Hello,
I’m planning on building a 4wd rover for the mapping of areas which are potentially hazardous. The maximum weight of the rover, including payload, is likely to be around 17kg, and I’m just wondering if the motor shafts of the planetary gearmotors (https://www.robotshop.com/uk/12v-84rpm-13471oz-in-hd-premium-planetary-gearmotor.html) can take the weight if directly driving the wheels, or will a means of support will be necessary?
Thanks in advance.
I am also interested in this topic - can anyone explain how we can measure the necessary force needed? For example, the motor mentioned here is “Stall Torque: 1,347.1 oz-in (97 kgf-cm)” - but how can we know what that means in practice? I found a page (https://www.orientalmotor.com/technology/motor-sizing-calculations.html) which explains some of this, but it seems rather complicated…
Thought I best clarify what I’m concerned about.
I know the motors have enough torque to drive the rover, I’m more concerned with the weight of the chassis bearing down on the shafts of the gear motors.
@JMoore Welcome to the RobotShop Community. 17Kg is likely way too heavy for the shaft of this motor to support, so you will likely either need to offset the output shaft (spur gears, timing belt etc.), or support it on the opposite side. The more important issue is that it is also likely too small and underpowered to properly move a 4WD rover of that weight. The stall torque of the motor is 1,347oz-in / 97 Kg-cm, therefore the maximum continuous torque is roughly one quarter (25%) that value.
@rkarlberg These two pages might help. Note that the torque you get as output is a worst case scenario… accelerating up an incline.
Thanks again! Those are helpful resources. My first order from RobotShop arrives today - should be a fun project.
Hi @cbenson, thanks for the welcome and reply! I figured as much, so will likely use a pillow block to support the wheel shaft.
As for the motor torque, I had previously used the Robotshop sizing tool and everything seems to be in order RSMotorSize|258x460. Am I missing something blaringly obvious?
Yes - The stall torque of the motor is 1,347oz-in / 97 Kg-cm, therefore the maximum continuous torque is roughly one quarter (25%) that value. Your applications needs 1200 oz-in of torque, therefore you need to find a DC gear motor which has a stall torque of 4800 oz-in and ~100rpm no load. For example the following combination comes close:
16:1 gear down:
The two together would provide an output of:
343oz-in x 16 = 5488 oz-in (stall) or ~1372oz-in continuous maximum torque
5310rpm / 16 = 332 rpm no load (far less under load)
Despite this, the calculations are for worst case scenario of accelerating up an incline, and 30 degrees is pretty steep. Given the motor you chose, the maximum angle it can climb will likely be a lot less.
