MCgunner
Member
Common down here in the south, done it myself. Now, google Ben Lily. He hunted black bear with a knife. HE was da man....
How much energy in ft lbs ?
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Common down here in the south, done it myself. Now, google Ben Lily. He hunted black bear with a knife. HE was da man....
Kachok
Member
? Where do you get that from? I just crunched the numbers again on my metric calculator and came up with 6.63624 kg m/s or exactly 48.00173599 lb ft/s. Where on earth did you come up with 0.76? Do you honestly think a 240gr bullet moving at super sonic speed would have less then 1lbs/sec momentum, I would think common sense would be screaming from somewhere.
Here plug in the number for yourself. http://www.calctool.org/CALC/phys/newtonian/momentum
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When calculating the energy of a bullet in flight, its velocity parallel to the axis of the barrel is much greater than its velocity normal to the barrel. Since g is usually normal to the barrel we have to account for the fact that 1lb (mass) isn't the same as 1lb (force) i.e. g = 0 so divide the mass of the bullet (lb mass) by 32.2 to get lb (force). Since we also assume that the bullet's momentum is greatest in the direction of the bullet's flight we need to assume that g = 0 in this case also i.e. divide by 32.2 ... that's all I'm saying.
Kachok
Member
We are measuring at the muzzle so the g factor if not relevant at all so why bring it up? Mabey if we were talking about shooting 3,000 yards, but there is no such talk here. I still don't see how that could possibly make a notable difference much less reduce 48 lbs ft/s of momentum to .76!
I always calculated momentum as: projectile weight x projectile speed / 225,120 = momentum
55 gr. x 3100 / 225,120 = .76 momentum
In my previous example involving arrows, I cited a 400 gr. arrow @ 260 FPS which has 60# KE
400 gr. x 260 fps / 225,120 = .46 momentum
Using my calculation, a 95 gr. bullet @ 875 fps (pocket 380) produces a momentum of .37
A snub 38 special shooting a 130 gr. bullet @ 850 fps = .49 momentum
What unit are you using here? Your method seems absurd to me...
Kachok said:
Ask yourself this ... measured at the muzzle, is the kinetic energy and momentum of a bullet shot into the ground (straight down) the same as the energy and momentum of a bullet shot horizontally?
Velocity is a vector ... it has both direction and speed. The velocity vector for something that is "stationary" is in the direction of g only (we all have angular velocity on earth due to rotation but our velocity relative to the earth is axial only). The velocity vector for a moving bullet is along the bullets path AND in the direction of g. However, we discount the vector in the direction of g for a bullet moving at any significant velocity because it's much less (may even be negative) than the velocity in the direction that the bullet is moving in.
Kachok
Member
1858 what part of at the muzzle was I unclear about? Bullet drop has zero to do with that. I am still waiting to hear where you got this 0.76 number, I have tried common conversion mistakes and I cannot make that load come up at that number.
Kachok,
Your energy calculation for a 55gr bullet @ 3100fps is as follows:
(55gr*1lb)/7000gr = 0.00786 lbf
k.e. = mass*(velocity^2)*0.5
lbm = lbf*s^2/32.1 ft
k.e. = 0.5*(0.00786 lbf/[(32.1 ft/s)^2])*3100 fps^2 = 1176.1 ft-lb
Your momentum calculation is as follows:
Momentum = mass*velocity
M = (0.00786 lbf/(32.1 ft/s^2))*3100 fps = 0.76 lbm-ft/s
If you place a 55gr bullet on a table, the force it exerts on the table is 55 lbf which is the same as lbm because the force is in the same direction as g. By definition, 1lbf is the force required to give 1lbm an acceleration of 32.1 ft/s^2. If g = 32.1 ft/s^2, then 1lbf = 1lbm.
