Some crazy stuff about velocity and energy relationships...

This is a discussion on Some crazy stuff about velocity and energy relationships... within the General Firearm Discussion forums, part of the Related Topics category; Preface : This is long, but only because I use a lot of examples to explain a very simple calculation. This would fall in the ...

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    Lead Moderator Array Tangle's Avatar
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    Some crazy stuff about velocity and energy relationships...

    Preface:
    This is long, but only because I use a lot of examples to explain a very simple calculation. This would fall in the category of being useful if you need it, and simply non-essential if you don't. I've been needing something like this for a while now just to help out with estimations at the range. It would be helpful if you are working up loads, especially for hunting, where you may be making a decision based on energy values.

    I'll use this to calculate percent changes in energy to predict energy shifts when changing barrel lengths. Ok, that being said here we go.

    Introduction:
    Sometimes I want to know, how much change in energy I would get if the velocity of a bullet increases by some percentage. Let's recall here that energy is a function of the square of the velocity. So a 10% increase in velocity makes a more significant increase in energy Ė but how much?

    Oh, BTW, if you're thinking you can subtract the two velocities and square the difference and calculate the energy, you can't. Well, you can do that but it won't give you the right answer. Just as an example,

    A change in velocity from 3,000 to 3,300 fps. The difference - 300 fps. The energy represented by 300 fps is, assuming a 55 gr bullet weight, is 11 ft-lbs. But that is no where near correct. The energy of a 55 gr at 3000 fps is 1100 ft-lbs; the energy of a 55 gr at 3300 fps is 1329 fps; the gain in energy is 229 ft-lbs, not 11 ft-lbs!
    So there was my challenge. I wanted an easy way to calculate the percent change in energy for a percentage change in velocity without having to square big velocity numbers and then calculate the percent change from that. Shoot thatís even hard to say!

    For example, letís say Iím at the range and I shoot a bullet from an AR and get, just for example convenience, 3,000 fps. I shoot the same bullet through a bolt gun with the same length barrel and I get a velocity of, again for example convenience, 3,150 fps. Thatís a 5% change in velocity. But by what percentage does the energy change? Is it enough to be significant? Is it negligible???

    The straightaway approach would be to calculate the energies of two bullets (the same type and weight) and find the percentage difference. In fact, since the bullet weight is the same, you donít really need to find the kinetic energy Ė just square the two velocities and calculate the percent difference. It sounds so easy when you say it fast like that! Well, I donít want to square 3000 and 3150 at the range and then try to divide one ginormous number into another. I need something I can approximate in my head or with simple paper and pencil calculations.

    So there it is, how do I determine the percent change in energy from the percent change in velocity? I got to playing around with the math using a spreadsheet and noticed something very interesting! You can pretty easily calculate the percent change in energy from the percent change in velocity - and it meets the pencil and paper math requirement!

    The gory (but easy) details:
    I think an example would be better right now than to try to explain the process in general procedural terms. So, letís take the very example I used above. Granted these numbers, 3,000 and 3,150 were chosen for example simplicity, but ANY numbers work. The hard part is you have to know what the percent change in velocity is. Well, Iíve made that easy by picking numbers, and Iíll get back to the percent calculation a bit later on.

    OK, as I stated earlier, an increase from 3,000 to 3,150 is a 5% change, so our task is to calculate the percentage the energy will increase from the 5% velocity increase.

    The solution:

    The first thing we do is convert 5% to a decimal:

    5% = 0.05 - and some other examples: 12% = 0.12; 22.3% = 0.223, etc.

    So now we take the 0.05 and double it, which gives us 0.1. Believe it or not, weíre almost done. The next step is to take the percent, 5%, in its decimal form, 0.05, and square it, i.e. 0.05 x 0.05. That gives us 0.0025. The last thing we do is add the latter value, 0.0025, to the former, 0.1, which gives us 0.1025. Well, there is one more step, convert this to a percentage, which is simply 10.25%

    The mathematical equation is: 2 x 0.05 + 0.05 x 0.05

    Simply stated, multiply the percent in decimal form by 2, then multiply it by itself (square it) and add the two numbers - it's that simple!
    Notice we could do this with no more than a pencil and paper Ė we wouldnít even need a calculator! And what we see is that a 5% change in velocity (for a given bullet) will increase its energy by 10%. So the percent of energy change is double the percent of velocity change! In fact in this case we could simply neglect the 0.025 and just call it 10%. Although, as the percentage gets larger, it's more than a doubling effect.

    But for changes up to 20%, you can simply double the velocity percentage to get a pretty good estimate of the energy. I.e. a 20% increase in velocity would give a 40% increase in energy - the error from only doubling up to 20% is 10% or less. Beyond 20% the error starts to increase too much to just double - you need to add the square thingy.


    Letís do this with 10%, i.e. the velocity changes from 10,000 fps to 11,000 fps Ė thatís a 10% increase. Actually, the actual numbers donít matter at all, just the percent change between them. So,

    10% example:
    10% = 0.1 ; we double the 0.1 to get 0.2; then square the 0.1 to get 0.01 and add 0.2 + 0.01 and we get 21%. From this we can see itís a bit more than a 2x ratio. If we did not do the square and add part, we would have 20% which is only a 5% error from the exact 21%.

