Can some one explain energy to me please?


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Eric F
September 3, 2009, 08:16 AM
I was running some figures for my 50-90 sharps on some web based calculators by pluging in velocity and bullet weight to get a energy value. All fine and dandy then I looked at some boxes of other ammo I have laying around and foung 3 boxes of the same chambering with 3 diffrent bullets types but they hade the same weight and velocity but the energy values were way diffrent.

Is energy dependant on bullet type? Do diffrent manufacturers use diffrent formula to develop a energy value? What gives?

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lykoris
September 3, 2009, 08:22 AM
W= weight in grains
V= velocity in feet per second

W x V x V(i.e. velocity squared) = a given number, say Y

divide Y by 450240 and you get the fpe value


so for example 40 grain bullet travelling at 1050 fps

would have a value of

40 x 1050 x 1050 = 44100000/450240 = 97.95 fpe (rounded up) or multiply by 1.3 to get the value in Joules

lykoris
September 3, 2009, 08:24 AM
by fpe I mean ft-lbf - foot pound force

Eric F
September 3, 2009, 09:20 AM
So then how can 3 diffrent manufacturers advertise the same bullet weight and same velocity but have diffrent energy values?

deadin
September 3, 2009, 09:29 AM
So then how can 3 diffrent manufacturers advertise the same bullet weight and same velocity but have diffrent energy values?

By having a faulty calculator?? By creating their own laws of physics??
Check and see if they are listing the energy as Muzzle Energy, 100 yard energy or other different ranges.

However, the one I like the most is "Creative Marketing":eek::D

lykoris
September 3, 2009, 09:42 AM
how can 3 diffrent manufacturers advertise the same bullet weight and same velocity but have diffrent energy values?

People get it wrong. Lyman has the value as 450400 rather than 450240, maybe the former is easier to remember than the latter ?

they also use ft/lbs which from my understanding is incorrect it should be fpe or ft-lbf....I didn't know (used to use ft/lbs all the time) until I read a thread on a British shooting forum with engineers arguing over the correct symbol and how the 'correct' symbol has been replaced with an incorrect one.

I must reread it again or even post it up here.

lykoris
September 3, 2009, 10:35 AM
for those interested, the thread I mentioned above which is an explanation of why ft/lb is incorrect and fpe or ft.lbf is the right term

http://www.airgunbbs.com/forums/showthread.php?p=3216809

rcmodel
September 3, 2009, 11:18 AM
ft/lb is the American term that has been the accepted norm in expressing muzzle energy for as long as I can remember.
What the British airgunners want to call it is bloody well their own business.

Now, what puzzles me is torque.

From the time I was a teenager in the 50's and first became interested in cars and motorcycles, engine torque was given in ft/lb's.

Now I see in the latest car & bike mags that it has become lb/ft.

What's up with that?

As to the OP's question?
Bullet shape has no bearing at all on muzzle energy.
It has a large bearing on downrange or retained energy however.
But it is not used in the equation to figure it.

rc

lykoris
September 3, 2009, 12:07 PM
it is an imperial measure so I would presume they know what they are talking about....being English and all.

there is a firearms section to that forum also and whether you are talking about a 16 grain pellet or 168 grain berger bullet, muzzle energy is muzzle energy. No need to knock airguns.

I think the problem is the / sign and lack of f...although I believe it can be written ft-lb

ft/lb vs ft-lbf to describe foot pound force.

perhaps it's all semantics :D

Steve C
September 3, 2009, 04:41 PM
Energy is Mass x Velocity squared. ftlb or FtLbf is correct as used. Reason being is that a pound (lb). is not a measure of mass but a measure of weight or actually force which is a measure of gravitational force on a unit of mass. In the English system of measurement the term for a unit of mass is called a Slug. A Slug is the unit of mass that is accelerated at the rate of one foot per second per second when acted on by a force of one pound weight. The Lbf just indicates that the lbs are units of force.

The constant 450240 of the equation in the 1st responding post converts grains (avoirdupois weight) to pounds. If one does all the math using all the proper units of measurement, most units cancel oul leaving Feet x Pound or foot pounds (FtLbs) as the remaining unit of measurement.

Now, what puzzles me is torque.

The physics definition of torque is "the measure of a force's tendency to produce torsion and rotation about an axis, equal to the vector product of the radius vector from the axis of rotation to the point of application of the force." The simple explanation is its the rotating or twisting force, in a simple model like a wrench the torque is calculated by multiplying he distance in ft. between the center of the bolt being turned and the point where the force is being applied by the lbs of force applied , thus ftlb of torque.

lykoris
September 3, 2009, 05:28 PM
this might be helpful also (I'm no engineer though)

McGraw Hill - 2nd edition

Dictionary of Engineering

foot-pound
foot-poundal

http://img35.imageshack.us/img35/9157/footpound.jpg (http://img35.imageshack.us/i/footpound.jpg/)

DBR
September 3, 2009, 08:54 PM
Also, to confuse things further:

FPE (foot pounds of energy) is the ability to do work. In the case of a bullet; crush tissue, expand the bullet etc.