Your energy calculation for a 55gr bullet @ 3100fps is as follows:
(55gr*1lb)/7000gr = 0.00786 lbf
k.e. = mass*(velocity^2)*0.5
lbm = lbf*s^2/32.1 ft
k.e. = 0.5*(0.00786 lbf/[(32.1 ft/s)^2])*3100 fps^2 = 1176.1 ft-lb
Your momentum calculation is as follows:
Momentum = mass*velocity
M = (0.00786 lbf/(32.1 ft/s^2))*3100 fps = 0.76 lbm-ft/s
If you place a 55gr bullet on a table, the force it exerts on the table is 55 lbf which is the same as lbm because the force is in the same direction as g. By definition, 1lbf is the force required to give 1lbm an acceleration of 32.1 ft/s^2. If g = 32.1 ft/s^2, then 1lbf = 1lbm.
Kachok
Member
Oh well first of all did you not read my post about a 240 gr bullet? I was talking about the 44 magnum having 48 lb ft/s but even with the 223 I still come up with 24.35804702 lb ft/s not 0.76. You might want to run your math through that calculator I posted for you As I said The relationship between energy, mass, and momentum is not linear only mass/momentum is linear in terms of speed but the energy is dramatically different.
As I said The relationship between energy, mass, and momentum is not linear only mass/momentum is linear in terms of speed but the energy is dramatically different.
Direct relationship is a mathematical principle with a specific definition.
http://www.learner.org/workshops/algebra/workshop7/index.html
http://en.wikipedia.org/wiki/Direct_relationship
Mass is directly related to both kinetic energy and momentum because the relationships between them fits the mathematical definition for "directly related".
Again, you don't get to redefine terms that already have well-accepted mathematical definitions.
Momentum and kinetic energy both have a direct linear relationship with mass because their relationship fits the mathematical definition of direct linear relationship.
One of the things that makes math so valuable is that there are standardized definitions that don't allow ambiguity.
When you're talking about mathematics, you can't just come up with terminology out of the blue and pretend it means something specific.
For example, you made up the idea that there can be two things of "primary importance" in a single calculation. The general definiton of "primary" doesn't allow such a thing since in any group there can be only one thing that is primary, and second, there is no accepted concept of "primary importance" that applies to mathematical calculations. So the concept doesn't mean anything in in the general sense nor does it mean anything in the mathematical sense. That's the kind of thing I was referring to.
People who want to talk about quantities related to and defined by science of physics can't reasonably turn around and complain when people oblige them by talking about physics and physics equations and it's ridiculous to pretend that it's a negative thing to bring those topics into such a conversation.
The whole point of physics and physics equations is to help bring understanding which means that it's a GOOD thing to get into the details of how the equations work. If understanding isn't actually the point, we might as well just make a poll and vote--we can all pretend that the physical world is controlled by the majority opinion.
Kachok
Member
OK you got me I am no PhD in mathmatics, but how exactly is the relationship between speed mass and energy all a direct relationship? So I increase the speed two fold and get an energy increase of four fold, but momentum has not increased that much, how exactly is that linear???
Heavier projectiles will have a higher momentum for a given energy level then lighter faster ones, this is a well established fact and no amount of spin will change that, and last I checked that was my point to begin with, or did you not read that part?
Heavier projectiles will have a higher momentum for a given energy level then lighter faster ones, this is a well established fact and no amount of spin will change that, and last I checked that was my point to begin with, or did you not read that part?
Well, I've no idea how to answer how much energy is in foot pounds, and it seems from that rule of thumb of 1000 ft lbs for deer, not many can.
Anyhow, I like shooting black powder in a Kentucky rifle. A .50 cal round ball with about 80 grains of BP gets that 180 grain ball going around 1800 fps which creates around 1300 ft lbs.
Plenty right? Seems so, but it drops off quickly due to the bullet being a round ball. Even so, the people that hunt deer with these guns report the ball usually passes clean through a deer out to 100 yards.
I don't know what the energy is at 100 yards other than being well below 1000 ft lbs. Still seems to work though...
So maybe its not so important.
BTW about that rule of thumb of 1000 ft lbs....does that strike anyone else as alot for an animal thats basically human size?
Anyhow, I like shooting black powder in a Kentucky rifle. A .50 cal round ball with about 80 grains of BP gets that 180 grain ball going around 1800 fps which creates around 1300 ft lbs.
