Crossbow physics

1. Apr 7, 2017

pevi70

Hi everybody, congrats on this great forum, it was completely new to me.

I am working on this project to measure the physics parameters of crossbows. My father is a member of this guild that uses crossbows for shooting contests. The goal is to shoot small discs from 12 meter high poles with arrows that consist of wooden cilinders (20 cm x 2 cm). The crossbows are often handcrafted and differ quite a lot in shooting speeds.

Last year I built a speed tester and I think that I can use arrow velocity to calculate some crossbow characteristics which would allow some sort of comparison between these handbuild crossbows.

The following parameters would be interesting to calculate and I would really appreciate if you guys would have a look at the calculations.

The following data are available:
Velocity of the arrow (m/s), constant v.
Mass of the arrow is 100 grams, constant m
Tension distance 0.192 m, constant d.

1) Spring constant, k (N/m)
The potential energy of the spring is Up = 1/2 * k * d^2
This is converted to kinetic energy when the spring discharges Uk = 1/2 * m * v^2
The spring constant can be calculated from Up = Uk or

k = m * v^2 / d^2

2) The tension force F (Newton)
Hooke's law states, F = k * d

With k known and d as well, the tension force can be calculated..

3) The Power of the crossbow (this value is more relatable than a constant k)

The amount of energy that is transferred can be derived from the amount of kinetic energy devided by the time the spring needs to transfer this energy to the arrow.

The Uk = 1/2 * m * v^2 corresponds with the kinetic energy of the arrow.

The time can also be derived but I am a bit stuck.

The acceleration is dependent on the tension distance in time.
Could you help me here?

Regards,
Peter

Last edited: Apr 7, 2017
2. Apr 7, 2017

Staff: Mentor

Welcome to the PF.

Fun project! The reason that energy is the easiest way to calculate the velocity of the arrow leaving the bow (due to the energy in the compressed "spring" action of the bow before release) is because the force is not constant during the acceleration period on the bow. To get the velocity as a function of distance during the launch phase, you need to integrate the variable force through the full distance of the launch. It's not a hard integral, but it's easier to just use PEi = KEf (initial spring stored Potential Energy equals the Kinetic Energy of the arrow leaving the bow).

There are also some losses due to friction, and you may just have to calculate those based on the ideal velocity leaving the bow versus what you measure with your instrument.

BTW, what are you using to measure the velocity of the arrow?

3. Apr 7, 2017

Dr.D

Much of your discussion is predicated on the spring being linear (such as your potential energy expression). Have you ever checked the linearity? I would suggest this as an important thing to check early on, before you invest too much time and effort on a bad assumption.

4. Apr 7, 2017

Staff: Mentor

Good point! I'd thought early-on to recommend he plot out the spring constant versus pull distance, but then latched on to just using the 1/2 k x^2 potential energy number. But I didn't check to see if that still applies if k is a function of x. I'm off to check now...

EDIT -- Nope, 1/2 k x^2 looks to be the PE only for the linear spring constant case. Good catch!

5. Apr 7, 2017

Dr.D

@ Berkeman - the expression PE = (1/2) k*x^2 comes from integrating k*x over zero to x, which shows exactly why it it tied to linearity.

6. Apr 7, 2017

Staff: Mentor

Yep, I just reminded myself of that on a Post-It with a quick calculation. Do you have any guesses how non-linear a simple bow like that might be? I know a compound bow will be very non-linear, but is a simple compound bow very nonlinear?

7. Apr 7, 2017

8. Apr 7, 2017

pevi70

Hi Berkeman and Dr D,

I think you are right that the spring constant is not linear because the extension of the spring will be changing as the angle between the rest state and the extended state changes. This just got a lot more complicated.

Do you have any recommendations how to calculate the time it would take to transfer the energy to kinetic energy?

9. Apr 7, 2017

Staff: Mentor

Did you see the plot I posted in Reply #7?

If the spring force is indeed linear, and friction is negligible, then you just use simple kinematics to calculate the velocity of the arrow versus time during the launch phase. If the spring is non-linear or friction is not negligible, then you will need to use an integral to get an answer with better accuracy. Are you familiar with integrals yet?

10. Apr 7, 2017

pevi70

I saw your graph after I posted the reponse. I guess I need to look into it in more detail to understand what is happening exactly. I might need to add that the bows have a metal arch that hardly changes form and that the arrow is propelled by a spring. Not exactly the same as a regular bow, I guess.
I am familiar with integrals but any help would be very much appreciated!

Last edited: Apr 7, 2017
11. Apr 7, 2017

Dr.D

I don't understand the curve Berkeman posted (I'm not sure I am reading the axis labels correctly, for one thing). If this is force versus draw distance, why is the force a max and dropping to zero as the draw increases for the conventional bow? That does not make sense to me.

12. Apr 7, 2017

Randy Beikmann

It looks like the data are plotted so that "x=0" is at the full draw position. For the conventional bow, that is at maximum force. But the compound bow is down to about 10 pounds there.

13. Apr 8, 2017

pevi70

I had another look and I think I can estimate the extension of the propelling spring versus the distance of the arrow acceleration path. I ignore the potentially added effect of the metal bar to which the spring is attached.

It looks like this:

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Last edited: Apr 8, 2017
14. Apr 8, 2017

pevi70

I think that this graph can correct and allow to calculate the spring constant.

k = m * v^2 / d^2 would become k = m * v^2 / (f(0.175))^2 = 0,1 * v^2 / (0.084)^2

I guess it can also be used to calculate the acceleration along the arrow's travel path. Do you perhaps have a hint on how to approach that?

15. Apr 8, 2017

Nidum

To get an accurate answer to this problem you will need to include the mass dynamic terms in the spring equation .

16. Apr 8, 2017

pevi70

If you would replace the F = k * x by F = k * x' for the corrected extension, I get to

F = k * x' = m * a. The distance of the spring is 17.5 cm and the acceleration changes with the distance from the reststate.

So d = 1/2 * acceleration * t^2 and acceleration = k * x'(t) / m. I know the speed v of the arrow when it is leaving the spring. Unfortunately I don't know how to continue.

Did you refer to this, Nidum?

17. Apr 8, 2017

Nidum

A leaf spring which is under load behaves differently depending on how fast the load is released .

At slow release speeds it is just a simple spring .

At high release speeds , as in a crossbow , the spring becomes a complex mechanism where the forces required to accelerate the self mass of the spring become significant . These forces can modify the action of the spring considerably .

Several ways of dealing with this problem theoretically but they are all rather messy for springs with non simple geometry .

Easiest way to get useable answers for cross bow springs like in picture would probably be to use a lumped mass approximation .

18. Apr 8, 2017

pevi70

I guess I can weigh the spring. Do you happen to know how I can calculate or derive the time the bow needs to propel a 100 gram arrow over the 17.5 cm, in the case the spring weight is not an issue?

19. Apr 8, 2017

Nidum

The motion of the arrow for the problem as in #18 can be described by a relatively simple differential equation . This equation can be solved to give the actual answers that you want

Please advise whether that approach would useful to you or not .

There are other ways of solving the problem available including numerical approximation and graphical methods .

Last edited: Apr 8, 2017
20. Apr 8, 2017

pevi70

I would appreciate it if you could help with the differential. I have no clue as what do next to be able to calculate the time. Thanks!