Thread: Some contemplations on overdrive

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Originally Posted by JoelGrimes

You guys think about all the angles! Since a fluid does not compress, why not build a drag car with the engine set to run at max torque (after warm up) and put the power through to the drive wheels hydrostatically, using a higher volume pump than the motor, and use a positive control valve controlling bypass, linkaged to the pedal on the right, you could theoretically turn the rear tires 300 + mph as soon as you got the go light! can you imagine the flying parts? Seriously though, I wondered about the logic of gear ratios as a youngster when I discovered that sprocket ratio on a bicycle was affected by the throw length (stroke) on the pedal cranks. Am I getting it?

I think there has been an issue of parasitic loss in the hydrostatic drives. But in theory if you could put the engine at the peak of the torque curve and vary the multiplication to keep it at peak that would be a good thing.
You bicycle analogy is a good simplification of crankshaft stroke.

FWIW Ford does have a constant variable transmission in a few of it's smaller cars. ZF builds it.

A poster on another board offered this. This is the kind of stuff I was looking for.

You knew this was going to happen. I had to put some numbers behind this discussion.

A typical helical gearset is 98-99% efficient. In a direct drive top gear there are none, in an OD top gear there are two. For the sake of discussion, assume 99% for the input shaft gearset and 98% for the OD gearset. These assumptions are supportable given that the input set is typically quite robust and has a relatively low reduction whereas the OD gearset is typically rather weak and is being used inefficiently as a speed increaser.

A hypoid bevel gearset is 92-96% efficient; efficiency decreases as ratio increases. For the sake of discussion assume 95% for a 3.08 gearset and 94% for a 3.55. These are supportable assumptions given the range of 1.5:1 to 6:1 normally encountered.

A Ford 300 at 2000 RPM produces approx. 130 HP under full load and WOT at a BSFC of approx. 0.5 lbs/HP-hour. At cruise under some unspecified but significant load assume for arguments sake that it produces half the HP (65HP) and that the BSFC is 10% worse or 0.55 lbs/HP-hour. Assume further that the gearing is such that this occurs exactly at 60 MPH.

If this engine were to be coupled to the direct drive transmission with the numerically lower rear end, it would deliver:

65 x 0.95 = 61.75 HP to the rear wheels.

It would consume:

65 x 0.55 = 35.75 lbs of fuel per hour which is also 35.75 lbs of fuel per 60 miles

At 6.25 lbs per gallon this would be:

5.72 gallons per 60 miles or 10.49 MPG, arguably what one could expect with a heavy load.

If this same engine were to be coupled to an OD tranny with a numerically higher rear end, to produce the same 61.75 HP at the rear wheels, it would need to produce:

61.75 HP / (.94 x .98 x .99) = 67.7HP at the flywheel

That would result in a consumption of:

67.7 x 0.55 = 37.23 lbs of fuel per hour which is also 37.23 lbs of fuel per 60 miles

At 6.25 lbs per gallon this would be:

5.96 gallons per 60 miles or 10.06 MPG.

So running at highway speeds under a significant load, the difference would be 0.43 MPG. At lighter loads it would be proportionally less, meaning about half as much difference at cruise running empty or about .20-25 MPG.

This treats only the effect of transmission losses and ignores all the other sources of inefficiencies such as bearing losses, seal drag aero and rolling resistance, etc. assuming that they are equal in both cases.

Obviously YMMV both literally and figuratively.