I know that foot pounds are a unit of torque. I know that PSI is a unit of pressure. I know that one needs more information than the foot/lbs to get the PSI.
Is area all that’s required to convert ft/lbs to PSI?
Let’s say I have an object with a sectional area of 64 square mm with a weight of 150 grains and a velocity of 1000 feet per second for a torque of about 330 ft/lbs.
If it hits something immobile, what will the PSI be? Is more information required?
So far you’ve described something with a momentum and kinetic energy. I’m not sure where you’re getting the torque figure from. A torque is change in angular momentum and I don’t see it in your example.
You’re a little confused. Torque is a measure of change in angular momentum, a measure of the force combined with moment arm needed to change the state of rotation of some system. It is the a-pplied force multiplied by the length of the moment arm times the cosine of the angle, more elegantly expressed in the mathematical form of a vector (“cross”) product. http://en.wikipedia.org/wiki/Torque
Pressure is force per unit area. Even if we’re talking about the same force in both cases, you need to know the physical setup – including the area the force is exerted over, the moment arm transmitting the torque, and the relevant angle – before you can make any sort of expression relating the (non equivalent) quantities.
Unfortunately there’s no easy answer for the pressure. Pressure is just force divided by area. You’ve got the area, but the force depends on the details of the target that’s slowing down the object (should I just call it a bullet, since that’s what it sounds like?). If you know exactly how much the target deforms (distance the bullet goes between contacting the target and coming to a complete stop), then you can get average force (average force = 1/2 mass *velocity squared /deformation distance), but that’s still an average, and the peak force is probably higher. And the average pressure (force divided by cross-section) is also probably lower than the peak pressure as well.
Unfortunately, there’s no easy way to tell how far the target will deform in advance.
No worries. If we’re talking about your object with said speed, mass and presented area impacting an immobile object (say, a massive wall that is perpendicular to the first object’s path) then the pressure (PSI) exerted will depend on the elasticity of your object. If it’s very rigid (high elasticity) the pressure needed to decelerate it will be high. Conversely, if your object is not rigid (low elasticity) the pressure will be lower.
This is because the time needed to deform the object is much less for the rigid object (higher acceleration) than for the non-rigid object (lower acceleration.) Both object’s impacts will be over the same area, given your conditions, but the forces generated (and thus the PSI) will be different.
What if you presume that it comes to a stop in 12 inches of flesh? Is the deformation supposed to be in inches or feet since we’re talking about both foot pounds and inches?
Presuming the deformation distance is supposed to be in inches:
If we presume 150 grains at 1000 fps coming to a stop within 12 inches:
75 grains X 1000 X 1000 = 7 500 000. Divided by 12 inches= 625 000.
Then divide 625 000 by the number of grains in a pound (7000)= 892.
That’s 892 for a 9mm circle (divided by Pi times radius squared) means (multiplied by 25.4 squared) 9059 PSI, have I made a mistake somewhere?
If the deformation is supposed to be in feet:
75 X 1000 X 1000 = 7 500 000 devided by 1 foot: 7 500 000.
7 500 000 divided by 7000= 10 714.
10 714 from a 9mm circle to 25.4mm square = 108 656 PSI.
Physics doesn’t care about what units you use, only that you use them. The distance it takes to come to a stop isn’t “12” or “1”, it’s “12 inches” or “1 foot”. For instance:
No. That second line should be
75 grains X 1000 fps X 1000 fps / 12 inches = 625000 grains feet squared per inches second squared. If you wish, you could convert the inches in the denominator to feet (or convert the feet in the numerator to inches) and partly cancel out.