Tbone38

04-09-2010, 00:03

Is there a calculator somewhere to convert your bullet speed and bullet weight into ft. lbs of energy? If not, what is the formula? I could have sworn that I seen a link to a calculator somewhere on this site.

View Full Version : Ft. Lbs. of energy???

Tbone38

04-09-2010, 00:03

Is there a calculator somewhere to convert your bullet speed and bullet weight into ft. lbs of energy? If not, what is the formula? I could have sworn that I seen a link to a calculator somewhere on this site.

Brian Lee

04-09-2010, 00:11

Bullet weight in grains x velocity (in FPS) squared, divided by 450400.

Example:

1200 FPS x 1200 FPS = 1440000

180 grain bullet x 1440000 = 259200000

259200000 / 450400 = 575.488 Foot Pounds of energy.

Example:

1200 FPS x 1200 FPS = 1440000

180 grain bullet x 1440000 = 259200000

259200000 / 450400 = 575.488 Foot Pounds of energy.

eaglesfan55

04-09-2010, 00:21

any idea of where the constant of 450400 comes??

Tbone38

04-09-2010, 00:37

Bullet weight in grains x velocity (in FPS) squared, divided by 450400.

Example:

1200 FPS x 1200 FPS = 1440000

180 grain bullet x 1440000 = 259200000

259200000 / 450400 = 575.488 Foot Pounds of energy.

Thanks man. I appreciate the info.

Example:

1200 FPS x 1200 FPS = 1440000

180 grain bullet x 1440000 = 259200000

259200000 / 450400 = 575.488 Foot Pounds of energy.

Thanks man. I appreciate the info.

Brian Lee

04-09-2010, 00:55

You asked:

"any idea of where the constant of 450400 comes??"

The constant would have to vary depending on the units of measure you are using for your bullet weight and velocity, and you'll notice that when using 450400, I specified using feet per second, and grains (one pound = 7000 grains).

First off, notice the similarity of the formula to it's more generic version popularized by Albert Einstein.

E=M(C squared)

E is your muzzle energy.

M is your bullet weight.

C is your velocity.

So in physics they use the same formula to calculate the energy of sub atomic particles which, in Einsteins formula, were assumed to be moving at the speed of light. But Einstein's version is the unit-less version of the formula, because it still has not added a constant at the end to convert the answer into a meaningful number.

But back to bullets:

Suppose you were measuring your bullet velocity in miles per hour, or furlongs per fortnight, or nautical miles per hour, or whatever. Then suppose you wanted to measure your bullet weight in metric tons, or some old-time British unit like Stones.

You could do the formula using any units you wish, as long as your constant was the correct one to convert the answer into the units you wanted your answer displayed in.

450400 just happens to be the one that works right using Feet Per Second, Grains, and Foot/Pounds of impact energy. But you could come up with any other one you need easily.

Lets just change out bullet velocity from FPS into Yards Per Second.

Our original example in my first post becomes:

400 Yards PS x 400 Yards PS = 160,000

and

160,000 x 180 grains = 28,800,000

Because we divided our velocity unit by 3 (from feet to yards) and we were using velocity squared in our formula, we need to divide our constant by 3 squared, which is 9.

450400 / (3 squared) = 50,044.4444444 = Our new constant for use when velocity is measured in Yards Per second.

so to finish our formula :

28,800,000 / 50,044.4444444 = 575.488 Foot Pounds of energy.

So as you can see, Einstein was calculating the energy present in sub atomic particles, in exactly the same way we calculate the energy in bullets hitting a target.

"any idea of where the constant of 450400 comes??"

The constant would have to vary depending on the units of measure you are using for your bullet weight and velocity, and you'll notice that when using 450400, I specified using feet per second, and grains (one pound = 7000 grains).

First off, notice the similarity of the formula to it's more generic version popularized by Albert Einstein.

E=M(C squared)

E is your muzzle energy.

M is your bullet weight.

C is your velocity.

So in physics they use the same formula to calculate the energy of sub atomic particles which, in Einsteins formula, were assumed to be moving at the speed of light. But Einstein's version is the unit-less version of the formula, because it still has not added a constant at the end to convert the answer into a meaningful number.

But back to bullets:

Suppose you were measuring your bullet velocity in miles per hour, or furlongs per fortnight, or nautical miles per hour, or whatever. Then suppose you wanted to measure your bullet weight in metric tons, or some old-time British unit like Stones.

You could do the formula using any units you wish, as long as your constant was the correct one to convert the answer into the units you wanted your answer displayed in.

450400 just happens to be the one that works right using Feet Per Second, Grains, and Foot/Pounds of impact energy. But you could come up with any other one you need easily.

