Can any one out there help with the formula to calculate impact force, and what amount is required for elk size game and deer size game?
I'll bet Dr. Ashbey would know.

Thankyou!!

T-Bone:
That would be next to impossible to calculate without knowing the velocity of the arrow at impact. and a arrow slowes down quite fast, so every yard the velocity would be different.
Bob

And the impct force will differant depending on the shapness of the point and the firmness of the tissues it hits....

Yep...WAY to many variables. Best way is to look at one of the ratios Dr. Ashby has already set up that are based on momentum prior to impact.

T-Bone, those are firearm questions and you can't compare rifle and bullet measures to archery broadhead principles and expect any meaningful results. We have zero impact compared to firearms-we kill solely by cut and bleeding, firearms have knock down power and bleeding is secondary. You might find what you are looking for on a regular hunting forum.

Maybe T-Bone means kinetic energy. Speed times mass = ke. I cant remember the formula off the top of my head.

I can tell you this from experiance this year.....

a 47# Bear Grizzly.....shooting a 2016 tipped with a green Bear Razorhead.....that hits a deer high in the neck striking the vertabrea at 22 yards.....will knock the deer right off it's feet.

My point is this.....

If the impact of an arrow from a 47# bow hitting the neck vertabrea will knock a deer off it's feet,I don't think you should have any worry about the "power" of an arrow hitting the vitals. You will have more than enough penetration and a good sharp broadhead will do the rest.

Fred bear put out a chart I once saw that demonstrated impact force for different bow eights and arrow grains.  What I am looking for is the formula for which arrow impact force is calculated.

the formula for Knetic energy is mass X velocity squared divided by 450,240.

So Feet per second X feet per second X weight of the arrow and divide that number by 450,240

Momentum is the amount of force it takes to stop an object. Mass X velocity. I think that may be what you want.

Its simple, basic physics.....F=MA
Force equals mass times acceleration
If you know the mass of the arrow and how fast it flies, then you know the force in which it will fly...but that is in a perfect vacuum.
To understand the distance you will get if you know the acceleration use s=1/2atsquared+vot
s being distance....a is acceleration, t is time vo is initial velocity and t again is time....time in this case is the time it took for your arrow to leave the bow and impact the target...initioal velocity in this case will be 0.

hope this helps a bit.  

I'll give it a shot. This isn't 100% correct as there are numerous variables, such as friction, arrow oscillation, arrow deceleration, etc, that should be quantified. But this will give you an idea of what you're after.

Impulse = change in momentum,
therefore:
Force X Time = Arrow mass X Arrow velocity
(assuming arrow stops in target)
rearranging and correcting for units:
Force = arrow mass X arrow velocity divided by time divided by 225,400.
where Force is in lbs, arrow mass is in grains, arrow velocity at impact is in ft/s and time the force is acting upon the target is in seconds.

I would assume time of force on the target to be in the neighborhood of 0.005 seconds. Faster arrow speeds equals less time and slower arrow speeds equals more time.

So for a 650 gr arrow going 165 fps with a time of impact of 0.005 sec you get 95 lbs of impact force.

I have no idea what the guidelines of impact force would be for elk or deer.

Try this web site and read Dr. Ashbys reports 1,2 & 3 (but stick with Kinetic Energy rather than Momentum):

www.archeryexchange.com/information/info_pages/kinetic/kinteic-calculator.shtml  -

Note: just do a search for "arrow kinetic energy" and much information will pop up.