Home Forums Campfire Forum Math Question;

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  • richard roop
    Member
    Post count: 106

    I know that there are guys here that are way better than me at math, so I thought I’d toss this out.

    Long time ago in a land far far away, I was hanging out at Tom Jenning’s (the compound guy) shop in Burbank.  I made a comment that it seemed like one could figure the foot pounds of energy that an arrow has and the bow’s draw weight in pounds and calculate the percent of efficiency to compare one bow to another.  Needles to say, Tom was really down on the concept, used a couple of engineering phrases and changed the subject.

    Sooooo ………….. Where am I wrong in my thinking ?????

  • Matt B
    Member
    Post count: 3

    The stored energy would be the “area under the curve” if you graph the draw weight versus length. So, if you, hypothetically, started at zero force and drew (power stroke) 24″ and ended with 50# draw weight, the area under the curve would be 50 ft-lbs. (25# average X 2′) This would be a very rare situation, but if you actually had this bow draw force curve, your formula would work.

    If you had a bow scale, you could take a measurement at each inch or two of draw length, graph it to your draw weight, and make a pretty simple graph and calculate the area under the curve. If that’s too much math, if you did it on square graph paper, with pounds on the vertical axis and inches on the horizontal, you could just count the squares under the graph line and be pretty darn close to the stored energy.

    Then you could shoot the bow through a chronograph and calculate the kinetic energy and get efficiency from that.

    I hope this helps.

    Matt

  • rgrist
    Member
    Post count: 35

    y=f(x) then integrate to get the energy under the force / draw curve.

  • richard roop
    Member
    Member
    Post count: 106

    Way more math than I can wrap my mind around.

    What I was implying was that, for example, my Super Kodiak, 60# @ 28″ shoots a 500 grain arrow 182 feet per second.  VxVxWT in grains divided by that 450,240 number  gives me 36.78 foot pounds. Divided by 60 pounds of draw weight ……….. the bow is 61% efficient.

    On the other hand, my 60# Pearson Mercury Hunter shooting the same  500 grain arrow at 174 feet per second comes out at 33.6 foot pounds of energy and is 56% efficient.

     

    No ??

    • This reply was modified 2 months, 3 weeks ago by richard roop.
  • Matt B
    Member
    Post count: 3

    Not really, but in personal, practical terms it probably will work for you to compare performance of different bows.

    A bow with a lower brace height will have a longer power stroke, so to most of us, it would seem “more efficient” because it shoots the same arrow faster with the same draw weight. In reality, it has more stored energy and may or may not be more efficient.  “Efficiency” requires comparing the stored energy (area under the force-draw curve) to the delivered energy (KE of arrow, just how you calculated it). But, to compare bows, all with the same draw length (you shooting) you can use your method. Should probably call it a “comparison quotient” or something rather that “efficiency” just to be accurate and not draw criticism from mathematical types.

     

     

  • richard roop
    Member
    Member
    Post count: 106

    Cool !!!

    Thanx.

  • RalphRalph
    Moderator
    Post count: 2499

    I got lost when 2+2=5………….:-))

     

    I shoot longbows cause I never was any good at rithmatic……………

  • richard roop
    Member
    Member
    Post count: 106

    Yeah, if it hadn’t been for my slide rule I never would have survived High School New Math.

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