# Differential gearset math

1. Oct 17, 2003

### one_raven

I am trying to solve gear ratios for a differential gearset when one of the output shaft's rotation is limited...

Let's say:
I have a power split device with an input (A), output1 (B) and output2 (C).
C has a speed limiting mechanism attached to it.
Without limiting the speed of C, ratio A:B is 2:1 and ratio A:C is 2:1.

Let's apply a 300 RPM input to A without the speed limiting device engaged.
B will be spinning at 150 RPM.
C will be spinning at 150 RPM.

Engage the speed limiter and slow C down to 75 RPM.
How do I find the speed of B?

My guess is C(before limiter) - C(after limiter) + B(before limiter)
150-75+150 = 225
So, the functional output gear ratio A:B after limiting C to 150 RPM would be 300/225 or 1.3333:1

I think this is right so far.
Please correct me if I am wrong.

If that is right, what happens if A:B does NOT equal A:C?
If A:B = 3:1 and A:C = 2:1...
If we apply the same 300 RPM input(without the limiter) we will have:
B = 100 RPM
C = 150 RPM

If we slow C down to 75 RPM again how do we solve for B?
If we apply C(before limiter) - C(after limiter) + B(before limiter) we get:
150 - 75 + 100 = 175 RPM @ B
Therefore A:B after the limiter is engaged is 1.714:1
Is this right?
It feels like there is missing something.

2. Oct 19, 2003

### one_raven

I tried working with the ratios rather than the speeds and came up with this:
To get A:B after the speed limiter is set...

Take the inverse of A:C (before limiter)
Subtract the inverse of A:C (after limiter)
Add the inverse of A:B (before limiter)
Invert the sum.
For the first example, that would be:
1/((1/2)-(1/4)+(1/2))
1/(.5 - .25 + .5)
1/.75
1.333333:1
So, output of B (after the limiter) would be 225.
That agrees.

Now, if you apply that to the second scenario:

1/((1/2)-(1/4)+(1/3))
1/(.5 - .25 + .333333333)
1/.583
1.715:1
So, output of B (after the limiter) would be 175.

It agrees there too, but I am still not sure I am taking everything into account.

3. May 20, 2004

### Kenneth Mann

Equation check

I compared your equation to a generic equation for planetaries, which is:

(P-1)(Sr/Ss) = P(Sc/Ss) - 1

and which, for a ring-gear planetary:

P = (Nr = Ns)/Ns

If the author of these equations was correct, then the following was derived:

(Sr - Sc)/(Sc - Ss) = Ns/Nr

If so, then your directions will be reversed. I'd like to know which is correct.

4. May 20, 2004

### NateTG

I think it depends a bit on the nature of the set up.

Consider, for example, that according to your post:
A:B=A:C=2:1

So if B is limited to 75 rpm, then the ratio can be reversed to get that A is going at 150 rpm, so that C is going at 75 rpm as well. (But that's not typical behavior for a differential.)

For a conventional differential, the rpm relationship is more like:

$$A = k_b B + k_c C$$

Where the constants are determined by gear ratios.

The actual RPM on each of the arms is determined, among other things, by the load on each arm. That's, for example, why normal cars have so much trouble if one of the drive wheels is in the mud, and why off-road vehicles have differential locks.

Now, if you have a differential where $$k_b=k_c$$ then you can start looking at the relationships that you describe, i.e. the example you give, where the RPM on one side are limited to 75 rpm, gives the additive result that show.

For the latter example, you should have:
$$A = \frac{3}{2}B+C$$
So slowing C by 75 rpm, results in an increase of 50 rpm for B, or a decrease of 75 rpm for A.

5. May 20, 2004

### Kenneth Mann

Differential vs reducer/increaser

I'm not sure I understand your assumptions. In a differential (particularly a planetary type), the relationship of one input/output to another is directly proportional (such as 5::1) only if the third is held fixed. Otherwise the relationship is a more complex one between the three gear units, and won,t be continuous between any two of them.

For a conventional ring gear planetary (ie., with an outer ring gear, an intermediate planetary carrier, and an inner sun gear) the relationship is essentially:

(Sr - Sc)/(Sc - Ss) = Ns/Nr (or Rs/Rr)
Where:

Sr = the RPM of the Ring (annulus) gear
Sc = The RPM of the Planetary gear Carrier
Ss = The RPM of the Sun gear
Ns = The number of Sun Gear teeth
Nr = The number of Ring gear teeth
Rs = The radius of the Sun gear
Rr = The radius of the Ring gear

When operating as a differential unit (ie. with none of the shafts or internal gears held fixed), not only does the relationship above hold, but also the relationship is a differential one (ie., one input/output RPM equals the difference of the other two, such as C = B - A, as an arbitrary example). This can be illustrated with the above equation simply by rearranging its terms to get the following differential relationship:

NrSr = (Nr = Ns)SC - NsSs

Or to illustrate that the 'N' terms are constants, we can substitute:

KSr = K'Sc - K"Ss

We have a simple differential, which to show the other two inputs/outputs, can also be written:

K"Ss = K'Sc - KSr

K'Sc = KSr + K"Sc
- - - - - - - - - - - - - - - - - - - - - - -
To define and determine any simple case relationships between any two of the inputs/outputs, the third must be held fixed, thus for the simple case relationship the Sun and the Carrier, we have:

(Nr + Ns)Sc - NsSs (with the Ring fixed)
Ss = ((Nr + Ns)/Ns)
Therefore:
Ss = ((Nr + Ns)/Ns)Sc (Obviously, in this case, the Sun must . always turn faster than the carier, . and in the same direction.)

Likewise, if we hold the Sun fixed, we get:
NrSr = (Nr = Ns)Sc
Sr = ((Nr + Ns)/Nr)Sc

Sr = (1 + Ns/Nr)Sc (In this case, the ring turns faster . than the carrier, and in the same
direction)

Or, if we hold the carier fixed, we get:

NrSr = -NsSs

Nr = -(Ns/Nr)Ss (Thus, since Ns < Nr, the ring must turn