# 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!

21. Apr 8, 2017

### Dr.D

The graph posted by the OP in #13 is not clear as to what it represents. Both axes are labeled with units of cm, i.e. length units. What is this graph supposed to represent?

Regarding the inertial effects of the spring mass, there are many models available using a series of lumped masses distributed along the length of the spring to represent the spring itself. These can be quite accurate if enough discrete masses are used.

22. Apr 8, 2017

### Nidum

Let us consider a simple coil spring acting directly on the arrow for now . We can look at the more complex problem of an actual bow spring later if you wish .

While the arrow and spring are still in contact the spring force is causing the arrow to accelerate .

If x represents the distance travelled by the active end of the spring and by the arrow from the initial cocked and ready to fire position what is the value of the spring force at x and how can this spring force be equated to the acceleration of the arrow at x ?

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

### Nidum

Hint : It's essentially the same maths as for the well known block and spring on a frictionless table problem .

24. Apr 8, 2017

### pevi70

@ Dr D, I was trying to show how I could correct for the fact that the path of the arrow is not the same as the extension of the spring. I calculated for each cm along the path of the arrow, the length of the spring and from that the extension of the spring. As you can see there is a polynomial one can fit to correct. In the graph, the horizontal axis shows the arrow path and the vertical axis the corresponding extension of the spring.

@ Nidum, I was thinking of using F = k * x' = m * a for this. Acceleration would then be a = k * x' / m.

I guess the force in time and with that the acceleration should have the same shape as this graph:

When the spring reaches the rest state, no acceleration is present.

EDIT: I guess x(t) = 1/2 * a * t^2, but I dont know how to do this with a changing acceleration

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

### Nidum

You need to write down the differential equation and solve it to get the answers that you want analytically .

Last edited: Apr 8, 2017