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Let's say I shoot 180 fps with a 400 grain arrow. I choose a new shaft and tune it, so that I get a 350 grain arrow. Is there a rule of thumb how many fps adds to the 180?

I'm searching for something like the draw weight rule, which says: "Add 5% of the poundage given at 28" per inch draw length."

  • Not sure there is a general rule, even the draw weight rule has a tons of different versions. When I hear stuff that extrapolates from IBO or ATA as starting point is all referred to compounds and even that is not consistent. As a matter of fact in traditional bows type of string and brace height has its own effect on speed. Heard things like 1.5 fps every 10grains but not sure how accurate it would be. At the end compounds are the ones that felt the need for speed, trads usually shoot slow and heavy and care less about the numbers so theres a bit less data – Erik vanDoren Apr 18 '16 at 14:04
  • BTW there are calculators online and most take a 1fps every 3 grains but its not even proportional as higher the arrow weight the more than 3grains you need to see that 1fps difference and its only on compounds. Remember that a compound can afford light arrows, trads dont, and once you use the same arrow weight the speeds are much more similar – Erik vanDoren Apr 18 '16 at 14:28
  • I guess the bow length is also something to consider as bows have a ideal peak load based on the draw length but this will be a constant in this case. – Desorder Apr 18 '16 at 22:48
  • I my world, fps stands for frame per second. I doubt it is what is meant here. – njzk2 Apr 19 '16 at 18:10
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    In other peoples world (but in mine neither =) fps is feet per second. – flawr Apr 19 '16 at 19:00
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I do not know any rule of thumbs but I do have some knowledge in physics and math, so lets see what we can do with that=)

Let's assume the energy that is put into an arrow is independent of the weight of the arrow. If we would want to model this too, it would be highly dependent of construction of the bow, and not useful for any rule of thumb. And as long as the weight of the two arrows we want to consider is sufficiently similar, this assumption will still provide good enough results.

Let's call the mass of the first arrow m1 and the mass of the second one m2, similarly v1 denotes the velocity of the first arrow, v2 the velocity of the second one. As the kinetic energy stays constant we have the equation:

0.5 * m1 * v1^2 = Ekin = 0.5 * m2 * v2^2

When we solve this for v2 we get:

v2 = sqrt( m1 / m2) * v1

where sqrt denotes the square root. If this formula is not simple enough we can approximate further:

If x is close to 1 then sqrt(x) is close to 1+x/2 (second order taylor approximation*), so we can simplify the formula to:

v2 = (1+0.5 * m1/m2) * v1  = (m2 + 0.5 * m1)/m2 * v1

Even this approximation of the coefficient is quite nonlinear so it is not possible to make a statement x grains equal y fps. In the following graph you can see the relationship between the ratio of the two velocities v2/v1 and the ratio m1/m2 of the masses.

As you can see my rule of thumb is less than 10% off the "exact" formula if the new arrow is at least half as heavy and at most twice as heavy as the old one. But if you consider arrows with a weight difference of a factor two, then the initial assumption is probably get us a way bigger error.

Example

In your example we have

m1 = 400 (units do not matter, as they'd cancel out)
m2 = 350
(v1 = 180fps)

Using the rule of thumb: In this case m1 and m2 are not really that close, so this result might be a bit off compared to the other one:

With the above values we have:

v2 = (1 + 0.5*(400-350)/350) * v1 = 1.071 * v1

That means the (muzzle-) speed of the new arrow will be about 7% greater than the one of the old arrow.

Using the "correct" formula

In this case we get:

v2 = sqrt(400/350) * v1 = 1.069 * v1

That would again mean the new arrow is about 7% faster than the old one. By the numbers you could guess that the approximation is about 3% off the "exact" value, as predicted.

Bottom line

Again looking at the graph above, we can conclude that if the weights are close enough, I suggest the rule of thumb the relative difference in velocity is going to be about half the relative difference in mass. What do I mean by that:

If you have a weight difference of 10% (as long as the weights are close enough, it does not matter which one you consider as 100%) then the difference in velocity is going to be about 5%.

Again checking with the "exact" formula (considering m1 = 100%)

v2 = sqrt(m1/m2) * v1 = sqrt(1.1) * v2 = 1.048 * v1 so that is about 5%

Alternatively if we consider m2 = 100%

v2 = sqrt(m1/m2) * v1 = sqrt(0.9) * v2 = 0.948 * v1 so that is about 5% too.
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    +1 for applying physics and maths... If you know that, you can figure out anything... :) – Desorder Apr 19 '16 at 20:38

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