 at 7:52:48 AM UTC-4, Poutnik wrote:
> > On 08/25/2015 12:48 PM, hdblenner@gmail.com wrote:
> > > 
> > > "Again, I don't want to try to snow the panel with a lot of equations. I think it is important here to point out that there is a significant difference between kinetic energy and momentum. As you see on top of this exhibit F-303, the energy, one-half mass times the velocity squared, is an expression of what shall we say, the destructive capability of the projectile, and as we all know from our familiarity with Einstein, that energy is conserved. Also momentum is conserved. But in this case, the conservation of momentum is slightly different from the conservation of energy. "
> > > "Conservation of momentum is a vector quantity, that is, it has direction. If a projectile were moving along and then struck another object, then both of those objects would move off with exactly the same momentum that the first object had coming in. In other words, the linear momentum, the product of the mass and velocity, is conserved and the direction is conserved.
> > > Let's apply both of these to a hypothetical bullet that is striking a head and losing some velocity."
> > > "Now, the next line labeled momentum lost, all I have done is taken the product of the mass-this is 162 grains divided by 7,000 - which gives us the mass of the bullet in pounds. Multiply that mass of bullet in pounds times 800 - feet per second, the velocity lost, and we have a quantity, an unusual quantity, 18.4 pound feet per second of momentum which has been deposited by the bullet."
> > > 
> > > I find multiple errors. 
> > > 
> > why not use few formulas, instead of msny words, and SI units, that I
> > hoped are used in imperial regions at least withing scientific community ?
> > 
> > (162/7000 ) pounds  * 800 feet/s = linear momentum cca 18.5 pound x feet/s
> > 
> > Where did you find multiple errors ?
> > If the bullet is trapped by e.g. a block
> > used to determine the bullet speed,
> > the block gains initial momentum
> > 18.5 pound x feet/s
> > 
> > -- 
> > Poutnik ( the Czech word for a wanderer )
> > 
> > Knowledge makes a great man humble, but a small man arrogant.
> 
> The weight, w, of an object equals its mass, m, multiplied by the
> acceleration of gravity, g. This result follows from the law of motion,
> f = m a, where the acceleration, a, is the acceleration of gravity, g, and
> the force, f, equals the weight. 

  Gravitation was not mentioned in the text you cited, so weight is irrelevant.

> So a proper calculation of momentum would have divided the weight by the
> acceleration of gravity and multiplied this quotient by the speed. In this
> case the dimensions of the momentum become [ pound / ( foot per second
> squared) ] foot per second = pound second, which is a proper unit for the
> action of momentum.

  No. Momentum is *defined* as inertial mass M times velocity L/T.

  As others have said, you are confusing inertial mass with the force a mass exerts due to gravitation.


  Mark L. Fergerson

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