## SIMPSON GEARSET RATIOS

As with any gear train, the gear ratios of a Simpson gearset are the number of rotations of the input shaft required for each rotation of the output shaft in a particular gear. The output shaft of an automotive transmission is connected to the driveshaft and differential; in an automatic transmission, the input shaft is usually driven by the turbine of the fluid coupling or torque converter. If we call the velocity of the input shaft **V _{I}** and the velocity of the output shaft

**V**, the gear ratio therefore equals:

_{O}**V _{I} / V_{O}**

(For these calculations, it’s important to specify **velocity** rather than **speed** because the *direction* of rotation is also significant. A positive velocity signifies rotation in the same direction as the input shaft (i.e., forward) while a negative velocity signifies reverse rotation (i.e., backward.))

The ratio of any gear train depends on the relative velocities and numbers of teeth of the various gear elements. For two meshed gears, the ratio is the teeth of the driven gear divided by the teeth of the driving gear. For example, if a gear with 25 teeth drives a gear with 40 teeth, the gear ratio will be 40 ÷ 25, or 1.6:1, meaning that the driving gear must turn 1.6 times for each complete rotation of the driven gear.

If the planet carrier is stationary while the annulus drives, the rotation of the annulus drives the planet gears forward at a ratio equal to planet teeth divided by annulus teeth. The planet gears then drive the sun gear backward at a ratio equal to sun gear teeth divided by planet teeth. The net ratio is the product of those two ratios:

**Annulus Velocity * (Planet Teeth / Annulus Teeth) * -(Sun Gear Teeth / Planet Teeth) = Sun Gear Velocity**

As mentioned above, the negative sign indicates a reversal of direction. You’ll also notice that the two planet teeth values cancel out; planet teeth divided by planet teeth equals 1. Therefore, we can drop those values and simply write:

**Annulus Velocity * -(Annulus Teeth / Sun Gear Teeth) = Sun Gear Velocity**

Conversely, if the sun gear drives while the carrier is stationary:

**Sun Gear Velocity * -(Sun Gear Teeth / Annulus Teeth) = Annulus Velocity**

For both of the above equations to be simultaneously true, the velocity of the planet carrier must equal:

**(Annulus Velocity * Annulus Teeth + Sun Gear Velocity * Sun Gear Teeth) / (Annulus Teeth + Sun Gear Teeth) **

That’s for a simple planetary gearset, but this equation still applies to compound planetary gearsets. In a Simpson gearset, the velocity of carrier A (which we’ll abbreviate **V _{CA}**) must equal:

**V _{CA} = (Annulus A Teeth * Annulus A Velocity + Sun Gear Teeth * Sun Gear Velocity) / (Annulus A Teeth + Sun Gear Teeth)**

At the same time, the velocity of carrier B (which we’ll abbreviate **V _{CB}**) must equal:

**V _{CB} = (Annulus B Teeth * Annulus B Velocity + Sun Gear Teeth * Sun Gear Velocity) / (Annulus B Teeth + Sun Gear Teeth)**

For the sake of brevity, we’ll abbreviate “sun gear teeth” as **ST**, “annulus A teeth” as **AAT**, and “annulus B teeth” as **ABT**.

In a Simpson gearset, annulus B is always affixed to the output shaft. Therefore, the velocity of annulus B (which we can abbreviate **V _{AB}**) must always equal V

_{O}. Typically, carrier A is also affixed to the output shaft or else affixed to annulus B. Either way, the velocity of carrier A (V

_{CA}) must also equal V

_{O}. However, in variants where annulus A is permanently connected to the input shaft and carrier A is connected to the output shaft through the forward clutch, V

_{CA}only equals V

_{O}when the forward clutch is engaged.

### FIRST GEAR

In any iteration of the Simpson gearset, both the forward clutch and the low/reverse brake(s) are engaged in first. Therefore, in first gear, the velocity of annulus A (which we’ll call **V _{AA}**) equals V

_{I}and the velocity of carrier A (V

_{CA}) must equal the velocity of annulus B and the output shaft (V

_{O}). Since carrier B is held by a brake in first, the velocity of carrier B (V

_{CB}) equals 0.

