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I am currently designing a ground based digging drill, that hopefully will be accepted by a commercial entity. I am currently working on the preliminary design of the robot and am hoping to move to the modeling phase soon, however I am slightly stuck on my Torque calculations. Can anyone please confirm if my calculations below are correct?

Background:

The drill will be relatively automatic, small, and simple. This autonomous digging drill, will be an updated model of a larger similar model, that I am using for my design process, which requires 2 motors to operate the drill: One to rotate the drill head, and one to push the drill head into the material.

Info

The larger drill design has a diameter 2x greater than the new design (25cm and 12.5cm respectively) and they both have simple right cone profiles. To properly scale and select the appropriate motor for this updated drill, I need to utilize the original design to make new estimates.

Data:

The large drill has a torque requirement 1000-1200Nm. Since the diameter of my drill is ½ of the original drill, I believe my drill will have a torque requirement of 500 – 600Nm based on T = Fr.

The large drill requires 750 – 1000N to push into the material. For this force estimate I thought I should use P = F/A (where A is the surface are of the cone (without the base)) as both drills should apply the same pressure on the material.

I get that the force requirement for my drill is 187.5 – 250N.

Can anyone please confirm if this sounds correct?

I just need to know if I am on the right track here.

Thank you for reading

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Very roughly, ignoring the nails or any twist in the angle of the blade, if your cutting cones are just two or three blades that cut the dirt basically acting as angled knives then the torque required is dispensed over all the length of the angle.

In reality, the attack angle of the blades is twisting and the blades are of a spiral shape which makes it easier to cut.

For any point along the length of the blade, the torque is $$\tau= R*dr*\sigma\tau$$

and for the entire drill

So $$\tau=\Sigma (\frac{RdR*t}{2})=R^2/2 *\tau$$

Therefore for half R, you need 1/4 torque.

For the vertical force your assumption in correct.

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  • $\begingroup$ Thank you so much for this in-depth reply. This is exactly the feedback I needed. $\endgroup$ – gunter Sep 16 at 20:26

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