    20% example:
    Letís do 20%: 20% = .20; .20 x 2 = 0.4 or 40%. We could stop there, but do we want to, or should we do the square and add? Letís do, just to see. 0.2 * 0.2 = 0.04; adding 0.4 + .04 gives us the exact figure of 44%. Leaving off the square and divide in this case would result in a 10% error.

    It's not always gonna be that easy, but as long as we can stick with two digit approximations, e.g. 7.5% 13%, IOW we don't get carried away with 7.53%, 13.3%, etc. we'll still be within the pencil and paper capability.

    Just a wee bit harder example:
    I will do one a bit harder. I've clocked a 110 gr 6.8 SPC from a 16" barrel at 2600 fps. I have a Bison Armory barrel on the way that's only 11.5" long and I've been told I'll get about 2300 fps from it. Let's look at the implications. The percent change in velocity is 13% or 0.13. Doubling the number is no problem - it's 0.26 and then all I have to do is - - hey, wait, maybe I'm done!

    This is a velocity change just a little over 10%. We saw the error by ONLY doubling 10% only produced an error of 5%. So the error produced by stopping here is gonna be about 6%. So I could stop here with an energy difference of 26%. But, I won't stop with doubling, just to prove I can do this without a calculator - you watch while I do this using only pencil and paper 0.13 * 0.13 = - - - - - - - - 0.0169! Adding 0.26 + 0.0169 we get right at 27.7% - pretty close, the approximation of 26% would produce a 6.5% error. But for thinkin' purposes, 26 - 28%, does it really matter?

    So there's my simple solution to seeing what effect a percent change in velocity has on the percent change in energy.

    Oh, I almost forgot - how do you get that percent change in velocity if you have two velocities? Well, you're likely gonna need a four-function calculator for that. Divide the lower velocity into the higher velocity and subtract "1" from that. The remainder is the percent difference in decimal form.
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    Lead Moderator Array Tangle's Avatar
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    Quote Originally Posted by ccw9mm View Post
    Is it something in the Chattanooga water, you think?
    Hey, I'm back in school - I have to think academically now - that's what I'm doing - you know, snatching practical application from the mouth of math
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    Quote Originally Posted by Tangle View Post
    Hey, I'm back in school - I have to think academically now - that's what I'm doing - you know, snatching practical application from the mouth of math

    Volume = is always the largest knob on the stereo
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    If you dropped 2 bullets off of Lookout Mountain..........
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    I like my bullets like I like my women: big and slow.
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    Quote Originally Posted by Tangle View Post
    Hey, I'm back in school - I have to think academically now - that's what I'm doing - you know, snatching practical application from the mouth of math
    Well, you sent up a parabola there. Much volume underneath those "slices" of area. Quick! What's the formula expressing the area under the curve for the one you launched, in the OP?

    Or, volume of a cylinder, as in how much air you pushed out of the way at the various velocities ...



    Just kidding. Glad to see you hittin' the books. Always a good thing.
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    Pretty cool - I like stuff like that. I was just gonna say "isn't that what a spreadsheet app is for?" - since almost all are gonna need the calculator on their phone for the velocity diff anyway.

    Only issue is that your velocity calc only works for increasing velocity. While going from 2300 to 2600 is a 13% change, going from 2600 to 2300 is only an 11.5% change. Or is it for your energy estimation calc, you always go from low to high for the formula to work?

    Dang it, Tangle (bet you've never heard that before ). You got me playing. If you get a velocity difference over 20%, just add the 10's digit to the percentage and you'll get reeeaaaaalllll close to actual w/o having to square anything.

    3000 -> 3800: 26.7% difference, doubled is 53%, add 5 for 58% vs actual of 60%.
    3000 -> 4200: 40% diff, doubled is 80, add 8 for 88% vs actual of 96. Little more off, but I'm thinking we're getting out of the realm of reloading small arms; or just double the 10's digit.

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    Lead Moderator Array Tangle's Avatar
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    Quote Originally Posted by ccw9mm View Post
    Well, you sent up a parabola there. Much volume underneath those "slices" of area. Quick! What's the formula expressing the area under the curve for the one you launched, in the OP?

    Or, volume of a cylinder, as in how much air you pushed out of the way at the various velocities ...



    Just kidding. Glad to see you hittin' the books. Always a good thing.
    Just kidding??? You get me all ready to tackle some more "practical math" () problems and now you're just kidding. Well that's sure let the air out of my sails, the powder out my ammo; the...
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    That's a good quick approximation.
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    Quote Originally Posted by Tangle View Post
    Just kidding??? You get me all ready to tackle some more "practical math" () problems and now you're just kidding. Well that's sure let the air out of my sails, the powder out my ammo; the...
    Yeah. You're a student. You got other things to worry about than some silly taunt from someone you don't even know.


    How about an easy one: the number of anti's in any given population.

    Hint: the number you think it might be, x2.
    Your best weapon is your brain. Don't leave home without it.
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    You lost me right after...