We don’t yet know the velocity of the sun gear (which we’ll abbreviate as **V _{S}**), but we do know that:

**V _{CA} = (AAT * V_{AA} + ST * V_{S}) / (AAT + ST)**

Since the velocity of carrier A is the velocity of the output shaft:

**V _{O} = V_{CA} = (AAT * V_{AA} + ST * V_{S}) / (AAT + ST)**

We also know that:

**V _{CB} = (ABT * V_{AB} + ST * V_{S}) / (ABT + ST)**

The velocity of annulus B also equals the velocity of the output shaft, so we can substitute V_{O} for V_{AB} as follows:

**V _{CB} = (ABT * V_{O} + ST * V_{S}) / (ABT + ST)**

The low/reverse brake is engaged in first gear, so carrier B cannot move and V_{CB} is 0. Therefore:

**(ABT * V _{O} + ST * V_{S}) / (ABT + ST) = 0**

Solving for V_{S} gives us:

**V _{S} = -ABT / ST * V_{O}**

Because gearsets A and B share a common sun gear, we can substitute this result for V_{S} in the first equation above. If:

**V _{O} = (AAT * V_{AA} + ST * V_{S}) / (AAT + ST)**

… and:

**V _{S} = -ABT / ST * V_{O}**

… then:

**V _{O} = (AAT * V_{AA} + ST * (-ABT / ST) * V_{O}) / (AAT + ST)**

We can then simplify and solve for V_{S}:

**V _{S} = V_{AA} * -(ABT / ST) / (1 + ST / AAT + ABT / AAT)**

With the forward clutch engaged, the velocity of annulus A is the velocity of the transmission input shaft, so we can substitute V_{I} for V_{AA}:

**V _{S} = V_{I} * -(ABT / ST) / (1 + ST / AAT + ABT / AAT)**

Note the negative sign, indicating a reversal of direction. Since the velocity of the input shaft is positive (forward), the sun gear’s velocity must be negative (backward).

Since we determined earlier that V_{S} equals -ABT / ST times V_{O, then:}

**V _{O} = V_{S} * -ST / ABT**

The sun gear’s velocity is already negative, so the two negative signs cancel each other out, making the velocity of annulus B positive. In other words, the sun gear’s backward rotation causes annulus B, carrier A, and the output shaft to rotate forward.

We can again substitute this result into the earlier equation, giving us:

**V _{O} = (AAT * V_{I} – ABT * V_{O}) / (AAT + ST)**

… which simplifies to:

**V _{O} = V_{I} / (1 + ST / AAT + ABT / AAT)**

Therefore, the gear ratio in first gear is:

**V _{I} / V_{O} = 1 + ST / AAT + ABT / AAT**

(We could also simplify this 1 + (ST + ABT) / AAT, but the above will help to illustrate an important point later.)

### SECOND GEAR

In second gear, the forward clutch remains engaged, but the low/reverse brake releases and the intermediate brake engages, which brings the sun gear to a halt and prevents it from turning backward.

As before, the velocity of carrier A — and thus the velocity of the output shaft — must equal:

**V _{CA} = (AAT * V_{AA} + ST * V_{S}) / (AAT + ST)**

However, with the intermediate brake engaged, V_{S} = 0. Therefore:

**V _{CA} = AAT * V_{AA} / (AAT + ST)**

… which simplifies to:

**V _{CA} = V_{AA} / (1 + ST / AAT)**

Since the forward clutch is engaged in second, V_{AA} equals V_{I} and V_{CA} equals V_{O}. Therefore:

**V _{O} = V_{I} / (1 + ST / AAT)**

… and the gear ratio in second gear is:

**V _{I} / V_{O} = 1 + ST / AAT**

You’ll notice that we haven’t said anything about gearset B. However, we know that the velocity of annulus B (V_{AB}) equals the velocity of the output shaft and that the sun gear is stationary. Therefore:

**V _{CB} = ABT * V_{AB} / (ABT + ST)**

… and:

**V _{CB} = V_{AB} / (1 + ST / ABT)**

Since annulus B rotates forward in second gear, V_{AB} is a positive number. Therefore, V_{CB} is also positive, meaning the rotation of annulus B attempts to rotate carrier B forward at reduced speed. If carrier B is held by a one-way clutch in first, the one-way clutch will allow carrier B to rotate idly forward. If carrier B is held instead by some other type of brake, that brake must be released for the shift to second.

