Formula for converting ft lbs to lbs per square inch

Triggermortis

Well-Known Member
Is there a formula for converting ft lbs to lbs per square inch?
I.E for the sake of the question, if a given bullet produces say 1000 ft lbs of muzzle energy, what would the figure be in lbs per square inch?
Would it be 1/144th of the ft lbs figure?
 
ft/lbs is not a pressure measurement. Its an Energy measurement and cant be converted into a pressure measurement.

what is it you are trying to do?
 
As stated the two measurements are for different things, there is no (readily available) direct correlation between ME and chamber pressure. I assume this is what you are looking for.
 
that formula is for converting between units of energy i.e. 'lbs of force per square foot' to 'pounds of force per square inch' then yes e.g. on a torque driver; but not converting energy to pressure as the others have already stated.
 
Is there a formula for converting ft lbs to lbs per square inch?
I.E for the sake of the question, if a given bullet produces say 1000 ft lbs of muzzle energy, what would the figure be in lbs per square inch?
Would it be 1/144th of the ft lbs figure?

So if a .250" diameter bullet is carrying 100ftlb energy, you want to know what pressure in lb sq in it hits the target at ?
Is it not as simple as working out the surface area of the bullet nose, dividing it into a square inch and X by the 1000ft lbs ?
And no applied mathematics is not my strong point :oops:

Neil. :)
 
So if a .250" diameter bullet is carrying 100ftlb energy, you want to know what pressure in lb sq in it hits the target at ?
Is it not as simple as working out the surface area of the bullet nose, dividing it into a square inch and X by the 1000ft lbs ?
And no applied mathematics is not my strong point :oops:

Neil. :)
Exactly Hornet, "what pressure in lb per square inch it hits the target at" If it is known for the purposes of this question that a bullet produces 1000ft lbs (1000lbs of energy per 1 square foot ?), there must be a way of calculating the equvilent figure per square inch? stands to reason doesn't it?
This is why i thought that as there are 144 square inches in a square foot, the figure would be 1/144th of 1000, I.E for the 1000ft lbs quoted, 6.944 lbs per square inch?
or have i got it wrong :?:
 
For your calculation, 1000 lbs per square foot of muzzle energy is 6.944 lbs per square inch of muzzle energy.

BUT, this NOT the "pressure in lb per square inch it hits the target at" because muzzle energy (i.e. the bullet's kinetic energy at the muzzle) is not pressure; so you cannot calculate it using your formula.

To calculate the energy at the target, you would need to factor in velocity and bullet weight.
 
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Is'nt a straight forward kinetic to potential energy conversion? Good old Newtons laws. You would need to know other factors within the calculation other than just the ft/lbs though.
 
It would be a very complicated calculation, as the shape of the round would affect how large the cross section is at various degrees of penetration.
Pedantic point, as the impact point is not enclosed the pressure will be 1 bar, that is why force units are used, as pressure is not applicable.
 
Tone, I will sort something out for you, although lord only knows why other than extending the arcane knowledge of that peashooter of yours. There is a big mixture of terms in the above thread and not all are applied correctly.

However as has been stated by the mighty Bewsh, it is a calculation. Muzzle energy as above is calculated in simple terms as 0.5 x mass of object x (velocity of object)2[squared]. It's imperial measurement is ft lb (note not foot squared lb).
Pressure = Force/Area, and as Limulus says, Newtons Second Law applies. This in simple terms says Force = mass x acceleration (for those that know, I know this isn't correct, but getting into a conversation about rate of change of momentum is probably not going to get far - happy to have it though).

So, by substitution, Pressure = (mass x acceleration)/Area.

Now we need to make some assumptions to make this easier and let's pick the point in time as the bullet strikes, i.e. the surface area of a ballistic tip is very small. Now in essence dividing by a very small number gives you a very large pressure, so your calc, wasn't quite rght anyway - you needed to divide by (1/144 in your case), which is the same effect as multiplying by 144. Now pick the distance at which you wish to fire the bullet at the target and we can extrapolate an acceleration from that (and the calibre obviously).

I'm going to do this in modern (SI) units rather than imperial, as it comes out in the wash a bit easier.

Assume 243 (specially for you Tone). Assume bang on the legal limit for deer at 1700 ft lb or 2305 Joules. Assume 100gr bullet = (64.8mg x100) = 6.48g = 0.00648kg. Assume muzzle velocity derived from energy. Assume surface area of tip of bullet = 1mm square = 0.000001 m2.

the calculation then effectively derives 2305/0.000001 = 2,305,000,000 Pascals - so - a LOT then. In old money, 1 psi (pound per square inch) = 6,895 Pa, so 334,300 psi at full muzzle velocity.

HAPPY NOW?!
 
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Tone, I will sort something out for you, although lord only knows why other than extending the arcane knowledge of that peashooter of yours. There is a big mixture of terms in the above thread and not all are applied correctly.

