Foot Lbs Energy

[lonne]

How can foot pounds of energy be calculated at the target?

[southernxpress]

If I understand your question correctly.... that is very simple. Most any ballistics program will obviously tell you drop, but usually just one or two columns over to the right it will tell you the Ft. Lbs for that given distance as well.

[rcdinaz]

Yes, it is on all of the tables and there are many calculators online... or just download shooter on your phone and enter your info so you don't have to calculate the velocity at the estimated distance you are shooting.

 

Energy = .5 * weight * velocity^2 / 7000 / 32.175 or Energy = weight * velocity^2 / 450450

 

taken from here: http://www.tresterscustomguns.com/calcnrg.html

 

weight is bullet weigh tin grains.

[reganranch]

Physics and math will get you there.

[lonne]

Thanks - I'm aware that this number is calculated in ballistics tables. Much like a chronograph measures muzzle velocity of bullet, are there any feasible ways to measure the foot pounds of energy impacted at the target? :-)

 

Thanks,

Lonne

[lonne]

I guess some silly ways might be, put a chronograph at the target (don't hit it!) and measure the velocity, and then determine foot pounds that way. Or, shoot a watermelon, and if it blows up, a watermelon requires n (say 100) foot pounds to do so, so it's at least >= n (100). Just curious. Probably not any easy way, and not really necessary, but was just wondering if there were any good ways to do this. Maybe doing with the chronograph is actually quite scientific as long as one didn't hit it :-)

[pwrguy]

I don't know if there is something that can measure it, I doubt it. If you know the velocity and bullet weight the math don't lie?

[lancetkenyon]

The "ft/lbs" column in a ballistics program takes into consideration the velocity at target and gives you the kinetic energy at ranges. Check out www.jbmballistics.com . It will have a column at you can see what the projected energy will be at different distances. Just make sure you put in good data. Garbage in=garbage out.

[rcdinaz]

I don't think we are getting the question? If you have bullet weight and muzzle velocity then you know the velocity and energy for a given distance. Gravity is a constant so velocity changes a consistent rate.

 

Energy is based on mass and velocity... what part is missing that you need...

[lancetkenyon]

I think velocity drops off at different rates due to ballistic coefficient (the ability to fight drag) and bullet weight.

 

Not gravity, which is usually consistent, except when I trip carrying a heavy pack. Then it seems to increase exponentially. I could be wrong though.

[rcdinaz]

yes you are correct drag has a big effect.

 

Now that they put the numbers on almost every box of ammo/bullets I assume everyone has access to the BC. I just wasn't sure what problem the OP was trying to solve. There may be something specific but I think just about every ballistic app out there can give the answer at a specific distance to a pretty accurate level.

[rclouse79]

kinetic energy = 1/2 m v^2. To have the units work out in ft lbs you would need to take the grains of your bullet, multiply it by (0.000143lbs/grain) to get the weight in lbs. Then use the equation mass equals weight/gravity. Divide the mass of your weight in lbs by 32ft/s^2 to get the mass with units of (lbs)(s^2) / ft. You then need to know the velocity of your bullet at the target in ft/s. Plug the mass and velocity into K = 1/2 mv^2 and all units cancel except for ft lbs.

[lonne]

Apologies, the question I posed wasn't quite clear.

 

I understand math, physics, ballistics tables, and thank you all as some of your answers are very informative!

 

The question should simply have been:

How can foot pounds of energy be measured at the target?

 

I understand the inputs of ballistic coefficient, bullet velocity, wind, altitude, temp, pressure, bullet grain, humidity, etc. But what if the result is higher or lower, as it naturally won't match exactly. Was just curious what the scientific way of measuring the actual foot lbs of energy delivered (without destroying the target, if possible, pun aside)? If there are easy ways to do. Surely there is some way science has verified the data in the charts?

 

It sounds like knowing the actual velocity at the target would be useful. I suppose if one knew the drops at various distances for the same bullet they could plug it into a multiple equations and try to solve for the variable, or even a diff eq of sort.

 

Just curious if folks have found published bullet drops to be exactly accurate, or some marketing hype and not as accurate as touted, or better than advertised. Along with bc's.

 

Again, apologies my question wasn't clear and actually sounded like I was asking for a ballistic table. I guess I was wondering a way to verify the table and/or the drops plus minus that we see.

 

rclouse79, rcdinaz, lancetkenyon and pwrguy, THANKS!

 

Lonne

[rclouse79]

A ballistic gel manufacturer might be able to give you some data on the average frictional force the gel will apply to different projectiles. If you are able to find this out and the bullet stops inside the block, the change in kinetic energy will be equal to the work done on the bullet by the block, or (average frictional force) x (stopping distance) = initial kinetic energy.

If you are a really good shot, I think it would be easier to shoot your bullet through a chronograph 500 yards away and use k=1/2 m v^2. That could end up being an expensive experiment, but it would be fun.

[lancetkenyon]

I would say the only way to truly get your energy on target would be to do as you suggested, and place a chrony out at distance and shoot through it. Then do the math. But to get close, you would need to rely on a ballistic program unless you shot and got on target, then go out and place the chrony. Don't hit the chrony.

 

This would also be a good way to verify your ballistic program by checking remaining velocities at known distances. But if the drop is accurate, I would think the remaining velocities would be close, meaning the kinetic energy on target would be close too.

 

I go by the rule of 1250+ ft/lbs on target for elk, and 1000 ft/lbs of energy for deer. Per my ballistics, I am pretty confident at taking a deer with my .25-06 Ackley shooting a 115gr. Berger VLD out to over 1K yards, which I know still has approximately 1150, but the conditions better be perfect, but ETHICALLY, I would not take the shot much over 800. I know I can first round hit out to 800, with light-medium winds. But even a 2mph miscall on winds will push that bullet 8" at 1K.