### THIRD GEAR

In third gear, the forward clutch remains engaged, but the intermediate brake is released and the direct/reverse clutch engages.

Once again, the velocity of carrier A equals:

**V _{CA} = (AAT * V_{AA} + ST * V_{S}) / (AAT + ST)**

… and the velocity of carrier B equals:

**V _{CB} = (ABT + V_{AB} + ST * V_{S}) / (ABT + ST)**

With the forward clutch still engaged, V_{AA} continues to equal V_{I}. However, with the direct/reverse clutch also engaged, the sun gear is also linked to the input shaft, so V_{S} must *also* equal V_{I}. Therefore:

**V _{CA} = (AAT * V_{I} + ST * V_{I}) / (AAT + ST)**

… and so:

**V _{CA} = V_{I}**

Since the velocity of annulus B is the same as the velocity of carrier A:

**V _{CB} = (ABT * V_{I} + ST * V_{I}) / (ABT + ST)**

Therefore, V_{CB} also equals V_{I}. Since both carrier B and the sun gear each have a positive velocity (rotating forward), any one-way clutches attached to those elements will unlock and remain unlocked in third.

Both carrier A and annulus B are connected to the output shaft, so their velocity equals V_{O}. Consequently, in third gear:

**V _{I} = V_{O}**

The gear ratio in third is therefore 1:1 — direct drive.

In Simpson’s split torque variants, the direct/reverse clutch does not connect the sun gear to the input shaft, but rather to the torus housing of the fluid coupling or torque converter. This causes the sun gear to rotate at engine speed, which is always somewhat faster than input shaft speed due to hydraulic slippage in the fluid clutch. The velocity of carrier A and the output shaft will therefore equal:

**V _{O} = (AAT * V_{I} + ST * Engine Speed) / (AAT + ST)**

For example, if the sun gear has 39 teeth and annulus A has 81 teeth, output shaft velocity will be:

**V _{O} = (81 * V_{I} + 39 * Engine Speed) / 120**

… or:

**V _{O} = 0.675 * V_{I} + 0.325 Engine Speed**

Therefore, the sun gear receives 32.5% of engine torque, effectively demultiplying any slippage in the fluid coupling or torque converter by that amount. For instance, if the engine is turning 2,000 rpm and there’s 5% slip in the converter, the input shaft will turn at 1,900 rpm. The output shaft will therefore turn at about 1,933 rpm. (Our earlier article provides more explanation of the split torque concept than you probably wanted.)

What about gearset B? The sun gear rotates at engine speed and annulus B rotates at output shaft speed. The velocity of carrier B is:

**V _{CB} = (ABT * V_{AB} + ST * V_{S}) / (ABT + ST)**

Since V_{AB} equals V_{O}:

**V _{CB} = (ABT * V_{O}) + ST * engine speed) / (ABT + ST) = (1 + ABT / ST) * V_{O} + (1 + ST / ABT) * engine speed**

We could simplify this further, but we’ll spare you and just say that carrier B will rotate idly forward, turning somewhat slower than the sun gear, but faster than the output shaft.

### REVERSE

For reverse, the forward clutch releases and both the direct/high clutch and low/reverse brake engage. With the direct/high clutch engaged, the sun gear now rotates with the input shaft, so V_{S} equals V_{I}. As in first, carrier B is stationary, so V_{CB} equals 0.

