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.
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Tbone38
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.
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.
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.
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).
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.
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.
We never got to finish that course in high school... the chalk kept breaking on the cave wall. :crying:
Jack
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:
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
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.
The whole 450400 is just rounding the product in the denominator (the # at the bottom).
Recall: KE = (1/2)×(Mass)×(Velocity)^2 where V^2 just means V×V or velocity times velocity.
So using the KE formula we can already see that the 2 in (1/2) is already in the denominator so what else is down there?
Well, let's look at M (Mass). When we use the above formula, we plug in bullet weight in grains but grains isn't mass it's force, we call it weight. So we need to do some conversion to get it into mass so we can use the formula KE = (1/2)×(mass)×(velocity)^2
Therefore, we change grains to pounds and we know there are 7000 grains in a pound or 1lbs/7000 gr. So this conversion gets our grains to lbs but it still isn't mass, to get to mass we have to convert lbs to mass by dividing by the acceleration of gravity (g), more specifically, (g = 32.174 but 32.17 is close enough. (not using SI, of course)
So now our KE equation looks like this:
KE = (1/2)×[(bullet weight in grains/7000×32.17)]×(Velocity)^2
Look at all the numbers in the denominator (bottom), it's 2×7000×32.17
When you multiply them all together you get 450380 which is close enough to 450400.
For the sake of curiosity, using g = 32.17142857143 will get you much closer to 450400.
If you like rounding g = 32.2 will give you 450800 as a constant in the denominator.
There's nothing mysterious about 450400, it's just a convenience to keep us from messing up calculations during the conversion process. Nothing more, nothing less. Einstein would be proud of the simplicity.
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Recall: KE = (1/2)×(Mass)×(Velocity)^2 where V^2 just means V×V or velocity times velocity.
So using the KE formula we can already see that the 2 in (1/2) is already in the denominator so what else is down there?
Well, let's look at M (Mass). When we use the above formula, we plug in bullet weight in grains but grains isn't mass it's force, we call it weight. So we need to do some conversion to get it into mass so we can use the formula KE = (1/2)×(mass)×(velocity)^2
Therefore, we change grains to pounds and we know there are 7000 grains in a pound or 1lbs/7000 gr. So this conversion gets our grains to lbs but it still isn't mass, to get to mass we have to convert lbs to mass by dividing by the acceleration of gravity (g), more specifically, (g = 32.174 but 32.17 is close enough. (not using SI, of course)
So now our KE equation looks like this:
KE = (1/2)×[(bullet weight in grains/7000×32.17)]×(Velocity)^2
Look at all the numbers in the denominator (bottom), it's 2×7000×32.17
When you multiply them all together you get 450380 which is close enough to 450400.
For the sake of curiosity, using g = 32.17142857143 will get you much closer to 450400.
If you like rounding g = 32.2 will give you 450800 as a constant in the denominator.
There's nothing mysterious about 450400, it's just a convenience to keep us from messing up calculations during the conversion process. Nothing more, nothing less. Einstein would be proud of the simplicity.
There are multiple ballistics apps out there as well as google search for formulas. I hate math. Energy numbers are an interesting diversion but in the real world of hunting or SD, doesnt mean a whole lot unless talking extreme ends of the energy argument.
In a handgun wouldn’t you say energy is only going to effect penetration? And most will do the job?
And as you pointed out correctly placement is going to trump most things such as caliber and velocity? And talking purely handgun cartridges.
Actually no, energy doesnt really mean much in penetration, its bullet design. A 38sp with 158gr LRN will likely go 24"+ in gel or thru a person. A 357mag with 125gr JHP will have 2X+ the energy & penetrate maybe 14".
FWIW, bullet placement with rifles isnt any diff. A near miss with a 458win mag isnt better than a perfect hit with a 30-30 as long as the bullet construction gets you to vitals. I helped dress a big bull elk with an old bullet wound in its shoulder caused by a 130gr BP 270 that didn't penetrate thru to vitals, bullet flattened like a quarter. People have been chasing energy numbers for decades & it often leads to choosing a weapon they cant shoot or using the wrong bullet thinking energy gets it done.
On my 2nd trip to Africa I wanted a cape buffalo stomper, so had a 458lott made. I couldnt shoot it effectively from every shooting position. So I sold it & had a 404jeffery made. It hammers everything including 2000# buffalo. Much less energy than the 458lott but i can shoot it well from any position needed. Shot placement is king with handguns or rifles. Power is useful but only if the bullets & shooter are capable.
FWIW, bullet placement with rifles isnt any diff. A near miss with a 458win mag isnt better than a perfect hit with a 30-30 as long as the bullet construction gets you to vitals. I helped dress a big bull elk with an old bullet wound in its shoulder caused by a 130gr BP 270 that didn't penetrate thru to vitals, bullet flattened like a quarter. People have been chasing energy numbers for decades & it often leads to choosing a weapon they cant shoot or using the wrong bullet thinking energy gets it done.
On my 2nd trip to Africa I wanted a cape buffalo stomper, so had a 458lott made. I couldnt shoot it effectively from every shooting position. So I sold it & had a 404jeffery made. It hammers everything including 2000# buffalo. Much less energy than the 458lott but i can shoot it well from any position needed. Shot placement is king with handguns or rifles. Power is useful but only if the bullets & shooter are capable.
Yeah I agree and I have several apps on my phone that I carry with me, not because I hate math but just because I'm lazy,... but that's just me.
Basically, I was responding to the question above asking where does that constant 450400 come from in the formula members where quoting. It didn't seem like anyone really knew and at least one member seem to be making stuff up without backing anything up... and no one really called him out on it, even after someone resurrected this thread in 2021.
I thought I'd just try to add some clarity to the issue. In case someone else besides me ends up using the search feature. 😆
Btw, that's a picture of Einstein, the dog from Back to the Future. Since someone in this thread was quoting Einstein's E=mc^2 which has little relevance to this topic.
I’m seriously asking. In simpleton terms, 158 gr 357 swc and same bullet in 38 spl. Is the higher muzzle energy or kinetic energy going to cause deeper penetration in one over the other?
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