However as has been stated by the mighty Bewsh, it is a calculation. Muzzle energy as above is calculated in simple terms as 0.5 x mass of object x (velocity of object)2[squared]. It's imperial measurement is ft lb (note not foot squared lb).
Pressure = Force/Area, and as Limulus says, Newtons Second Law applies. This in simple terms says Force = mass x acceleration (for those that know, I know this isn't correct, but getting into a conversation about rate of change of momentum is probably not going to get far - happy to have it though).

So, by substitution, Pressure = (mass x acceleration)/Area.

Now we need to make some assumptions to make this easier and let's pick the point in time as the bullet strikes, i.e. the surface area of a ballistic tip is very small. Now in essence dividing by a very small number gives you a very large pressure, so your calc, wasn't quite rght anyway - you needed to divide by (1/144 in your case), which is the same effect as multiplying by 144. Now pick the distance at which you wish to fire the bullet at the target and we can extrapolate an acceleration from that (and the calibre obviously).

I'm going to do this in modern (SI) units rather than imperial, as it comes out in the wash a bit easier.

Assume 243 (specially for you Tone). Assume bang on the legal limit for deer at 1700 ft lb or 2305 Joules. Assume 100gr bullet = (64.8mg x100) = 6.48g = 0.00648kg. Assume muzzle velocity derived from energy. Assume surface area of tip of bullet = 1mm square = 0.000001 m2.

the calculation then effectively derives 2305/0.000001 = 2,305,000,000 Pascals - so - a LOT then. In old money, 1 psi (pound per square inch) = 6,895 Pa, so 334,300 psi at full muzzle velocity.

HAPPY NOW?!

Si you have far to much time on your hands just reading that made my head hurt :eek:
 
Not sure about that CV...!!

There's one assumption in here that I over simplified that has an important factor difference - and that is the factor of hoe much the target moves as a result of the strike. I'm going to say 1cm and this reduces the pressure by a factor of 100 - to 3,343 psi - still one hell of an amount at point pressure.

E&OE and all that - apologies for the rather obvious mistake.....
 
Is there something fishy about this, EtR?
Is not the acceleration you refer to in this case a deceleration, or rate of negative of change of velocity?
The pressure if the target is a sheet of lavatory-paper would the be next to nothing, as the change in velocity would be negligable.
If the target is 12" of plate-armour, it would be considerably higher as the bullet would stop.

I hypothesise that if it were a deer, it would be nearer the bumfodder end of the scale.

Disclaimer: might all be wrong.

It would be a very complicated calculation, as the shape of the round would affect how large the cross section is at various degrees of penetration.
Pedantic point, as the impact point is not enclosed the pressure will be 1 bar, that is why force units are used, as pressure is not applicable.

If you want another pedantic point to go with that one, it isn't the shape of the round so much as the shape of the bullet which will make the difference you refer to, perhaps?
;)
 
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Tone, I will sort something out for you, although lord only knows why other than extending the arcane knowledge of that peashooter of yours. There is a big mixture of terms in the above thread and not all are applied correctly.

However as has been stated by the mighty Bewsh, it is a calculation. Muzzle energy as above is calculated in simple terms as 0.5 x mass of object x (velocity of object)2[squared]. It's imperial measurement is ft lb (note not foot squared lb).
Pressure = Force/Area, and as Limulus says, Newtons Second Law applies. This in simple terms says Force = mass x acceleration (for those that know, I know this isn't correct, but getting into a conversation about rate of change of momentum is probably not going to get far - happy to have it though).

So, by substitution, Pressure = (mass x acceleration)/Area.

Now we need to make some assumptions to make this easier and let's pick the point in time as the bullet strikes, i.e. the surface area of a ballistic tip is very small. Now in essence dividing by a very small number gives you a very large pressure, so your calc, wasn't quite rght anyway - you needed to divide by (1/144 in your case), which is the same effect as multiplying by 144. Now pick the distance at which you wish to fire the bullet at the target and we can extrapolate an acceleration from that (and the calibre obviously).

I'm going to do this in modern (SI) units rather than imperial, as it comes out in the wash a bit easier.

Assume 243 (specially for you Tone). Assume bang on the legal limit for deer at 1700 ft lb or 2305 Joules. Assume 100gr bullet = (64.8mg x100) = 6.48g = 0.00648kg. Assume muzzle velocity derived from energy. Assume surface area of tip of bullet = 1mm square = 0.000001 m2.

the calculation then effectively derives 2305/0.000001 = 2,305,000,000 Pascals - so - a LOT then. In old money, 1 psi (pound per square inch) = 6,895 Pa, so 334,300 psi at full muzzle velocity.

HAPPY NOW?!

For those that are wondering what he looks like, I've just found a photo of Eric doing some light reading on ballistics:


article-2106025-0B1813B6000005DC-657_468x336.jpg
 
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