7

I am studying implementation of 9398 crawler reduction. Here is the stripped part, in order to see the gears and their relation - the image is made with the excellent tool which is LDCad.

enter image description here

My idea is to use this Lego design for other model and I need to know how would I calculate the effect of changing it a bit. My intention is to replace the two 16 teeth gears 94925:

enter image description here

with a pair of 12 and 20 teeth (32270 and 32269):

enter image description hereenter image description here

and to remove the last reduction pairs, the ones in portal axle:

enter image description here

so it would be as follows:

enter image description here

Would this become faster/less powerful or slower/more powerful? My aim would be to keep it pretty close to how it performs now.

Also, is there a way to combine gears in order to add a distance of half of a stud between the motor axle and the next or the second one? In the image below, I have moved the parts and created a gap between the gears, in order to point what I intend to move, but the contact must be assured by some gears, obviously by replacing the existing ones. Would be there possible?

enter image description here

Improve this question
1

This modification won't change the rate of reduction.

You can calculate the ratio of the angular velocities of two adjacent gearwheels as

i12 = w2/w1 = n1/n2

where w is the angular velocity and n is the number of teeth of the gears. Hence the original reduction reads as:

i12*i23*i34*i45*i56 = 12/20 * 16/16 * 20/28 * 12/20 * 12/20 = 27/175 = 0.1543

while the modified is:

i12*i23*i34*i45 = 12/20 * 12/20 * 20/28 * 12/20 = 27/175 = 0.1543

What you did is you simply removed the 16/16 and replaced the 12/20 gears, which results in the same reduction because the order of the multiplication is arbitrary.

To the second question:

You can change the 12-20 gears to 16-24 ones to fit the increased distance. In this case, the new reduction is

i12*i23*i34*i45 = 16/24 * 12/20 * 20/28 * 12/20 = 6/35 = 0.1714

Note that the new gears may intersect parts not shown in the figures.

Improve this answer
1

If the only change is replacing the two 16-tooth gears with a 12-tooth gear and a 20-tooth gear, then you will reduce speed by a factor 12/20 and increase torque by the same factor.

If you want to increase the distance between gears by 1/2 stud, then you can experiment with using bigger gears and offsetting the motor sideways so the gears don't touch vertically. This may induce some play in the gears.

Improve this answer
1
  • I would drop the portal axles, so the performance will be kept. The reason is for implementing this on other model with less axle height (42069). Well, I did it partially, but using a horizontal transmission and keeping a similar design, with only 2 reductions instead of 3 (as this model has). It performs acceptable - a little faster than 9398, but powerful enough to run over decent obstacles for its tracks. I couldn't find a combination of gears for that 0.5 stud displacement, though. The gears are usually too tight to work safely together.
    – mike
    Jun 22 '18 at 5:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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