Wednesday, 31 July 2013

Draw and Poundage Arithmetic

The 35# Hazel overdrawn against the 50# Yew underdrawn has intrigued me so lets do some sums (well it's hardly higher maths is it?)

35# Hazel with a 5" brace height drawing 28" that gives a 23" power stroke (28-5")
So assuming the draw weight increases evenly (a pretty fair assumption)
That gives 35/23 pounds per inch of draw = 1.52 (I'm just going to 2 decimal places, as there are plenty of approximations)
So overdrawing to 30" gives a power stroke of 25"
25" x 1.52 pounds per inch = 38 pounds at our overdraw.
So the energy put into the bow is the total from zero at no draw up to 38 at 25". The average of 0-38 is 19
We can call this 19x25  = 475 pound inches. An unusual unit but it doesn't matter if we do the same for the other bow.

50# Yew with 6.5" brace height drawing 28" gives a 21.5" power stroke (28-6.5)
That gives 50/21.5 pounds per inch of draw = 2.32
Now under drawing to 27" would give a power stroke of  20.5"
20.5 x 2.32 pounds per inch = 47.56 pounds at our underdraw
So the energy put into the bow will be the average from 0 - 47.56 which is 23.78 times our short draw.
23.78 x 20.5 = 487.49 pound inches

Well there you go, the two energies aren't much different, the lady is getting 475 and the gents getting 487.49

Now if the lady manages the odd shot with yet another inch of draw while the gents are tiring, she could well out range them!
It would give a poundage of  39.52 and an average of  19.76 times the draw of  26"
That would be a total of 513.76 !
That really shows how a little extra draw goes a long way!

Also the lower brace on the Hazel adds extra power contrary to what you might think, as it increases the power stroke.

Sorry if that bored you all to death!
I s'pose I should plot a graph showing where the two lines of force vs draw for the hazel over takes the Yew, but I can't be bothered.
You can all do that for your homework! ; )

Update:- I found this old graph I'd plotted of a longbow vs my Asiatic/Horsebow.
It confirms that poundage starts from zero at somewhere near brace height (if you extrapolate the curves back to zero)


  1. "Sorry Sir - the dog ate my homework..." ;-)

  2. I like this approach to determining energy storage, very straightforward. It also implies that it does not mater much how much the bow had to be bent to even get to brace height.
    You read a lot about "stored energy" in that context, but looking at this equation implies that this is probably only wasting the woods capacity to store energy. Energy put in to brace will only be released upon unstringing. So no more reflexed bows. What do you think about this?

  3. Cheers.
    I think the energy required to brace the bow shows up in the higher pounds per inch figure, and the final draw weight.
    The big assumption is a linear increase of poundage with draw, but that's not too bad for a self bow (especially a longbow).
    It's a rather rough and ready approximation, but I think it gives a feel for some of the factors involved.