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Math Majors- calculate weight of projectile

Discussion in 'General Gun Discussions' started by chestnut ridge, Apr 2, 2007.

  1. chestnut ridge

    chestnut ridge Well-Known Member

    Here is the data I have- projectile at 26,152 cm/sec or 858 fps.
    Kinetic energy of 3.07x10 (5) g-cm or 22.15 ft pounds.

    The (5) means 10 to the 5th power I think.
    I used to know how to figure this, but time does pass.
  2. KenRocks

    KenRocks Well-Known Member

    KE = (1/2) * m * (v^2)

    Cant find my calculator, but there you go :)
  3. Smellvin

    Smellvin Well-Known Member

    Should be about .9 grams (.898g to keep the three significant figures you have). This is the equivalent of 13.8 grains. The accuracy of your measurements obviously affect the accuracy of the calculation.

    By the by, you wanted a physicist -- not a mathematician. :)
    Last edited: Apr 3, 2007
  4. VARifleman

    VARifleman Well-Known Member

    0.878 gram You gained a decimal place somewhere...and to nitpick, grams are a mass, and thus g-cm is not an energy.
  5. mek42

    mek42 Well-Known Member

    You actually want a physics guy, but here goes.

    kinetic energy = 1/2 m v^2
    where m is mass and v is velocity

    rearranging gives m = 2 KE / v^2

    so (using the cgs units) m = 2*3.07E5/(26,152)^2 = 8.98 E-4 g
    = 0.0139 grains.

    Unless you're shooting fruit flies at 100 yards (and by that I mean using the fruit fly as a projectile) this doesn't seem right in a firearm forum, so let's assume that 26,152 is actually meant to be 26.152. Now the answer is 8.98 E2 g = 898 g = 13900 grains. That's one heck of a projectile.

    The end result is that I am unsure of the correctness of my answer because without knowing the context it seems to fail the reasonableness test.

    I hope this has been helpful somehow.

    Edit: Well it looks like others have posted since I started trying to figure this out and they disagree. It's late, I've had a crappy day and I'm probably wrong. Good luck.
    Last edited: Apr 3, 2007
  6. JohnKSa

    JohnKSa Well-Known Member

    About 13.6 grains. VARifleman is correct, that's about 0.878 grams

    KE (ft/lbs) = [weight(grains) x velocity(fps) x velocity(fps) ] / 450436

    22.15 = [weight(grains) x 858 x 858] / 450436

    22.15 x 450436 = [weight(grains) x 736164]

    22.15 x 450436/736164 = weight(grains)

    22.15 x 0.61187 = weight(grains)

    13.553 grains

    The 450436 is (2 x 7,000 x 32.174)

    2 is from the formula 1/2 x M x V^2

    7,000 converts grains to pounds

    32.174 converts pounds to slugs (unit of mass in the English system)

    You'll sometimes see 450250 or something similar in the denominator instead of 450436. That value is incorrect, but it turns out that anything that looks like 450xxx will give an answer to within a few percentage points of error so no one bothers to correct the value.
  7. VARifleman

    VARifleman Well-Known Member

    Just have to go and show all your work John...:p
  8. JohnKSa

    JohnKSa Well-Known Member

    How does that go...

    If you can't be first, at least show your work. (Hmmm... That doesn't sound right.) :D
  9. Hkmp5sd

    Hkmp5sd Well-Known Member

    I thought it was always show your work, at least you can get partial credit. :)
  10. Mal H

    Mal H Administrator

    John - I have to correct your absolute statement: "That value is incorrect ..."

    Most any value is correct within a certain range. It all depends on where you are talking about on the earth. At a latitude of 45 deg (N or S) your factor of 450436 is correct, but, for example, at the equator the factor will be approx. 449230. At the pole (N or S), the factor will be approx. 451600. The gravitational constant is a variable and not an absolute on the surface of the earth, and varies by latitude. It also varies by elevation, however, sea level is usually used as the standard elevation.
  11. chestnut ridge

    chestnut ridge Well-Known Member

    Thankyou one and all. The 13 grain answer is going to be what I was looking for. This involves ballistic injection of Naxcel antibiotic into feeder cattle.
    This is done by a air powered rifle that fires a pellet of 25 caliber; velocity
    and weight we know; into the flank of the sick steer. Look at
    www.solidtechah.com for information on the rifle. Look up ceftiofur
    sodium biobullet for articles on the antibiotic pellet.
    The outfit is well over 2000; I figure I can cobble up some type of squib
    load using a 6.5 rifle.
  12. shc1

    shc1 Well-Known Member

    Dang! Ats loud!
    Redneck (nobody said what kind of) physicist…:D
  13. 230RN

    230RN Marines on Mt. Curibacci

    Rifling on the delivery rifle?

    Just curious. Do you have any info on the form and twist of the rifling on that delivery rifle --if any?
  14. Jim Watson

    Jim Watson Well-Known Member

    A 6.5mm is .256" bore, .264" groove, not a good match if those are true .25" drug pellets. A real .25 cal with .257" groove diameter (.257 Roberts?) would seem indicated. A primer would probably be enough propellant.

    Or you could go air power for a lot less than $2000 if you don't need a repeater, there is such a thing as a .25 caliber pellet rifle. Not common, but Webley still makes them.
  15. Sniper X

    Sniper X Well-Known Member

    Let me guess, a formula for a .22ca pellet gun projectile?
  16. chestnut ridge

    chestnut ridge Well-Known Member

    The rational behind the "cow gun" is ease of treatment; and much less
    stress on the animal. This is designed to treat sick feedlot steers.
    Apparently the rodeo required to pen and headgate a recalcitrant steer
    causes higher mortality due to stress. By shooting the antibiotic pellet
    into the flank muscle or neck muscle; the animal just flinches as if a big
    fly has bitten. Then the antibiotic is absorbed so that the shipping fever
    sickened bovine has a greater chance of recovery.
    Any how, look at the www.solidtechah.com site for a video and much more
    than you ever wanted to know about shooting cows. They even have
    ballistic gelatin tests in slow motion. thanks to all.
  17. JohnKSa

    JohnKSa Well-Known Member

    Although, as you correctly point out, the value varies, there is an accepted convention just as there is an accepted value for other values (such as the radius of the earth, distance from the earth to the sun) that really have a range of correct values.

    While there is certainly at least one place on the earth where the gravitational constant takes the value 32.1607 (yielding 450250 in the denominator of the energy calculation) the convention is to use the "standard value of 9.806 65 m/s/s" -- equivalent to approximately 32.1741 in English units. That gives a value for the denominator of approximately 450436 or 450437.
    Last edited: Apr 12, 2007
  18. model4006

    model4006 Well-Known Member

    I like that...
  19. mek42

    mek42 Well-Known Member

    9.8 m/s^2 is not the gravitational constant, which is, in fact, a constant, not a variable.

    The gravitational constant is 6.67E-11 N m^2/kg^2.

    The 9.8 m/s^2 number comes from taking into account the earth's mass and an agreed upon normal distance fro the surface of the earth to the earth's center of mass. This value is commonly referred to as 'acceleration due to gravity'.
  20. JohnKSa

    JohnKSa Well-Known Member

    That's correct. I knew that and I still used the wrong term in my reply. :eek:

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