Been thinking about this, I guess wheeling automatics is a little different than what is considered "traditional" (though maybe not so much anymore). In my old 80 Series, I had adopted a two-footed driving method, for a variety of reasons, but one of them was to be able to rev the engine while braked, before tackling a steep climb, in order to lock up the torque converter and get more efficient power transfer from the crank and less heat buildup. Recently, though, I have heard that some people will include the torque converter's stall ratio (1.88:1 on my old A442F tranny) in the equation for crawl ratio. However, I am wondering if that is really the case (and what happens in reality)?
Crawl ratio is determined by (diff ratio) x (trans ratio) x (transfer case ratio). So in the case of the 80 (in stock form), it is:
4.11 x 2.95 (1st gear) x 2.488 (Lo range) = 30.17:1 crawl ratio
However, some people have said that I need to include the torque converter stall ratio in that equation as well. In the case of the A442F on my 80, that introduces a 1.88:1 stall ratio. So the end result is:
4.11 x 2.95 x 2.488 x 1.88 = 56.71:1 crawl ratio
That is actually a fairly significant difference! But is it true and do you actually see that kind of reduction when wheeling? Given my admittedly limited understanding of how torque converters work, I was under the impression that, while you may be utilizing that 1.88:1 increase in torque, that is made up for by the fact that the converter is slipping (to keep the engine from dying), causing power loss and heat buildup. I'm not sure if it is a one-for-one relationship between torque lost to slip and torque gained (as if through gear reduction) by the converter, but for the purposes of the argument, I'll assume they are relatively equal. Now, the A442F (and I assume every Land Cruiser automatic tranny since) utilizes a lockup torque converter, so if you rev the engine high enough, the converter locks up and you effectively get a solid mechanical connection between the crank and transmission, so no more loss of power to slipping. However, I believe you have now also eliminated the torque multiplication (or "gear reduction") effect of the converter, so wouldn't that mean that you should actually not figure in the converter's stall ratio in the equation for crawl ratio in any situation?
Sorry for the long-winded post and the "Auto Transmission 101" questions, just couldn't find a clear (to me) answer searching on MUD or Google.
Thanks,
Crawl ratio is determined by (diff ratio) x (trans ratio) x (transfer case ratio). So in the case of the 80 (in stock form), it is:
4.11 x 2.95 (1st gear) x 2.488 (Lo range) = 30.17:1 crawl ratio
However, some people have said that I need to include the torque converter stall ratio in that equation as well. In the case of the A442F on my 80, that introduces a 1.88:1 stall ratio. So the end result is:
4.11 x 2.95 x 2.488 x 1.88 = 56.71:1 crawl ratio
That is actually a fairly significant difference! But is it true and do you actually see that kind of reduction when wheeling? Given my admittedly limited understanding of how torque converters work, I was under the impression that, while you may be utilizing that 1.88:1 increase in torque, that is made up for by the fact that the converter is slipping (to keep the engine from dying), causing power loss and heat buildup. I'm not sure if it is a one-for-one relationship between torque lost to slip and torque gained (as if through gear reduction) by the converter, but for the purposes of the argument, I'll assume they are relatively equal. Now, the A442F (and I assume every Land Cruiser automatic tranny since) utilizes a lockup torque converter, so if you rev the engine high enough, the converter locks up and you effectively get a solid mechanical connection between the crank and transmission, so no more loss of power to slipping. However, I believe you have now also eliminated the torque multiplication (or "gear reduction") effect of the converter, so wouldn't that mean that you should actually not figure in the converter's stall ratio in the equation for crawl ratio in any situation?
Sorry for the long-winded post and the "Auto Transmission 101" questions, just couldn't find a clear (to me) answer searching on MUD or Google.
Thanks,