# Help calculating torque required

1. Jul 29, 2013

### basil4j

Hi All,

I'm having a bit of trouble remember basic physics and hoping someone can help If i've got the wrong forum, please feel free to move this post.

I'm designing a robot which resembles this one (Kind of, its a bit different but the basic mechanics etc would be the same so i've found this picture for illustrative purposes)

The wheels are attached to struts which can rotate through 360 degrees around a central 'hub'.
To rotate the struts, I intend to put a gearmotor in each strut, acting on a bevel gear within the hub. Excuse the paint drawing, but here is a plan view of one side of the robot.

Hope it makes sense, my CAD file is at home (this is a plan view, not to scale)

My problem lies in selecting an appropriate sized gear motor with enough torque to lift the robot as i've come up with 2 possible results and im not sure which (if any) are right!

Assume the following.
-No energy lost due to friction
-Length of strut 150mm (center of hub to center of wheel)
-Radius of central bevel gear 40mm
-Weight of robot 4kg, all 4 wheels in contact with the ground, so weight in each wheel 1kg.
-Gear ratio of bevel 'gearbox' 4:1

Here is my first force drawing for a single strut (from memory this is a third class lever?).
The strut needs to be rotated counter clockwise to lift the body of the robot.

Here is my train of thought...
Fe is the force provided by the motor acting on the bevel gear. (Assume the motor providing the force weighs nothing)
Fl is the weight of the robot acting on 1 wheel.

The motor is acting on the outside of the bevel gear, which is radius 40mm, so:
De = 0.04m
The strut is 150mm long so:
Dl= 0.15m

Fl = 1kg * 9.8 = 9.8N

Fe*De=Fl*Dl
Fe=(Fl*Dl)/De
Fe = 36.75Nm

Now is where my brain goes funny (well at least I think it does, it could have been funny from the start...)

The torque form the motor is acting at 40mm from the pivot point of the strut, therefore:
T=Fxd
T=36.75 * 0.04 = 1.47Nm

Now, my first question assuming, everything above is correct, is:
Do I need to take into account the gear ratio of 4:1? In other words, I could get away with a 0.36Nm motor and not a 1.47Nm?

Any my second question:
Can I treat this as just a simple lever, with Fl of 9.8N acting at .15m, and the gearmotor acting through an additional 4:1 ratio torque multiplier provided by the bevel gear.
This gives me a much nicer result of
T=Fxd = 9.8 x 0.15 = 1.47Nm
Through 4:1 gearbox = .36Nm.

Coincidently this is the same result as solution 1, but i'm not sure if solution 1 is correct!

The problem with this second solution is that i'm unsure of the effect of having the motor acting 40mm from the center, and not directly on the axis?

So my question is, are both of these results flawed? Have I got my force diagrams wrong?