As we noted above:

**V _{CB} = (ABT * V_{AB} + ST * V_{S}) / (ABT + ST)**

Therefore, since V_{CB} is 0, then:

**V _{AB} = -(ST / ABT) * V_{S}**

… and:

**V _{O} = -(ST / ABT) * V_{I}**

The gear ratio in reverse is therefore:

**V _{I} / V_{O} = -ABT / ST**

Note the negative sign, indicating a reversal of direction. As in first gear, holding carrier B stationary while the sun gear turns causes annulus B to rotate in the opposite direction. Since the sun gear is rotating forward, annulus B must turn backward.

The behavior of gearset A in reverse depends on the layout:

- In Simpson gearset variants where annulus A is permanently connected to the input shaft, annulus A will naturally continue to rotate forward at a velocity equal to V
_{I}. However, the release of the forward clutch disconnects carrier A from the output shaft. Since V_{AA}and V_{S}both equal V_{I}, just as in third gear, carrier A will rotate idly forward at input shaft speed. - In variants where disengaging the forward clutch disconnects annulus A from the input shaft, carrier A will rotate at the same velocity as annulus B. In this case, that means carrier A rotates backward as the sun gear rotates forward. Annulus A resolves that difference by rotating idly backward at a velocity equal to V
_{S}* (ST² / AAT² – 2 * ST / AAT).

### NEUTRAL

Most Simpson gearsets achieve neutral by releasing all clutches and brakes. With none of the elements driving, no torque can be transmitted to the output shaft.

In variants where annulus A is permanently connected to the input shaft, the forward clutch is disengaged in neutral to disconnect carrier A from the output shaft. Gearset A therefore tends to rotate idly forward as a unit. The rotation of the sun gear and the inertia of annulus B (which is connected to the mass of the output shaft) tend to rotate carrier B forward, so if there is a one-way clutch on carrier B, it remains unlocked, allowing carrier B to idle.

### GEARING SUMMARY

If your eyes rolled back in your head at all the math above, here’s a quick summary of a Simpson gearset’s ratios in each gear:

**First**: 1 + ST / AAT + ABT / AAT**Second**: 1 + ST / AAT**Third**: 1 (direct drive)**Reverse**: -ABT / ST

… where ST means “sun gear teeth,” “AAT” means “annulus A teeth,” and “ABT” means “annulus B teeth.”

## GEARING LIMITATIONS

We mentioned earlier in this article that it’s very common for both gearsets in production Simpson gear trains to use identical gears for reasons of production economy.

In a Simpson gearset, if annulus A and annulus B have the same number of teeth, then ABT divided by AAT must obviously equal exactly 1. The gear ratio in first gear would therefore become 2 + ST / AAT. As you can see, that will always be the ratio in second gear plus 1.

For example, early GM Turbo Hydra-Matic 400 (TH400) transmissions had a sun gear with 39 teeth, identical planet pinions with 21 teeth, and identical ring gears with 81 teeth each. Since 39 divided by 81 is 0.481, a TH400’s forward ratios are therefore 2.481, 1.481, and 1.000. Reverse is 81 divided by 39, or -2.077. We don’t have gear teeth counts for the early Chrysler TorqueFlite, but we would surmise it had a sun gear with 40 teeth, identical planets with 24 teeth, and identical ring gears with 88 teeth, which would yield the published ratios of 2.45, 1.45, 1.00, and -2.20.

Ratios like these were well-suited to American V-8s or big six-cylinder engines, especially with a torque converter to provide extra multiplication under load. Even “tight” torque converter with a modest 2:1 stall ratio gave these transmissions a robust maximum starting ratio of almost 5:1. Second gear, meanwhile, provided a useful passing gear for typical U.S. highway speeds (45 to 75 mph, 72–121 km/h).

However, gearing like this was less than ideal for smaller engines with less torque, which needed a shorter (higher numerical) first gear for decent acceleration off the line. A Simpson gearset with identical gears can’t get a first gear shorter than about 2.6:1 without using tiny planets on an enormous sun gear, and doing so makes second gear too short to be an effective highway passing gear.

