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Ft Lb Equation/Formula


Dekers
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I suspect many are familiar with the way to work out energy (ft Lb) for rifle ammo/pellets, but a recent thread has thrown question over the division figure.

 

Recently mentioned, or what I have just turned up with on a quick search are:-

 

450240

450395

450400

450436

There may well be more

 

How is that figure calculated?

 

So, anyone got a definitive and a source for this definitive?

 

:good:

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2 is the 1/2 transposed from the formula for energy.

 

The 7000 we all know.

 

I can't be precise, without checking - Googling '450436 constant' should enable you to find it - but at a conference in Paris 32.174 was internationally agreed as the standad for gravity (apparently this figure varies throughout the world but the set figure gives an international base).

So: 2 by 7000 by 32.174 = 450436.

 

Really, any of the given numbers are such a small difference as to be of no consequence. However, as in all things, if you have the precise information you can decide for yourself what tolerance is acceptable.

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..........................

 

Really, any of the given numbers are such a small difference as to be of no consequence. However, as in all things, if you have the precise information you can decide for yourself what tolerance is acceptable.

 

I can see for most practical purposes the different figures will bring little meaningful differences in results.

 

Just curious where all these figures have come from? :good:

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have a look here,reloadammo.com/footpound2.htm

I suspect many are familiar with the way to work out energy (ft Lb) for rifle ammo/pellets, but a recent thread has thrown question over the division figure.

 

Recently mentioned, or what I have just turned up with on a quick search are:-

 

450240

450395

450400

450436

There may well be more

 

How is that figure calculated?

 

So, anyone got a definitive and a source for this definitive?

 

:good:

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dekers,

 

It will be different constants of gravity which were used. The number of grains in a pound (7000) and the factor of 2 multiplier are exact quantities with no variation. The differences in the set of numbers above are different values of the speed of gravity from 32.16 to 32.174. As mentioned above, the real difference is negligible unless you're shooting long range.

 

thanks,
rick

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0.04 % is the difference between the highest and the lowest of the figures you quote and there can be no doubt the effect of the difference is miniscule as others have said. I'd pick the median and use that - personally.

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I suspect many are familiar with the way to work out energy (ft Lb) for rifle ammo/pellets, but a recent thread has thrown question over the division figure.

 

Recently mentioned, or what I have just turned up with on a quick search are:-

 

450240

450395

450400

450436

There may well be more

 

How is that figure calculated?

 

So, anyone got a definitive and a source for this definitive?

 

:good:

I'll have a go for fun (going almost back to first principles)

 

source for the data is Kaye and Laby tables published by NPL National Physics Laboratory. This is the ultimate source for measuremets in the UK. http://www.kayelaby.npl.co.uk/units_and_fundamental_constants/1_1/1_1_3.html

 

you probably know that Kinteic Energy (KE) = 0.5mv2 where m=mass and v=velocity

 

 

Now if we were happy to work with SI units, then KE in Joules would be - square brackets indicate the [units]:

KE [J] = 0.5 x m [Kg] x v [m/s] x v [m/s] and this would give an exact answer everytime with no room for variation and at the moment no reference to gravity is needed..

 

 

 

now if we want to use velocity in ft/s instead, the formula becomes (1ft is 304.8mm):

KE [J] = 0.5 x m [Kg] x 0.3048 x v [ft/s] x 0.3048 v [ft/s]

 

 

now if we want to use mass in grams as it is more convenient than Kg:

KE [J] = 0.5 x 0.001 x m [g] x 0.3048 x 0.3048 x v [ft/s] x v [ft/s]

 

 

now if we want to use mass in grain. This is the first place where variation possibly sets in, but NPL define 1gr as exactly 0.06479891g, so substituting in, we get:

KE [J] = 0.5 x 0.001 x 0.06479891 x m [gr] x 0.3048 x 0.3048 x v [ft/s] x v [ft/s]

 

 

now the fun comes if we want to have the answer in ft lbf. For the minute I will run with the NPL figure of 1 J = 0.737 562 ft lbf, substituting in we get:

KE [ft lbf] = 0.737562 x 0.5 x 0.001 x 0.06479891 x m [gr] x 0.3048 x 0.3408 x v [ft/s] x v [ft/s]

 

tidying up we get:

KE [ft lbf] = 0.00000222006742007 x mass [gr] x vel [ft/s] x vel [ft/s]

 

 

or expressing as a reciprocal, to give your preferred format:

KE [ft lbf] = (mass x vel x vel) / 450436.77 :good:

 

************************************************************************************

 

where does the reference to gravity come from?

It is involved in the way ft.lbf (foot pound-force) is defined.

 

The SI unit of energy is Joule which is a specific name for Nm. This has dimensions for length (metre) and for force (Newton). In SI this is straightforward as a Newton is a Newton whether on the earth or the moon.

In the imperial system, the unit of energy is ft lbf. Again this has dimensions for length (foot) and force (pound-force). Now though the pound-force isn't an absolute unit, it is a relative unit ie it is the force exerted by gravity on a mass of 1 pound. This will actually vary from one place to another as local gravity varies slightly.

 

Note that the "f" is commonly dropped when writing the units,

so instead of ft lbf it is usually written ft lb (It is not ft/lb which means a completey different thing.)

 

We can convert from J to ft lbf as follows:

 

substitute 1 foot = 0.3048m we get:

1 ft lbf = 0.3048 x m lbf

 

substitute 1 lb =0.453 592 37 kg we get:

1 ft lbf = 0.3048 x m x 0.45359237 x Kgf (where Kgf is kilogram force which in Newtons is Kg x gravity constant )

 

Although local gravity varies by approx 0.5% from the equator to the poles, g is defined by NPL for use in calculations as exactly 9.80665 m/s2 (approx 32.174 ft/s2) Substituting in, we get:

1 ft lbf = 0.3048 x 0.45359237 x 9.80665 x m x N

 

tidying up

1 ft lbf = 1.3558179 Nm (Nm=J)

or taking the reciprocal, 1 J = 1/1.3558179 ft lbf

so 1J = 0.7375621 ft lbf which agrees with the figure quoted by NPL earlier.

 

****************************************************************************************

so 450436 or 450437 looks like a good value to use. Anything else probably comes from too much rounding up in the calculations and maybe using an out-dated value for gravity as suggested already by plenty of others.

 

Hope that helps rather than hinders :good:

(ps it is getting late so that's my excuse in advance for any typos etc :lol: )

Edited by HW682
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Why do we even need to know? with the exception of off ticket arguments is law its meaningless as is kinetic energy when we are talking terminals as if you get a strike through the bullet must also carry some energy with it. 450,400 for me because its easy to remember. if anyone can remember some of the formulas given you have both my admiration and at the same time sympathy!

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I think everyone agrees that all the different values are close enough for any practical purposes, but there is nothing wrong with having an enquiring mindset and wanting to know where they come from.

 

1/2 m v2 is really quite a simple equation and is all we would need to work with Kg, m/s and J. The rest of it is just working out conversion factors one step at a time to use different units that are more convenient and/or familiar.

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