Flagella dei

Russ Mitchell · Post by **Russ Mitchell** » Thu Apr 19, 2007 10:39 pm

I have to leave that to the experts... anybody?

I have my own question: I just derived the KE from the heavier Grozer bow there, the 57-pounder, and got averages of 83.2J, 85.9J, 85.6J, and 83.6J, respectively, with the four arrow weights documented.

EDIT: I'm an idiot. 41.6J, 42.9J, 42.8J, and 41.8J, respectively.

They show 70.8J, which is 1.24J/#, on a line with what Szollosy Gabor in Budapest has found with composites...

Can anybody tell how they derive their number? I'm bothered that mine and theirs would be so far off, but I don't have the calculus any more to derive the total area under the force curve for the second graph. And understanding how to derive this correctly is going to be important when I do some experimental archaeology and chrony my own bows later this year.

tnx...

Adam Karpowicz · Post by **Adam Karpowicz** » Fri Apr 20, 2007 10:03 am

Russ, your numbers are confusing.

The 42J you list for the KE is too low. I Think you meant 42ftlb. Otherwise the efficiencies, taking the 70.8J for stored energy, would be much too low. 42ftlb is about 57J, so the efficiency of your arrow is 57/70.8=80.5%, about right for a not-too-light arrow.

The stored energy of 70.8J was taken from the force-draw curve. They probably measured the weight every 1-2" and added all segments for the total.

1.24J/lb is possible, but not particularly good for a composite. especially for a longer one.

Is this bow a true composite ot a fiberglass replica?

Adam

Russ Mitchell · Post by **Russ Mitchell** » Fri Apr 20, 2007 1:19 pm

My numbers are 1/2mv^2 based on the arrow velocity and mass. (Converted to metric.) Did I mess up the math?

How are those numbers summed under the force curve? I get measuring by inch or half-inch, but what I don't get is the formula, since simple summing gets you (a ton of pounds force)*(28 inches draw).

Adam Karpowicz · Post by **Adam Karpowicz** » Fri Apr 20, 2007 4:24 pm

You need to keep your units the same for calculations.

A ton? try to divide by 12 - there is 12 inches to a foot.

Adam

Russ Mitchell · Post by **Russ Mitchell** » Fri Apr 20, 2007 5:57 pm

I was speaking figuratively, Adam. And we can't keep the units the same: one measurement is in Imperial and the other's in Metric: how on earth does one compute kinetic energy in foot-per-second-grams?

EDIT: Okay, actually, I misinterpreted what you meant. Yes, the numbers come into line nicely... but I'm still surprised at the low KE...