    Some Crazy Stuff About Velocity And Energy Relationships.
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    Well, perfessor, as one engineer to another, if I know the velocity gain in percentage, then the energy gain is that percentage squared. A 10% gain in velocity is 110% of the inital value, or a factor of 1.1. That's easy enough to square in my head, (1.1)^2 =1.21, or a 21% increase in energy.

    Maybe I've just been running numbers in my head for too many years, but I can square 10% increments in my head pretty easily, and if I linearly interpolate between those numbers for 5% increments, I'm pretty darn close without hardware or software tools.

    Not taking anything away from your estimation method! I've just been working the mental calcs for a long time and do it differently.
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    Wow, Tangle, you sure managed to make a simple calculation more complicated and less accurate!

    Quote Originally Posted by gasmitty View Post
    Well, perfessor, as one engineer to another, if I know the velocity gain in percentage, then the energy gain is that percentage squared. A 10% gain in velocity is 110% of the inital value, or a factor of 1.1. That's easy enough to square in my head, (1.1)^2 =1.21, or a 21% increase in energy...
    Took the words right out of my mouth, Smitty!
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    Lead Moderator Array Tangle's Avatar
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    Quote Originally Posted by nedrgr21 View Post
    Pretty cool - I like stuff like that.
    Me too!

    Quote Originally Posted by nedrgr21 View Post
    I was just gonna say "isn't that what a spreadsheet app is for?" - since almost all are gonna need the calculator on their phone for the velocity diff anyway.
    Yeah, maybe, but I doubt many are gonna have a spreadsheet app - - - well, unless I post a copy of mine. But they wouldn't have it at the range anyway.

    Quote Originally Posted by nedrgr21 View Post
    ...Only issue is that your velocity calc only works for increasing velocity. While going from 2300 to 2600 is a 13% change, going from 2600 to 2300 is only an 11.5% change. Or is it for your energy estimation calc, you always go from low to high for the formula to work?
    Nedrgr, you have just drug us into the wacky world of percentages.

    The formula to convert the velocity percentage to energy percentage works fine regardless of whether the velocity percentage represents an increase or decrease in velocity. But the formula I gave to calculate the velocity percentage was one-way - for increasing velocities only.

    In view of the need/desire to go either way, this formula can be used for either:

    % diff = 100% x | value1 - value2 | ų reference value

    The "|" mean absolute value of value1 - value2

    In words this formula could be stated: 100% times the absolute value of the difference divided by the reference value. The reference value will be one of the two values.

    Percentage calculations are weird, sometimes confusing, and sometimes even frustrating. Let's look at an example:

    Let's say I have an 11.5" barrel that produces a velocity of 2300 fps and a 16" barrel that produces a velocity of 2600 fps and I want to know what the percentage difference is between the two. Well, you can get one of two answers, depending on which value is used as the reference. Let's say we want to know how much velocity we pick up, percentage wise, with the 16" barrel. That makes the 2300 fps the reference because we are comparing a value to 2300 fps. We use the % diff formula from above:

    % diff = 100% x | 2600 - 2300 | ų 2300
    % diff = 13% more velocity.

    That is, 2600 is 13% more than 2300.

    But let's do the same thing the other way - i.e. we start with the 16" barrel and want to know what percentage we lose going to the shorter barrel. That makes 2600 fps the reference value. Again, the formula; notice it's the same numbers as before, but the reference value is 2600 instead of 2300:

    % diff = 100% x | 2600 - 2300 | ų 2600
    % diff = 11.5% less velocity.

    At this point, we can use my little formula to convert either percentage to energy percentage. Even better, we could use your simplification of it you described and quoted here!

    Quote Originally Posted by nedrgr21 View Post
    ...Dang it, Tangle (bet you've never heard that before ). You got me playing. If you get a velocity difference over 20%, just add the 10's digit to the percentage and you'll get reeeaaaaalllll close to actual w/o having to square anything.

    3000 -> 3800: 26.7% difference, doubled is 53%, add 5 for 58% vs actual of 60%.
    3000 -> 4200: 40% diff, doubled is 80, add 8 for 88% vs actual of 96. Little more off, but I'm thinking we're getting out of the realm of reloading small arms; or just double the 10's digit.
    Absolutely correct!

    And there could be a threshold where under the threshold you use just the 10's digit and above the threshold you do the squaring thing. And, I'd say you've already identified the threshold - about 25%.

    And I agree, that 40% would be far out of the realm of reloading small arms, but it might not be for calculating percentages from cutting a barrel down from 24" down to 10".

    Update: 21-Aug-2015, 9:48 am
    After playing with this some more, it looks like it may not work correctly for a decrease in velocity. I'm checking now to confirm this.


    Ok, after some more checking, for increasing velocity you add the square. For decreasing velocity you subtract the square.

    That's a little more complicated, but still really easy. Add for increasing; subtract for decreasing. The values obtained by this method and the classic percentage difference method agree exactly! Well, out to 5 decimal places; I didn't check it any further than that.
    I'm too young to be this old!
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