Plenty right? Seems so, but it drops off quickly due to the bullet being a round ball. Even so, the people that hunt deer with these guns report the ball usually passes clean through a deer out to 100 yards.
I don't know what the energy is at 100 yards other than being well below 1000 ft lbs. Still seems to work though...
So maybe its not so important.BTW about that rule of thumb of 1000 ft lbs....does that strike anyone else as alot for an animal thats basically human size?
Kachok
Member
Well remember those figures are for quick clean ethical kills with a small caliber not the minimum amount of energy that has the potential to kill. And yes an old musket shooting round ball is well below recommended levels of energy at range however as I stated larger caliber heavier projectiles tend to kill more efficiently then smaller faster ones of the same energy that is why I don't put much faith in the minimum energy formulas unless you are only talking about small caliber weapons only.
weight x speed / 225120 = momentum
I got .76 momentum for the 55 gr. @ 3,100 fps
Absurd?
See next reply.
Loosedhorse
member
I'm not talking about mathematics. Again (missed it the first time, did you?) I'm talking about rifle ballistics. Perhaps you think you've wandered into a mathematics forum here?
I understand that, in order to "prove" me wrong, you have decided to claim that I am making up novel mathematical concepts by using the commonly understood English words like "important", "primary," and "secondary."
Despite your obtuseness, I am not and was not writing a math treatise--therefore, criticising me for writing a math treatise incorrectly is silly.
I was and am talking ballistics, in English. And I have given you all the examples you need to see exactly what I mean, rather than to continue to make stuff up. And given you all you need to see that I am right: if you want a certain bullet momentum, you can achieve that by concentrating on either bullet mass or velocity--and so many choose a heavier bullet. If you want a certain energy, you can achieve that with less recoil with a lighter bullet than a heavy one, so many people choose a lighter bullet.
And that is why, in ballistics, I (and others) don't consider bullet weight "equally" important to momentum and to energy. You, however, do (as you stated in post #31) and are free to do so.
I'm not sure how you got from your reasonable "it's all in how you look at it" at post #36 to a relentless need to claim falsely that I'm inventing mathematic concepts. What I am doing is simply stating principles that are well known to anyone who handloads or for other reasons looks at the momentum and energy actually obtainable for different bullet weights in the same cartridge.
Aren't you a caution! First giving me a math (or was it physics?) lesson, and now an English lesson. All in the Rifle Country forum!
When two different items are both equally important, and both more important than any other factors, you say they can't both be primary? Kinda like if the two lead runners in a race tie, they can't both be in first place, "because of the general defintion of 'first'"? Okay: whaddya got besides your opinion that says I can't call both of those items "primary"?
All of us here in "Rifle Country" await your exegesis on the legal and illegal uses of the word "primary"!
:banghead:
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weight x speed / 225120 = momentum
I got .76 momentum for the 55 gr. @ 3,100 fps
Absurd?
Momentum is a vector, not a unit. Mass*velocity = momentum. Dividing by 225120 is absurd. Unless you're converting to some unknown unit.
theCan said:
How about some dimensional analysis to convert a bullet with a weight of 55 grains to a mass in units of pound mass (lbm). Why do we need to convert the bullet's weight in pound force (lbf) to pound mass (lbm)? Because we neglect the acceleration of the bullet in the vertical direction due to gravity because its acceleration in the direction of flight is much larger. If the bullet is sitting stationary on a table, then its weight in lbf is the same as its mass in lbm if g = 32.1 ft/sec^2. Every ballistic program converts lbf to lbm for a bullet in flight to calculate the ft-lb of energy at a given distance from the muzzle.
In this example, pound force is represented as lbf.
55 grains*(1lbf/7000 grains) = 7.86 x 10^-3 lbf
But we need lbm rather than lbf, therefore ...
7.86 x 10^-3 lbf*(sec^2/32.1 ft) = 2.45 x 10^-4 lbm
You may notice that 7000 * 32.1 = 224,700 which is close enough to the value of 225,120 used by CDW4ME (different value for g).
