Greetings, mechanical engineers of Physicsforums,(adsbygoogle = window.adsbygoogle || []).push({});

I have been struggling with a particular problem calculating an ideal motor RPM for economical highway cruising.

The motor I am looking at is the Subaru EE20 Diesel motor. Its torque curve can be found here: https://subdiesel.files.wordpress.com/2011/05/powergraphsubaru.png

I am considering this engine for a swap in to my car for its great low-RPM grunt and economy. But with the stock 6 speed transmission, in top gear 75 mph comes at 3600 RPM -- just about the peak of the power band, and at around 290 N-m, or 80% peak torque of the motor. I have calculated the peak thrust at the wheels (from 1800-2400 motor RPM) to be around 1000 Newtons, while drag at 75 miles per hour (33.5 m/s) is around 700 Newtons. So clearly the motor running at that RPM for that speed (33.5 m/s) is sufficient to overcome drag on a perfectly flat plain. So why the heck is the overdrive gear set for such a high engine speed? Surely the 110 kW (148 hp) of power at that speed far exceeds what is required to move the car, isn't it? The car I would be swapping it in to weighs around 16.5 kN, or 3700 pounds. This is a rough estimate for my own weight and a bunch of gear in the back. Am I missing something?

Additional note: The final drive of the car is 4.444:1. I calculated that with a 3:1 final drive, I could travel at 73 mph at 2400 RPM, which is close enough for my needs... or is it? Again, I feel like I am missing something. I know diesel power bands are far narrower than petrol ones, but I have trouble accepting that Subaru would design a transmission to basically never exceed 80 miles per hour. The CVT, which I would much rather have considering the narrow power band, has ratios closely matching the 6MT as well and has the potential to achieve a far greater gear spread, so I am not sure what to think from their perspective.

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# Rolling motion mechanics & ideal economical gearbox ratios

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