Foot pounds of torque is a static force - like a weight on a lever.

rfwobbly
September 3, 2009, 10:59 PM
Now, what puzzles me is torque. From the time I was a teenager in the 50's and first became interested in cars and motorcycles, engine torque was given in ft/lb's. Now I see in the latest car & bike mags that it has become lb/ft. What's up with that?

Torque is nothing more than force acting at a distance. (The only gotcha is that the distance is always measured on a line through the center of rotation to a point where the force is "normal" (90 degrees) to the line.)

Being a simple product of multiplication, we all realize that 2x3 gives the same answer as 3x2. As go the digits, so go the units. Therefore "foot-lbs" is just as correct as "lb-feet", because both mean "pounds multiplied by feet".

The one thing it cannot be is "lb/ft", because that would mean "pounds divided by feet" or "pounds per feet", which is a decidedly different beast.

clem
September 3, 2009, 11:43 PM
Energy is what my grand daughters have, and what I don't have.:)

Mal H
September 3, 2009, 11:51 PM
Eric - what were the 3 energy values given on the 3 different ammo boxes, and what were the bullet weights and velocities listed?

People get it wrong. Lyman has the value as 450400 rather than 450240, maybe the former is easier to remember than the latter ?
Those values are both correct. :) It depends on where on the earth the person using the formula lives!

That factor is simply derived from the gravitation constant applied to the weight (mass) of the bullet when converted from grains to pounds, i.e., 2*g*7000. It is used only to make the math easier. The factor is about 450240 at 40 degrees latitude, and about 450400 at 45 degrees lat. It will be different at the equator than it is at the north pole. Different, but never wrong. Any factor from 450000 to 450600 won't make much difference in the calculated ft-lb of energy for most of North America. IOW, don't worry too much about that factor.

Remo-99
September 4, 2009, 12:03 AM
However, the one I like the most is "Creative Marketing"

Possibly the velocities stated on the boxes is just an approximate or rounded off figure and maybe the muzzle energy was calculated from actual velocities.

As previously said bullet shape/type doesn't alter energy stats at the muzzle, just velocities at differing ranges etc.(which in effect will change the energy stats at those ranges.)

don
September 4, 2009, 12:22 AM
The correct formular for kinetic energy is 1/2 mxvxv and the correct units are ft-lbs in the english system. Torque has already been defined but the correct units are lb-ft again in the english system. Source: Modern College Physics.

Eric F
September 4, 2009, 01:39 AM
Eric - what were the 3 energy values given on the 3 different ammo boxes, and what were the bullet weights and velocities listed?
I will have to get back home and look again. I will try to post again tomorrow.

lykoris
September 4, 2009, 04:10 AM
Those values are both correct. It depends on where on the earth the person using the formula lives!

that's very interesting Mal, thanks for explaining it to me as I thought it was a fixed constant in the equation but your explanation makes sense.

the great thing about forums is the input on things from members more knowledgeable than myself and learning something new

SSN Vet
September 4, 2009, 10:19 AM
Ahhhhhhhh!

not the pound force, pound mass, slug lecture again.....

all consequences of a system of measurement developed before Isaac Newton, when people didn't grasp the difference between mass and weight.

and once again illustrating how SI (commonly refered to as metric) is such a superior system.

TimRB
September 4, 2009, 11:23 AM
"and once again illustrating how SI (commonly refered to as metric) is such a superior system."

Right up until the point that people use mass units for force.

Tim

Jim Watson
September 4, 2009, 11:37 AM
Well, if they can't tell a kilogram from a Newton, that is not my fault.

But the metric system is French, so why do we bother with it? Everybody makes fun of all other things French.

Thomas Jefferson devised an American decimal system of weights and measures to go along with decimal money, but it did not catch on.

don
September 5, 2009, 12:25 AM
That factor is simply derived from the gravitation constant applied to the weight (mass) of the bullet when converted from grains to pounds, i.e., 2*g*7000. It is used only to make the math easier. The factor is about 450240 at 40 degrees latitude, and about 450400 at 45 degrees lat. It will be different at the equator than it is at the north pole. Different, but never wrong. Any factor from 450000 to 450600 won't make much difference in the calculated ft-lb of energy for most of North America. IOW, don't worry too much about that factor.
Please explain how the equator affects energy. The formular doesn't even use weight. It uses mass and velocity. Mass doesn't change regardless of where it is. Neither does velocity.

Spencer Hart
September 5, 2009, 02:09 PM
my head hurts after reading this:neener:

don
September 5, 2009, 04:47 PM
Mal H. I did not read your post correctly. My bad. You wrote about the factor and not the energy.

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