Crossbow Physics: Calculating Spring Constant, Tension Force & Power
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pevi70
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This is the set-up:
I can measure the velocity of the departing arrow. The mass of the rubberband spring is unknown and not practical to measure.
I would be interested to use the velocity and the other known parameters to calculate bow characteristics including the power of the bow.
Let me know if you need more information!
I can measure the velocity of the departing arrow. The mass of the rubberband spring is unknown and not practical to measure.
I would be interested to use the velocity and the other known parameters to calculate bow characteristics including the power of the bow.
Let me know if you need more information!
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pevi70
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The rubberband has a round shape (rod shape) with a diameter of about 5 mm. I will post a picture of the actual bow later today.
I have no equipment to measure extensions under different loads, so I don't have or can easily collect that information.
I have no equipment to measure extensions under different loads, so I don't have or can easily collect that information.
pevi70
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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!
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!
Science Advisor
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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.pevi70 said:I have no equipment to measure extensions under different loads, so I don't have or can easily collect that information.
Different crossbows are then ranked by their draw weight at full draw.
pevi70
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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.
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.
Science Advisor
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?
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.
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.
pevi70
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These bows are so strong, you need a special lever to charge the spring.
Science Advisor
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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).
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).
pevi70
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Here is a picture of the bow:
(The width of the wooden part of the bow is 4 cm)
(The width of the wooden part of the bow is 4 cm)
Science Advisor
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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.
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.
pevi70
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I think s=1/2vt only applies if the acceleration is constant.
pevi70
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OK, I thought it didnt apply. Thats great, thanks a million!
pevi70
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Thanks Nidum, what is it?
Do you agree with JBA?
Do you agree with JBA?
pevi70
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What does that mean?
And do you agree with JBA?
And do you agree with JBA?
Science Advisor
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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 ?
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 ?
pevi70
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Do you agree with JBA?
Dr.D
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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!
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