Kachok
Member
OK I gotta call BS on this 0.76 lb ft/s thing, I don't know where you get your formula from but you might want to double check it. THE formula for measuring momentum is p=mv it is real simple, weight in kg x speed in mps, nothing else needed. 1 kg is 15432.336gr so a 240gr bullet is 0.0155517609258896 kg and 1400fps is 426.72 meters per second. Now just multiply and we get 6.63624 kg m/s which if you convert to lbs is 48 lbs ft/s which both calculators and myself have said all along. If you wish to disprove the formula for linear momentum you should take that up with the guys at MIT or something but as far as I am concerned that is final.
I did my schooling overseas so I learned everything in metric, I don't know what kind of funky formulas you guys are talking about, but p=mv is the universally accepted formula for all science and it does not agree with yours.
I did my schooling overseas so I learned everything in metric, I don't know what kind of funky formulas you guys are talking about, but p=mv is the universally accepted formula for all science and it does not agree with yours.
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Kachok,
What you don't seem to understand with your "black box" calculator is that kg in the S.I. system is a unit of mass. The Newton (N) is a unit of force in that same system. No adjustment for gravity is needed in the metric system when using kg. However, in the Imperial system, there is pound mass (lbm) and pound force (lbf). They are only equivalent when g = 32.1 ft/sec^2. If you discount g as we do for a bullet in flight then you need to use lbm and not lbf.
This is the problem when you don't understand the basic principles involved. You can get a "black box" to spit out numbers wihout any real understanding. Calculating the momentum in S.I. units and then making a conversion to Imperial units without accounting for the difference in kg and lbm is the problem here.
What you don't seem to understand with your "black box" calculator is that kg in the S.I. system is a unit of mass. The Newton (N) is a unit of force in that same system. No adjustment for gravity is needed in the metric system when using kg. However, in the Imperial system, there is pound mass (lbm) and pound force (lbf). They are only equivalent when g = 32.1 ft/sec^2. If you discount g as we do for a bullet in flight then you need to use lbm and not lbf.
This is the problem when you don't understand the basic principles involved. You can get a "black box" to spit out numbers wihout any real understanding. Calculating the momentum in S.I. units and then making a conversion to Imperial units without accounting for the difference in kg and lbm is the problem here.
Kachok
Member
1 kg m/s is 7.23301385 lbs ft/s I don't see any notes in the conversion that imperial units have to be reduced because of gravity or any such thing.
I am going to just start talking in metric, Imperial gives me a fracking headache. OK so my 44 mag has 6.63624 kg m/s momentum but I cannot translate that into imperial, and since we know a kg m/s is always greater then a lb ft/s apparently neither can anyone else here since 6.63624 kg m/s cannot equal 0.76 lb ft/s.
I am going to just start talking in metric, Imperial gives me a fracking headache. OK so my 44 mag has 6.63624 kg m/s momentum but I cannot translate that into imperial, and since we know a kg m/s is always greater then a lb ft/s apparently neither can anyone else here since 6.63624 kg m/s cannot equal 0.76 lb ft/s.
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Loosedhorse
member
Confusion often comes from the fact that the lb is the imperial unit for both mass and force. One lb (force) = ~32.2 lb (mass) x ft/second^2. 1858 has it right.
So a unit for momentum can be either lb (mass) x ft/sec; or lb (force) x sec. And the numbers will be different depending on which lb you use.
One lb (mass) = ~0.454 killograms; but 1 lb (force) = ~4.45 Newtons. Similarly to above, momentum can be expressed either in units either of kg-m/sec, or N-sec.
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Kachok said:
I clearly stated that kg in the S.I. system is a unit of mass and therefore no adjustment for g is required. This is not the case for pounds mass and pounds force in the Imperial system and the specific case of a bullet and its energy in flight.
All you have to do is show a calculation for the kinetic energy of a bullet in flight using the Imperial system and the S.I. system to convince yourself of the difference between kg, lbf and lbm.
Kachok
Member
Yeah I erased that part when I re-read, sorry, but I am still waiting for someone to explain how 6.63624 kg m/s can be less then 1lbs ft/s that makes absolutly no sense at all kg m/s is a much higher unit of messurment.
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