Lets just change out bullet velocity from FPS into Yards Per Second.

Our original example in my first post becomes:

400 Yards PS x 400 Yards PS = 160,000

and

160,000 x 180 grains = 28,800,000

Because we divided our velocity unit by 3 (from feet to yards) and we were using velocity squared in our formula, we need to divide our constant by 3 squared, which is 9.

450400 / (3 squared) = 50,044.4444444 = Our new constant for use when velocity is measured in Yards Per second.

so to finish our formula :

28,800,000 / 50,044.4444444 = 575.488 Foot Pounds of energy.

So as you can see, Einstein was calculating the energy present in sub atomic particles, in exactly the same way we calculate the energy in bullets hitting a target.

Angry Fist

04-09-2010, 01:04

^^^ Thanx, Brian! How the hell did I miss E=mc2?

BobbyT

04-09-2010, 03:17

Once you know the ~450,000 constant, it's easy to do a rough calculation in your head to get in the ballpark if you just keep the most significant digit or two.

147 gr at 1200 fps:

1.5 * 12*12 / 450, or about 22/45 for a bit under 500 ft-lbs, say 470 (you know it isn't 50 or 5000).

200 gr at 1100 fps:

2 * 11*11 / 450, or 24/45 for about 550 ft-lbs.

160 gr at 3000 fps:

16 * 3*3 / 450, or about 150/45 for about 3300 ft-lbs.

147 gr at 1200 fps:

1.5 * 12*12 / 450, or about 22/45 for a bit under 500 ft-lbs, say 470 (you know it isn't 50 or 5000).

200 gr at 1100 fps:

2 * 11*11 / 450, or 24/45 for about 550 ft-lbs.

160 gr at 3000 fps:

16 * 3*3 / 450, or about 150/45 for about 3300 ft-lbs.

MinervaDoe

04-09-2010, 03:22

Is there a calculator somewhere to convert your bullet speed and bullet weight into ft. lbs of energy? .... I could have sworn that I seen a link to a calculator somewhere on this site.

Yup.

It's very cool.

http://billstclair.com/energy.html

Yup.

It's very cool.

http://billstclair.com/energy.html

BobbyT

04-09-2010, 03:22

Without going to the level of Einstein, if you remember high school physics, the formula for kinetic energy was:

KE = MV^2 / 2

with Mass in kg and Velocity in meters per second. So you could translate your bullet weight to kilos and your muzzle velocity to m/s, and get an answer in Joules with no need to scale it by a constant, but that would be pretty useless.

Take each constant to go from kg to grains, m/s to fps, and joules to ft-lbs, and you get ~450,000 (or 45 or 4.5 to make the math easier, since you're going to know the rough area to be in).

KE = MV^2 / 2

with Mass in kg and Velocity in meters per second. So you could translate your bullet weight to kilos and your muzzle velocity to m/s, and get an answer in Joules with no need to scale it by a constant, but that would be pretty useless.

Take each constant to go from kg to grains, m/s to fps, and joules to ft-lbs, and you get ~450,000 (or 45 or 4.5 to make the math easier, since you're going to know the rough area to be in).

Zombie Steve

04-09-2010, 07:36

http://www.beartoothbullets.com/rescources/calculators/php/energy.htm

...a simple one.

...a simple one.

PEC-Memphis

04-09-2010, 08:16

"any idea of where the constant of 450400 comes??"

The best answer is that it is a resultant of the units conversion so that when using FPS for velocity and grains (for mass, although strictly speaking grains is a measure of force rather than mass) the answer is in Ft*Lbs.

So as you can see, Einstein was calculating the energy present in sub atomic particles, in exactly the same way we calculate the energy in bullets hitting a target.

Um, not really.

The Newtonian form for kinetic energy is E(k) = 1/2*M*V^2

this is NOT E=M*C^2

This formula is the MASS to ENERGY (and vice-versa) calculation.

The "Little Boy" atomic bomb was equal to about 30,000,000 pounds of TNT. If this amount of TNT were detonated, all of the energy released would be chemical energy by an exothermic reaction - without any matter (mass of material) - being destroyed, eg. if all of the components were weighed before and after the reaction, the weight would be the same.

However, in the "Little Boy" atomic reaction (E=M*C^2), about 1/2 of a gram of U-235 (about 1/40th of a teaspoon of U-235, weighing about the same as 1/2 a teaspoon of water) of matter was converted directly to energy. Although about 140 lbs of U-235 was needed to achieve critical mass.

So you can see, they are really not the same at all.