For this reason, some later transmissions with Simpson gear trains (and some aftermarket replacement gearsets) sacrifice the cost advantages of identical gears in order to provide a shorter (numerically higher) first gear. For example, keeping the TH400’s stock gears for gearset B while replacing the gears of gearset A with smaller planet pinions of 15 teeth each and a smaller annulus of 69 teeth would make first gear:

**1 + 39 / 69 + 81 / 69 = 2.739**

… while second gear would become:

**1 + 39 / 69 = 1.565**

Reverse would remain -2.077:1 and of course third gear would still be 1:1. (These ratios, incidentally, are very close to the published gear ratios of GM’s light-duty TH200 and TH200C, for which we don’t have gear teeth counts.)

Some aftermarket gearsets go the opposite direction, using *larger* gears for gearset A to obtain a lower numerical first gear. The object is to provide closer-spaced gears to keep a highly tuned engine in its power band. For example, if the sun gear has 36 teeth, annulus A has 90 teeth, and annulus B has 66 teeth, first gear will be:

**1 + 36 / 90 + 66 / 90 = 2.133**

Second will be:

**1 + 36 / 90 = 1.400**

… and reverse becomes:

**-66 / 36 = -1.833**

With these ratios, engine speed drops only 34.4% on the 1–2 shift and 28.6% on the 2–3. By comparison, the stock gears of a Turbo Hydra-Matic 400 give an rpm drop of 40.3% from first to second and 32.5% from second to third. However, such a tall first gear would make for sluggish launches unless accompanied by a very short axle ratio and/or a torque converter with a high numerical stall ratio. A gearset like this is therefore intended mostly for racing.

Thank you for continuing this fascinating exploration of the technical details and history of automatic and semi-automatic transmissions. I’ve been interested in cars since 1985 when I was finally within reach of driver’s license age and I have read quite a bit of the engineering history of the automobile but I still had no idea of the variety of designs that the American manufacturers put on the road before getting to what I always thought of as the norm – three speed plus reverse, torque converter coupled automatic with the PRND21 shift pattern. I had read references to “slip and slide with Powerglide” but the only car I ever drove with an automatic that was different from the C4/C6/Turbo-Hydramatic/Torquflite experience was a VW Beetle with the semi-automatic transmission. I hope you do an article on pre-selectors someday too.

In fact, I was just looking at some stuff about the Wilson preselector, which is an ingenious device.

Part of the reason I’ve written so much about Hydra-Matic is that I was amazed at how the earlier versions have fallen into obscurity — which is amazing when you consider their production volume. Most mechanically inclined automotive people are probably still reasonably familiar with Turbo Hydra-Matic, TorqueFlite, and the C4/C6, but the older ones are now poorly understood (and some, notably the dual- and triple turbine torque converter transmissions, weren’t that well understood when new!).

My daily driver (by choice) is an ’89 Wrangler. Probably one of the last applications of the classic Torqueflite. Three speeds, no lock up, no overdrive, no electronics. Not all that efficient, but over 300k with a single tear down for new seals.

Another system that saw service with UK’s BMC and its later incarnations was the 4 speed AP (Automotive Products) automatic transmission.

When it was working properly it was an excellent transmission, making automatic versions of the original Mini and Austin 1100 (sold as the Austin America in the USA), and the Austin Maxi. The Mini and 1100 models suffered very little performance penalty compared with stick cars, the Maxi had a 5 speed stick shift, so the automatic suffered in comparison.

The designers managed to cram it underneath the engine as in the stick transmission. Even more remarkably, they persuaded it to work sharing the engine oil. However the latter may have proved not such a great idea, they had a very short lifespan, typically 30-40,000 miles. If it had its own separate lubrication I think that would have helped improve its durability.

It has long been confined to history now, like so many British innovations lack of investment and development meant it fell by the wayside.

I know of the AP automatic, and of the earlier AP Manumatic, but I confess I haven’t ever investigated its innards!