RP5 Mech/Gear/Speed Parameters

Robotics and electronics discussion Mechanical design and actuators
1

SOME RP5 INFO & MEASUREMENTS

After some recent discussions and questions on the Pololu with regards to the RP5 and possible modifications
that could be made to it, myself included, I thought it would be a good idea if we could fill in the gaps on the
performance on this great little chassis. The RPM and gear ratio doesn’t come with a tech data provided.

After having taken it apart and put it together again, I can share with you some findings which might be
a useful reference.

The RP5 gear train has a series of gear pieces (4 to be exact) which act as speed reducers and torque boosters.
Without describing in words, I will shortly place a scanned in neat sketch of the arrangement, teeth count, arrangement, etc.
There are three gear pieces with 12 & 50 teeth ratios as you work your way through from the motor
output shaft to the output shaft driving the wheels. The 47mm diameter plastic wheel is wrapped around with a
6mm rubber tread making the ground contact diameter approx 51mm. This assumes a new un-worn tread.
All measurements were done with a fairly high precision digital precision vernier caliper and great care was taken to reduce
measurement and parallax (eye) errors.

The gear reduction ratio works out at about (72.3 : 1) or 0.013824 times the motor shaft angular velocity.

The next step is to determine how motor RPM at the output shaft. I dont know the answer to that one, however through gear ratios
and basic mechanics equations, It can be worked backwards also (maths is relatively simple).

[Motor RPM x (0.24)^3 = Output RPM] is the basic equation for translating RPM from one end of the gear train to the other. (12/50=0.24)

That is not all though, a conversion from RPM to RPS is then required,followed by a formula for changing from angular (omega) to (velocity) using
the ground contact radius of the wheel/tread (0.0255).

I think up to here most would probably agree with me.

The next step is not certain, and I will have to revise. If anyone has input, please share.

Taking the design speed of 15 cm/s as per the spec sheet, and the radius of the wheels, I arrived at a calculated 25,535 RPM
in order to get 15 cm/s by working the formulae backwards. I still have to confirm this as it seems high. A sensibility check on
most toy motors would see you can get up to 16,000 RPM for a 6 volt motor. Hmmm! 25,500RPM seems too high!

I have used the formula : arc lenth (L) = radius * theta (full wheel circumefrence is approx 16cm which is right)
The derivative version is: L_dot (velocity) = radius * theta_dot (or angular velocity) - 5.88 radians/second (for 352 RPM)
The problem is here I think! It could be that I forgot to divide by Pi somewhere.

Initially it was estimated at 11,000 RPM however this only produced 6.5cm. If were to base it on full revolution per second only at the
output shaft end, it would be 4341 RPM (2.5 cm/s). And if we were to assume 25,535 RPM (the calculated figure to give 15 cm/s) then that
would translate into 5.88 revs/sec at the output shaft end. Both values appear to be at the extremes.

Note: Motor is always assumed running at 100% speed.

Any comments or corrections would be appreciated.

Cheers,
Carl

2

Hello,

5.88 radians/s is 56 rpm, not 352 rpm. So yes, by your numbers it would be about a revolution per second and 4000 rpm at the motor shaft. That sounds reasonable to me.

We say on the RP5 web page that the gear with a hole in it should be turning at 4 times the speed of the output shaft - are you saying they actually have a 50:12 ratio?

-Paul

3

Hi Paul,

Thank you for the correction. It now makes sense. Working it backwards, it is as follows:

15cm/s maximum speed (starting point)
using a mean wheel radius of 2.55cm and the formula [L_dot = radius x theta_dot],
we arrive at 5.88 radians/second (about the center of the drive wheel)
Converting 5.88 radians/second to RPM, that would be 56.15 RPM

You said that the approximate motor shaft output is 4000 RPM.

Using my counted&calculated gear ratios, 56.15 rpm divided by (0.24)^3 would yield 4062 RPM. (12/50 * 12/50 * 12/50 = 0.24^3)

4062 RPM (output) / 56.15 RPM (input) yields 72.53:1 gear ratio. So the derived figures are coherent and correct.

The 72.5:1 gear ratio is realized through 3 x 50/12 teeth speed reducers. So if my understanding is
correct, that would mean for every turn of the wheel (one with the hole in it), the motor output shaft would have to be
spinning roughly 72 times. This off course is the relationship between the motor shaft and wheel shaft only.

In order to confirm the 50/12 gear teeth ratio, I literally counted each tooth on each gear piece (small and large) carefully by marking each tooth
with a permanent marker to avoid mistakes.

Now that the maths has been proven, what does “or tracking motor rotation with a resolution of 4 counts per output shaft revolution.” mean?
As you have in your RP5 General description section. Isn’t it 72 counts ?

If the hole in wheel revolves once per 0.9333 seconds (56 RPM), it would be equate to 72.5 revs of the output shaft. Which would mean
you get a 0.933 x circumference of tread wheel for linear resolution (= 0.933 * Pi * 2 * 0.0255m = 14.9cm , SAY 15cm). At once revolution per
second, you would be travelling ~15 cm/s which would agree with manufacturer specifications.

Question: So how do you get 1.5 inches (3.8cm) linear resolution if you only spinning almost once per second?

Regards,
Carl.

4

Carl,
I think you and I are talking about different holes. One of the gears has a single, round hole in it. Unless you put the chassis back together differently, that gear meshes directly with the gear on the output shaft, so, according to your calculations, it should turn 12 times for every 50 rotations of the output shaft, right?

-Paul

5

Paul,

The hole on the last gear in the chain (on the output shaft) is the same one that we were talking about.

I only disassembled one side of the gears, then when I put them back, I returned them
exactly as per the other side which is as original. There was no need to disassemble both sides.
So there was no alternation to the original gear assembly in that respect.

I will have to time the rate at which this hole turns around. Although 1 revolution per seconds sounds about right,
and not 12 times for every 50 revolutions. According to my calculations, this same hole should be completing one
revolution just under 1 second.

Cheers,
Carl

6

I don’t understand why you seem to think that the gear with a hole in it will turn at the same speed as the final shaft.

-Paul

7

Hi Paul,

My apologies. That was because I was thinking that the hole was on the last gear piece,
however, when I checked it this morning I found that you were right. The
hole is not on the output shaft gear piece but the one before it. The hole does indeed
rotate about 4 times for every one output shaft rotation.

I calculated 233 RPM or about 4 revolutions per second (3.9 to be exact).

Thank you. That clears everything up and makes perfect sense. So that would mean you
could in fact get 1.5 inch or 3.8 cm resolution approximately, every time that hole passes
through an encoder say.

I will draw this up on CAD file up soon, along with an excel spreadsheet which performs the calculation.

Thanks Again,
Carl.