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I keep reading all over the internet that a torque converter multiplies torque at stall speed. Every explanation I have found sounded more like hand-waving than quantitative physical explanation. Essentially, the argument goes that the stator redirects flow which makes the torque converter "more efficient", thereby multiplying torque. This would not qualify as a demonstration to me.

As far as I understand it, the efficient slippage a torque converter allows lets the engine revolve faster and that alone can produce more torque due to the internal combustion engine output torque curve. But this is probably not the whole story: people and manufacturers insist that a torque converter does multiply torque.

Can anyone give a quantitative explanation of how torque can get multiplied in this context ? I assume this is physically possible since only energy conservation must be met in such a system. I expect the explanation to resemble somewhat the operation of a boost converter that can increase voltage to the expense of current.

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Since I wasn't satisfied with the answers here, I did some studying, and I thing I can explain, what enables the torque multiplication.

I will follow the fluid starting from entering the pump, and explain how it will apply torque to each of the parts.

First, the fluid will enter pump at the center, which will both move the fluid through through the pump, to the outer edge, as well as add rotation to the fluid in the direction of the pump. This will naturally apply a torque to the pump in the direction opposite to the rotation of the pump (this torque is the torque coming from the engine).

Now, the rotating fluid will move from the outer edge of the pump to the outer edge of the turbine and enter the turbine. In the turbine it's blades are shaped to reverse the rotation of the incoming fluid, while it is pushed to the center. This interaction will apply a torque to the turbine (and from there to the transmission) in the direction of the rotation of the incoming fluid (opposite to engine torque).

Last, comes the critical part to the torque multiplication, the stator. The stator is in the center of the torque converter, between the turbine and the pump. The stator blades will once again reverse the rotation of the fluid coming from the turbine, to match the rotation of the pump. This will apply a torque to the stator in the direction opposite of the one to the turbine (the stator is fixed in place, though, as it's name tells us, so it will not rotate). Exiting the stator, the fluid will repeat it's cycle.

So, what is it, that allows the mismatch between the input torque and output torque. It is the torque applied to the stator. At an equilibrium, all the torques in the system will add to zero, so the torque applied to the engine must be exactly the opposite of the sum of the torques applied to the stator and the output. Since the output torque and the stator torque are opposite, the output torque has to be larger than the input torque (by the torque applied to the stator to be exact)

Hope this helps someone to understand the black magic of torque multiplication. For me, at least, it was a mystery for far longer than I'm willing to admit.

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A torque converter has a pump driven by the engine and a turbine, which drives the remaining powertrain.

Fluid leaving the pump has velocity components in the axial and rotational reference frames (it's swirling). Increased swirling means that the fluid impacts the turbine with a higher angle of attack. The effect is similar to pitch angle vs. torque for a wind turbine, or attack angle vs lift for an airplane wing. When the fluid's angle of attack increases, the torque multiplier at the turbine increases(not absolute torque, but the torque ratio).

Stators are used in an attempt to control the rotation of the flow to achieve some effect. For example, to increase compression ratios in jet engines. See this page to view the torque ratio vs speed ratio/ efficiency tradeoff for heavy duty applications, where torque converters are very popular: http://jmclutch.com/site/book/export/html/7

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I had the same question!! And you're correct.

Some student at MIT put together a webpage with the energy/torque shares of the stator/impeller and turbine. It's like an extra impeller that utilizes the energy by fluid exiting the turbine, but disengages when inefficient.

https://web.mit.edu/2.972/www/reports/torque_converter/torque_converter.htm

The best description I found was actually in the graph below, describing the two phases as the "shovel" phase and the "tractor" phase. Essentially, the stator acts just like flaps on an airplane wing - the wing creates lots of lift (or torque in this case) at low speed, but the shape is too radical and would stall at higher speed. On an airplane, the flap retracts manually, but in the torque converter, the one-way clutch automatically removes the extra foil camber (or bucket, or shovel, or however you want to call it) at the high speed where the shape isn't efficient. GD genius.(In this analogy the thrust from the engines driving the wing through the air would be analogous to the engine driving the impeller through the fluid).

http://jmclutch.com/site/book/export/html/7

An interesting side note, the stator has way fewer vanes than the turbine and impeller (thus slightly more cross sectional area) - I presume this is to minimize the impedance of flow during the higher speed lockup phase due to the large camber of the vanes.

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I think the torque multiplier effect is related to the size and shape of the impellers like it is with gears with teeth (in a gearbox).

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  • $\begingroup$ Tourer should be tourque $\endgroup$ Nov 8, 2016 at 1:08
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    $\begingroup$ Then edit your answer. If you have multiple accounts, a moderator may be able to combine them for you. $\endgroup$
    – hazzey
    Nov 8, 2016 at 1:55
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It’s better to look at the power generated by the torque converter. The power out of the torque converter should equal the power into the torque converter multiplied by the efficiency of the torque converter. Generally, the torque converter input speed will be higher than the output speed due to the fluid coupling when the engine/motor is applying a positive torque. So the torque converter output torque is equal to the input torque times the torque converter efficiency at the particular operating point times the ratio of the input speed to the output speed of the torque converter (T2 = nT1w1/w2). So at very high speed differential, the torque output of the torque converter can be double the torque input. For example, the torque multiplication (T2/T1) is 2 if the speed ratio (w1/w2) is 10 and efficiency n is 0.2 or 20%.

Someone else provided this link: http://jmclutch.com/site/book/export/html/7

At the bottom there’s a graph that shows the torque multiplication factor vs. the speed differential in the torque converter. As the speed differential decreases, the torque multiplication goes to 1. Since there is always some loss in the fluid coupling, efficiency will be less than 1. This means that due to power loss from efficiency < 1, there will be a speed difference between input and output speeds from the torque converter.

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  • $\begingroup$ Are you sure "the power into the torque converter should equal the power out of the torque converter multiplied by the efficiency of the torque converter"? Shouldn't it be the other way round, the power out of the torque converter should equal the power in of the torque converter multiplied by the efficiency of the torque converter? $\endgroup$
    – Fred
    Jun 27 at 0:42
  • $\begingroup$ Yes you’re right it’s the other way around. That was an editing mistake, I’ll amend it. $\endgroup$
    – crews_der
    Jun 28 at 3:54

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