# How much (usable) potential energy is stored in a compound bow?

I have done a bit of reading about the energy stored in bows, but I haven't seen anywhere a description of how much energy actually is stored. Clearly there are many factors, bow design being dominant, but surely this value can be calculated. It appears that compound bows store the most energy, but how much is actually stored?

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You can draw the bow with force meter, noting the force at set intervals of distance. If you draw a graph where force is on the y-axis and distance on the x-axis, the energy stored is the area beneath the line. Mathematically, this would be $E=\int{F dx}$ with E = energy, F= Force and x the distance.

One part of bow design is actually making sure that this curve has a shape that comes close to filling a box within the constrains of the maximum length per the shooters arms lenth and so on, and the max strength - mongol compüound bows were built such that the leverage is shorter, and the required strangth higher, at the early part of the draw.

Now, for the usable energy, you would relax the bow and repeat the measurement - -you will probably see a curve slightly below the first one. However, to get an exact measurement you would have to relax the string (and measure) at the speed you would shoot at, because - and now I'm mostly guessing, because this is not my really my expertise - most of the energy lost will be lost due to internal frixtion and that is strongly dependant on speed.

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And we can get a very rough esitmate of that value by throwing some BoTE numbers at it. Assume that the bow requires a constant 300 Newtons to draw and that the archer's arms allows a 1 meter draw. You get 300 Joules. If a arrow masses 150 grams we can impart a velocity of $v = \sqrt{2 T / m} \approx 45 \text{ m/s}$. – dmckee Jan 21 '13 at 17:27