Crossbow Physics: Calculating Spring Constant, Tension Force & Power

  • Thread starter pevi70
  • Start date
  • Tags
    Physics
In summary, the conversation discusses a project involving the measurement of physics parameters of handcrafted crossbows used in shooting contests. The goal is to calculate the spring constant, tension force, and power of the crossbow through various calculations involving the velocity, mass, and tension distance of the arrow. However, the accuracy of these calculations may be affected by factors such as losses due to friction and the linearity of the spring. The use of energy as a means of calculating the arrow's velocity is suggested as it takes into account the variable force during the launch phase. The speaker also asks for assistance in determining the time it takes for the spring to transfer energy to the arrow.
  • #36
Despite the picture, quite some assumptions need to made: no recoil for example.
If you would assume that the propulsion is coming only from a mass less rubber spring, no friction exists, how would you calculate the time it would take for the arrow to cross the 17.5 cm (knowing the final speed)?

How could I approach that?

Thanks for your thoughts!
 
Engineering news on Phys.org
  • #37
pevi70 said:
I have no equipment to measure extensions under different loads, so I don't have or can easily collect that information.
Extension of the rubber band can be measured easily enough with a ruler. If you don't have a dynamometer on hand, you can measure draw force the way it was traditionally measured - hang the crossbow on a wall, and attach weights of known mass to the spring, measuring the distance the nocking point moves for each weight (hence the traditional nomenclature for bow power - 'draw weight'), as well as the associated extension of the rubber band.
Different crossbows are then ranked by their draw weight at full draw.
 
  • #38
Thats simply not practical. You need a special lever to lock the spring. I can measure the speed of the arrow.

I would really appreciate it if someone could help me how to calculate the time needed to propel the arrow by the spring in an idealized environment.
 
  • #39
?
Why would you need to lock the string? Get a bottle, fill it with sand, weigh it, then attach it to the string and measure how much it stretches. Then change the bottle for a different weight or add more bottles and repeat.
 
  • #40
These bows are so strong, you need a special lever to charge the spring.
 
  • #41
If you do it by hand, sure. But here you're just hanging weights on the string. You can hang as much as it takes. I really don't see the problem here.
I really don't have any experience with stretching strings, but if it's anything like traditional bows in efficiency, then you'll need something like 50-70 kg at full draw for 60m/s bolt (which you don't need, since we just want to find out how the string extension vs draw weight looks like - the first few cm of draw should suffice).
 
  • #42
Here is a picture of the bow:
(The width of the wooden part of the bow is 4 cm)

Crossbow.jpg
CrossbowZoom.jpg
 
  • #43
Since you can measure v (velocity at the arrow separation); and, for an object accelerating from 0 to v, s= 1/2vt, and with s and v then you can determine t (time of acceleration) and subsequently the acceleration a. If of any benefit, from that using F = ma you can determine the average effective bow force. (if, as you specified, you ignore friction and recoil.)

Edit: I just saw thr bow picture and are you sure there is no flexing of the crossbar? If there is as strong a pull force as you state, it is hard for me not to expect flexing in that component.
 
  • #44
I think s=1/2vt only applies if the acceleration is constant.
 
  • #45
What this will give you is the average acceleration for the entire travel, not the profile of the acceleration. For for a given succession of values regardless of linearity there exists an average value.
 
  • #46
OK, I thought it didnt apply. Thats great, thanks a million!
 
  • #47
bow spring v3.png
I estimated the dimensions but they won't be far out . 150 N on each arrow .
 
  • #48
Thanks Nidum, what is it?
Do you agree with JBA?
 
  • #49
FEA deflection estimate for the bow .

If deflection had turned out to be very small then it would have confirmed what you have said about bow being just a rigid bar .
 
  • #50
What does that mean?
And do you agree with JBA?
 
  • #51
Problem is that all responders - including myself - are now a bit uncertain as to what exactly we are trying to analyse .

So I'm trying to get a definite understanding of the mechanics of this cross bow . Is it the metal spring , the rubber rope or a combination of both that provides the driving force ?
 
  • #52
Do you agree with JBA?
 
  • #53
Nidum said:
So I'm trying to get a definite understanding of the mechanics of this cross bow . Is it the metal spring , the rubber rope or a combination of both that provides the driving force ?

Amen!
 

Similar threads

  • Introductory Physics Homework Help
Replies
29
Views
931
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
357
  • Mechanical Engineering
Replies
2
Views
3K
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
846
  • Introductory Physics Homework Help
Replies
6
Views
322
  • Introductory Physics Homework Help
Replies
3
Views
373
  • Introductory Physics Homework Help
Replies
10
Views
923
Back
Top