The best answer is that it is a resultant of the units conversion so that when using FPS for velocity and grains (for mass, although strictly speaking grains is a measure of force rather than mass) the answer is in Ft*Lbs.

So as you can see, Einstein was calculating the energy present in sub atomic particles, in exactly the same way we calculate the energy in bullets hitting a target.

Um, not really.

The Newtonian form for kinetic energy is E(k) = 1/2*M*V^2

this is NOT E=M*C^2

This formula is the MASS to ENERGY (and vice-versa) calculation.

The "Little Boy" atomic bomb was equal to about 30,000,000 pounds of TNT. If this amount of TNT were detonated, all of the energy released would be chemical energy by an exothermic reaction - without any matter (mass of material) - being destroyed, eg. if all of the components were weighed before and after the reaction, the weight would be the same.

However, in the "Little Boy" atomic reaction (E=M*C^2), about 1/2 of a gram of U-235 (about 1/40th of a teaspoon of U-235, weighing about the same as 1/2 a teaspoon of water) of matter was converted directly to energy. Although about 140 lbs of U-235 was needed to achieve critical mass.

So you can see, they are really not the same at all.

GioaJack

04-09-2010, 10:41

Without going to the level of Einstein, if you remember high school physics, the formula for kinetic energy was:

We never got to finish that course in high school... the chalk kept breaking on the cave wall. :crying:

Jack

We never got to finish that course in high school... the chalk kept breaking on the cave wall. :crying:

Jack

OJ

04-09-2010, 11:01

any idea of where the constant of 450400 comes??

The constant 450400 comes from 7000 ( grains in a pound ) multiplied by (g=32.17) by 2.

2000 comes from 1000 ( grams in kg ) multiplied by 2

Only first two decimal digits are used.

:wavey:

The constant 450400 comes from 7000 ( grains in a pound ) multiplied by (g=32.17) by 2.

2000 comes from 1000 ( grams in kg ) multiplied by 2

Only first two decimal digits are used.

:wavey:

El_Ron1

04-09-2010, 11:19

A fireball calculator would be useful.

chemcmndr

04-09-2010, 14:33

You could also use the free app for the iPod/iPhone called "Bullet Power Calc"

Mystic Knight

04-09-2010, 18:28

I don't mean to be a smar-tazz, but I don't understand this.:headscratch:

ft-lbs is the unit for torque (force X distance)

energy is measured in joules

http://www.physics.uci.edu/~silverma/units.html

ft-lbs is the unit for torque (force X distance)

energy is measured in joules

http://www.physics.uci.edu/~silverma/units.html

chemcmndr

04-09-2010, 23:26

I don't mean to be a smar-tazz, but I don't understand this.:headscratch:

ft-lbs is the unit for torque (force X distance)

energy is measured in joules

http://www.physics.uci.edu/~silverma/units.html (http://www.physics.uci.edu/%7Esilverma/units.html)

Well, you are correct, but let me explain it a little bit better. Both torque and energy have the same physical units of measure, but they mean two different things. The unit of energy "Joule" is defined as the amount of force (pounds) required to move an object a certain distance (foot). So, you could express energy in terms of foot-pounds (in ballistics) or you could express it in terms of Joules (S.I./metric system of measure). As a side note, the common unit of energy used in US engineering parlance is the BTU. Torque is defined as a force (pounds) applied perpendicularly to an object a distance (foot) away from a point of rotation. Foot-pounds of torque is the U.S. conventional unit. In the metric/S.I. unit system, torque is defined as a Newton-meter. Hope this helps.

ft-lbs is the unit for torque (force X distance)

energy is measured in joules

http://www.physics.uci.edu/~silverma/units.html (http://www.physics.uci.edu/%7Esilverma/units.html)

Well, you are correct, but let me explain it a little bit better. Both torque and energy have the same physical units of measure, but they mean two different things. The unit of energy "Joule" is defined as the amount of force (pounds) required to move an object a certain distance (foot). So, you could express energy in terms of foot-pounds (in ballistics) or you could express it in terms of Joules (S.I./metric system of measure). As a side note, the common unit of energy used in US engineering parlance is the BTU. Torque is defined as a force (pounds) applied perpendicularly to an object a distance (foot) away from a point of rotation. Foot-pounds of torque is the U.S. conventional unit. In the metric/S.I. unit system, torque is defined as a Newton-meter. Hope this helps.

dbarry

04-10-2010, 07:12

We never got to finish that course in high school... the chalk kept breaking on the cave wall. :crying:

Jack

Our cave wall only went to algebra II. Thank God, cuz I got a C in that! :rofl:

Jack

Our cave wall only went to algebra II. Thank God, cuz I got a C in that! :rofl:

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