Super simple overdrive question

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Old 11-24-2017, 11:33 PM
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Super simple overdrive question

Before the flaming begins, search as I may, I did not find an answer to this question:

I have a 1966 F100 with a C4 transmission and 3.25 pumpkin.

I am considering swapping in an AOD transmission and going with a 3.73 chunk.

The C4 has a 1.00 3rd gear ratio. The AOD's 4th gear is 0.67 "overdrive"

Is the math as simple as this: If I have a 3.73 third member with a 0.67 overdrive, is that essentially the same thing as having a C4 with a 2.50 gear (3.73 X 0.67 = 2.4991)? Am I missing something or is the math that simple?

other gear options with an AOD with 0.67 overdriven 4th gear...
4.11 gear X 0.67 O/D = 2.75 equivalent final drive w/a C4
4.30 gear X 0.67 O/D = 2.88
4.56 gear X 0.67 O/D = 3.05
4.88 gear X 0.67 O/D = 3.27 - you get the picture. I feel like I'm overlooking something, but would be REALLY thrilled if I were not....
 
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Old 11-25-2017, 12:35 AM
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Yes, the math is that simple.

However, the actual in-use action won't be "dead on" like one would expect with the math.

Because the C4 has torque-converter action in ALL gears. But the AOD, unlike every other auto trans, has effective torque converter lockup permanently in gears 3 & 4.

The AOD is a two-input transmission via concentric shafts. One shaft is driven by the output of the torque converter (has T-C action) and is used for gears 1, 2 & reverse. Shifting up to 3rd gear engages the Direct Clutch (also engaged in 4th gear too). The Direct Clutch skips using the output of the Torque Converter, and instead uses the shell of the torque converter, so it's effectively driven by the flex plate, no torque converter action at all. So both 3rd and 4th gears are direct-drive gears.

So what does this mean in driving? It means that in 3rd and 4th gears, if you happen to be low down in RPM, and want to accelerate, either you wait some for engine revs to slowly build up into the engine's power band, where it will start to pull decently, or you have to press it down enough to force a downshift into a lower gear (4th to 3rd to get gear ratio change, or from 3rd down to 2nd to get BOTH gear ratio change + torque converter action).

In old automatics (C4, C6, etc.), you always had torque converter action available, and a request for increased speed could be helped by the torque converter, did not always need a forced downshift.

In an automatic with a locking torque converter (AOD-E, 4R70W, E4OD, etc.), a request for increased speed first unlocks the torque converter, to let it have a go at it. If that doesn't do it, then the PCM will downshift to a lower gear.

So from having driven full-size Ford, Lincoln, and Mercury cars with 302s and AOD trans's for years, I would suggest not going too far out on the numbers towards the numerically-low overall drive ratios. You won't have a torque converter to help out in 3rd and 4th gears.
 
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Old 12-10-2017, 07:59 AM
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good gravy, that's some good tech!!

SOOOO much better than what I'm used to:
"google it"
"wrong section"
"watching"
"my friend's cousin's third husband's uncle had this same question - I don't remember the answer"

THANKS, torky!!
 
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Old 12-10-2017, 08:09 AM
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Originally Posted by Torky2
However, the actual in-use action won't be "dead on" like one would expect with the math.

So what does this mean in driving? It means that in 3rd and 4th gears, if you happen to be low down in RPM, and want to accelerate, either you wait some for engine revs to slowly build up into the engine's power band, where it will start to pull decently, or you have to press it down enough to force a downshift into a lower gear (4th to 3rd to get gear ratio change, or from 3rd down to 2nd to get BOTH gear ratio change + torque converter action).
Here's the plan to offset this:
truck has a mild cam in it: lopey & has a little hair on it, but not so much that it's too hairy and uncivilized for DD. But it DOES need a stall, which it has in the C4 - 2,800 RPM B&M unit, which works to a TEE

Since my 3rd grade math question evolved into concentric shaft talk, here's a real world app question:

The AOD is going in. I'm thinking with that kind of final drive advantage, I'm going to go with 4.11's instead of 3.73's and planned on using a stall in the AOD to help the cam. Is 2,800 still a good number, or has the dynamic changed?
 
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Old 12-10-2017, 01:40 PM
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If you needed a 2800 stall with the C4 the AOD doesn't change the dynamic that much all else being equal given that the 1st gear ratio is about the same between the two. Going to a deeper final drive might make it so you can reduce the stall seed some without giving anything up off the line though.

Another option to solve the low rpm bog inherent to the AOD is to tun a converter that is built to bypass the lockup function. The B&M holeshot 2400 is the first one that comes to mind, but I'm sure there are others. This type of converter runs both shafts off the impeller so you are always driving though the converter just like you would with a C4. You loose some of the benefits of the overdrive and create a lot more heat, but you can get some RPMs up in 3rd and 4th without having to force a downshift to 2nd.

I'd get a good auxiliary cooler if you go this route.

I had a holeshot 2400 in an otherwise stock '85 crown Victoria and actually liked it quite a lot. Turned the fun factor up quite a bit. Can't speak to longevity, though since I got t-boned 10k miles after having the trans rebuilt and the converter installed. Was fun to drive with the higher stall and shift kit, though
 
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