0
$\begingroup$

I am attempting to make my first project idea of using a DC motor as a winch motor in order to pull an object, I am using string as the rope to pull object I am attempting to "winch" (air quotations as I am not using a proper winch motor. My project is small scaled).

The current information I have (from my own calculations) is:

  1. The object I am attempting to "winch" will require a force of 367.999N
  2. The speed at which I want the object to move at is 0.39m/s
  3. The torque required to perform this act from a motor will be 310.959155N-m
  4. This object is to be moved over 0.845 metres

This object is attached to these wheels and is dangling from these wheels I should add which rolls along a hard plastic rail. The wheels are also plastic

However I will confess that I am new to electrical engineering and physics in general but still wanting to finish this project that I have started. Are there any equations or possible considerations I can implement that will gauge the required voltage and RPM of the DC motor in order to gauge how powerful a motor I will need to be able to "winch" this object? Any and all help would be appreciated or guidance on areas to study for a better understanding of what I should do from here.

$\endgroup$
4
  • $\begingroup$ How did you get to winch torque? from force to pull? If so apply the same "conversion factor" to your speed. (Look into what conversion factors are- makes engineering as simple as keeping track of units.) $\endgroup$
    – Abel
    Commented Jan 11 at 12:27
  • $\begingroup$ I got the winch torque from using T = F*r, force being the amount of newtons it would require to move the object in the first place which came out to 367.999N and the radius of the distance from where the object is being pulled from was 0.845m if I've understood it correctly $\endgroup$ Commented Jan 11 at 13:37
  • $\begingroup$ @AidanDocherty the radius of that equation is the radius of the gear or sheave on the motor. RPM is dependent on gearing $\endgroup$
    – Tiger Guy
    Commented Jan 11 at 14:47
  • $\begingroup$ @TigerGuy Thanks for letting me know, are there any other places I should place my focus on with figuring what I need to know? $\endgroup$ Commented Jan 11 at 15:15

1 Answer 1

0
$\begingroup$

.845m radius aka .845m/radian (from comments; question seems to suggest .845m is range of motion) would mean you have 845m/radian = .845m/radian×(2×pi radians/revolution). You may need to account for a spool radius when the spool is bare and when the spool has enough rope to affect the radius.

If you need .39m/s, apply that conversion factor to get .39/(1.69pi) revolutions/sec. Apply 60sec/min to get 4.41revolutions/min aka 4.41 RPM. This and the torque you calculated are at the spool. If you use a gearbox, you will need to adjust (more torque for gearbox losses, etc). You should also get a significantly more capable motor than you need- often double or triple the torque for rpm you need.

One thing to note in unit conversion is that the units of torque are not the same as the units of energy. Torque×radians = Energy = Force×Distance. Hiding the "unitless" radians is one of those peculiar things in engineering like measuring pressure in inches of fluid (and even designing equations that contain conversion factors around them) that makes "engineering" more confusing than it needs to be.

$\endgroup$
2
  • $\begingroup$ Thanks for this answer - I'll apply it as soon as I can, I was wondering where you learned all of this stuff? Education? I'm looking to expand what I can know concerning physics and electrical engineering $\endgroup$ Commented Jan 12 at 13:06
  • $\begingroup$ School is the standard route. Many will even keep a few textbooks from when they were in school. $\endgroup$
    – Abel
    Commented Jan 13 at 8:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.