# Acceleration force question

If I were to accelerate a straight shaft such as an arrrow to a speed of 260 fps over a distance of 24". The shaft weight would be 200 grains and the weight of the shaft equally distributed with no point weight. The length of the shaft would also be 24". What would be an equivalent amount of weight sitting on top of the arrow if the point were blocked against a table for instance? The bow has imparted about 30 ft #'s of kinetic energy into the arrow.

If I understand your question correctly, you are asking for the average force applied to the arrow during the period when the bowstring is accelerating it.

Going from 0-260 fps, its average speed is 130 fps; to cover the distance of 24", this requires 0.01538 seconds.

Accelerating from 0-260 fps in that amount of time works out to 16,899 ft/s[sup]2[/sup].

The mass of the arrow is 200 grains, or 0.0008873 slugs.

f=m*a, so f = 0.0008873 slugs * 16,899 ft/s[sup]2[/sup]

f = 14.99 pounds of force.

Note that this is the average. The force could be substantially higher at maximum pull (for a conventional bow) or near full release (for a compound bow).

The length of the arrow is not relevant for this calculation.

Did I understand your question correctly?

Maybe if I stated the actual problem it would be clearer. My hobby is archery flight shooting. This is a contest to see how far we can shoot an arrow. I use all wood primitive bows. The biggest challenege we face is comming up with arrows that are can maximise aerodynamics. We do this by making them the smallest diameter we can get away with. We als need to make them as light as possible for maximum launch speeds.

``````Unlike flight arrows traditional arrows which are much heavier and have substantial point weight can be determined by measuring spine. We do this by hanging a 2# weight on the center of the arrow that is hung at points 26" apart and then measuring the deflection of the arrow.

Flight arrows require a lot less spine because of no point weight. I am trying to find a way to get my arrows a skinny as possible without having them too flexable. I thought if I could find out how much force they were being subjected to I could place an equal amount of weight on top of the arrow while the point rested on the table and see how much it flexed, as if I were trying to push the point of the arrow into the table. We do this strictly by feel right now and by just finding out what seems to work best. If the arrow breaks which does happen  with some regularity there is a good chance it will go through my hand. I have witnessed this 3 times now and want to avoid it but I still need to push the limits a bit.

So, I guess I am trying to find out how much resistance to moving the arrow is giving the bow string.``````

Then yep, you’ve got your answer, sort of: 14.99 pounds, on average. For your wood primitive bow, the force during the launch sequence will not be constant: it will be much higher at the moment you release the string, and close to zero near the end of the launch sequence. It’s difficult to quantify the actual force-versus-time or force-versus-position curves without knowing:

-the length of the bowstring
-the spring constant of the bow
-the per-unit-length mass of the bow string (for very low-mass arrows, this may actually matter)

A pretty reasonable first-order approximation would be to assume the force varies linearly: zero force at zero pull, and 30 pounds at 24" displacement. A more sophisticated analysis will consider the above-listed factors.

For design purposes, you’ll probably want to make an arrow that can tolerate something close to 30 pounds of deadweight when you stand it up on a table.

I was thinking some more about this, and realized that this is not an accurate simulation of what’s happening when you launch an arrow.

In the above test, you have 30 pounds of weight pressing down on the butt of the arrow, and the table is pressing upwards on the nose of the arrow with 30 pounds of reaction force.

In a real arrow launch, at maximum pull the bow is pressing on the butt of the arrow with 30 pounds, but the reaction force (due to the mass of the arrow shaft) is distributed throughout the body of the arrow. This means that testing it with 30 pounds of deadweight concentrated at one end is an unnecessarily severe test; in order to pass this test, the arrow will be stronger (and therefore heavier) than it needs to be to survive a real bow launch.

If you want a more accurate test, you’ll need to find some way to apply a load that is distributed along the length of the arrow, with the butt of the arrow against the tabletop. Maybe an approximation that uses three ten-pound weights affixed at 1/3-height, 2/3-height, and the top of the shaft. or four 7.5-pound weights. The more weight/length partitions you use, the more accurately you will simulate the forces induced by an actual launch, although at some point the tediousness would exceed the value provided by any further increment in accuracy.

I see what you mean, the bows are actually 50# draw weight but allowing for efficiency only 30# is actually going into the arrow. There may not be an accurate way to pre-determine the actual needed spine beyond the trial and error we are doing now. The finished shape of the arrow is actually a slight barrel taper that allows it bend in a nice even arc, and also reduces weight on the front which further reduces need for spine. Thanks for giving it an honest shot.

In both cases, you will have stresses distributed throughout the body of the arrow. The test is accurate, to within the limits you stated earlier.

Here’s a useful equation to relate acceleration to distance travelled while accelerating at some constant acceleration.
v^2 - u^2 = 2as
( which is derived from two more
v=u+at
and
s=ut+1/2 at^2
)
where,
v- final velocity
u- initial velocity
a- acceleration
s- distance travelled