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I have a crossbow with a 220lb draw pull. I am designing a mechanism to draw the string back 12" as quickly as possible using a 1/2" ball lead screw connected to a 12 to 24v DC high torque motor and guided by two linear slides. I need to know how many KG of torque will be needed to draw the 220Lb string back in around two seconds? Or is it even possible?

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    $\begingroup$ torque isn't measured in kg; this is a basic physics/mechanical engineering question, not really electrical engineering. $\endgroup$ Mar 30, 2020 at 1:05
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    $\begingroup$ You will have better luck on Engineering SE, this SE is for Electrical Engineering design, yours is mostly a mechanical problem. They will need to know the lead (distance traveled per revolution) of the ball lead screw, and more information on the motor. Motors have torque/speed curves, torque is not independent. $\endgroup$
    – Mattman944
    Mar 30, 2020 at 1:32
  • $\begingroup$ I'm voting to close this question as off-topic because it's a purely mechanical question, nothing to do with electronics $\endgroup$
    – Neil_UK
    Mar 30, 2020 at 4:59
  • $\begingroup$ Energy = force x distance. Power = Energy/time. Energy here is ABOUT F x D = 100 kg x g x 0.3m = 100 x 9.8 x 0.3 ~= 300 Joules. To do that in seconds = E/t = 300/2 = 150 Watts. Say 200 - 250 Watts to be safe due to losses. Not trivial. Motor spec will depend on lead screw pitch (gearing) but Wattage is as above. For an eg 1mm pitch thread motor must turn d/p = 300mm/1mm = 300 turns in 2 seconds = 9000 RPM. With a 12V motor that is a motor with kV (RP/Volt) of RPM/V = 9000/12 = 750 kV. ... $\endgroup$ Mar 30, 2020 at 10:15
  • $\begingroup$ RC model motors with 250 W rating and 750 kV speed are commonly available. A brushed motor would probably be cheaper. RPM at 12V may be quite a lot below 9000 RPM - adjust lead screw pitch to suit. $\endgroup$ Mar 30, 2020 at 10:15

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I have a crossbow with - 220lb (100 kg) draw pull.
- 12" (0.3m) draw distance - 1/2" (12.5 mm) ball lead screw - 12 to 24v DC high torque motor - guided by two linear slides.

I need to know how many KG of torque will be needed to draw the 220Lb string back in around two seconds?

You really want to know motor power.

Energy = force x distance.
Power = Energy/time. Draw force = 220 lbf = 100kg = 980 newton.
Draw stroke = 12" = 300mm = 0.3M g = 9.8 (or 10 :-) )

Energy here is ABOUT F x D
= 100 kg x g x 0.3m = 100 x 9.8 x 0.3 ~= 300 Joules.
To do that in 2 seconds Power = energy/time = 300 J / 2s = 150 Watts.
Say 200 - 250 Watts to be safe due to losses.
Not trivial.

Motor spec will depend on lead screw pitch (gearing) but Wattage is as above.
For an eg 1mm pitch thread motor must turn draw / pitch = 300mm/1mm = 300 turns in 2 seconds = 9000 RPM.
With a 12V motor that is a motor with 'kV' (= RPM/Volt) of
RPM/V = 9000/12 = 750 RPM per volt = 750 kV.

RC model motors with 250 W rating and 750 kV speed are commonly available.
A brushed motor would probably be cheaper.
For a 250 Watt brushed motor RPM at 12V may be quite a lot below 9000 RPM - Adjust lead screw pitch to suit.
eg a 3mm pitch lead screw will require
300 / 3 = 100 turns in 2 seconds = 50 turms/second = 3000 RPM (as you'd expect).

Cordless drill:

As Solar Mike noted - a cordless drill may be a good solution.

Power ratings well above 150W - 250W are available.
These are usually not the cheap ones :-(

Determine loaded chuck speed in RPS.
eg 140 RPM (low speed setting) = 2.333 RPS Winding drum circumference = draw distance / time / RPS = 300mm / 2 / 2.333 = 64 mm
~= 20mm dia drum.

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Battery:

Using 12V battery at 250 Watt, I = P/V ~= 250/12 (or less) ~= 20A
This is within the capability of good quality power rated 18650 LiIon cells.
4 x 18650 Lion = 12 - 16 V.
More V = less current.

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