Foot Pounds

[SA John]

Can anyone explain what this unit of measurement means in some practical way that is easy to understand. I understand PSI, but foot pounds I can't visualize as it relates to doing work.

SA John

 

[Bolt308]

wikipedia is your friend.

http://en.wikipedia.org/wiki/Foot-pound_force

 

[KMaurer]

Wikipedia gives you the translation coefficients, but not the simple definition. A foot-pound of energy/work is the energy required to (or the work done by) raising a 1 pound weight 1 foot against the (standard sea-level) force of gravity.

Ken

 

[Another_David]

A simplified explanation:

Take a wrench one foot long and put it on a bolt. Then take a one pound weight and hang it on the far end of the wrench. The apparatus would be applying 1 foot pound of force on the bolt.

The heavier the weight the more foot pounds applied. Also, the longer the wrench, the more leverage you have and the foot pounds applied would also increase.

There are complicating factors, but that's the basic concept.

Diagram of bolt, wrench and weight:

<> ]-------- [1 lb.]

A better diagram:

http://mdmetric.com/tech/torqcht2.htm

 

[EddieCoyle]

A simplified explanation:

Take a wrench one foot long and put it on a bolt....

You're describing torque not energy, torque is measured in pound/feet, not foot pounds (even though it is often misexpressed as foot/pounds).

Ken's definition of foot/pounds of energy is accurate.

 

[dsm]

Take a wrench one foot long and put it on a bolt. Then take a one pound weight and hang it on the far end of the wrench. The apparatus would be applying 1 foot pound of force on the bolt.

That's a foot-pound of torque which is a completely unrelated thing. Sometimes called a pound-foot to avoid confusion.

The other post had it right, it's the energy needed to lift a pound of weight one foot.

I posted something here previously that while true from an energy perspective, ignored the effect of momentum and so was somewhat misleading. I decided I better retract it.

 

[SA John]

KMaurer's explaination is easy to picture. But does that mean a pistol round that generates 500 foot pounds has the juice to lift a 500 pound object one foot straight up if the object were shot from directly underneath and the target were able to absorb 100% of the energy? That seems hard to believe when a .44 mag round generating that kind of energy at 200 meters sometimes has trouble dropping a 50 pound steel silhouette target. And all it has to do is tip it over - not lift it. I believe KMaurer's example is the correct answer, it just doesn't jibe with what one observes in the field. What do you think? Is much of the energy dissipating as lead splash on the steel target? I have also watched bowling pins hit with rounds generating way more than 500 pounds strike pins solidly, stay in the pin, and fail to move it the requisite two feet off the table. Again, the pin took the whole energy load and didn't go very far at all, and they don't weigh anything like 500 pounds.

SA John

 

[Dick]

My physics courses are too far in the distant past for me to fully explain the difference (and I'm sure someone will be along shortly to do a better job of it), but your observations are correct and are due to the difference between energy and momentum. It's momentum that determines the knockdown ability of a projectile, and momentum increases linearly with velocity while energy increases as the square of the velocity. The high published energy for rifle or pistol bullets is due primarily to their high velocity rather than their mass, so for a given projectile energy the momentum is always far less.

 

[dsm]

KMaurer's explaination is easy to picture. But does that mean a pistol round that generates 500 foot pounds has the juice to lift a 500 pound object one foot straight up if the object were shot from directly underneath and the target were able to absorb 100% of the energy?

It does indeed have that amount of energy, and I had posted something almost identical to this as an example. But I decided to retract it later because the "absorb 100%" of the energy premise is so unrealistic that I thought that even though theoretically correct, it was too misleading an illustration to help anyone visualize the energy.

The problem is the vast differences in masses between the projectile and the 500 pound object, which makes it exceedingly unlikely that the energy can be transferred efficiently. Complete transfer of the energy would require a perfectly elastic collision and as the difference in mass between the objects increases, this becomes increasingly complex to have happen because of physical material limitations.

The difficulty comes from the need to accellerate the heavy weight. You would need a mass comprised of some kind of impossibly elastic super rubber to be able to "convert" the high-speed, low-mass energy of the projectile to a low-speed, high-mass energy in the weight by absorbing the fast motion of the projectile and releasing it slowly into the weight. This is because a heavy weight is hard to accellerate quickly.

Another way to picture it, rather than a rubber mass, would be a 500 pound weight with a very long spring under it, maybe 30 or 50 feet long. If the projectile was shot into the spring, the long travel of the compression of the spring would store and smooth the energy of the projectile and make the transfer to the mass above it more efficient as the other end of the spring slowly applied increasing accelleration to the mass. With a perfectly frictionless spring of exactly the correct stiffness and length, you could lift the mass one foot with the energy of the projectile. Sort of. Theoretically. Maybe if you added some gearing. This is why I retracted this earlier. Sigh.

The human mind has an innate understanding of the effects of momentum and so it makes it hard to use this as an example to understand the amount of energy in the projectile. That's why I retracted this sort of example from my original post. Maybe with the longer-winded explanation it's not so misleading.

 

[SA John]

dsm: That is not a bad example you give with the spring loading. It's kinda like gearing down a motor to turn a huge weight. Can't do it all at once, but it can work up to it over time. I did have one man tell me, years ago, that if the target and the bullet were able to collide with absolutely no distortion or bounce, then you would get the full transfer of energy. He seemed to think the energy loss was primarily in the powdering of the bullet. Think of how much force it takes to virtually atomize a solid chunk of lead and copper. And any denting of the target means more energy going into moving steel molecules around (huge energy drain) and not moving the whole mass. Still, it's funny to watch a popper target that can be knocked over with a breath of air, fail to fall to a bullet with 200 foot pounds of not-so-energetic energy.

SA John