So this is a math question. Is IT possible that at 4000 rpm when the sled starts rotating the driveshaft, there is roughly 1000 ft lbs / 1600 nm? As im playing with the idea of electric direct drive this number will crush that idea.
Torque output at drive shaft?
- Thread starter Melfjord
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olymon
Well-known member
Depends on what you start with at the motor and multiplication of clutches and gearing. I really don't see it. What you need is a custom torque transducer on the shaft (I make variations of these at work) then you can play with clutching and gearing to maximize torque. Big $$$$ though
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Im using the current engine to figure out what i need. At 4000 rpm (clutch engagement) at the engine, the sleds moves at 3kmh?(walking speed) That will turn the driveshaft at 113 rpm with a driver at 7cm radius. 4000/ 133 = 30 witch is the factor needed to get the torque at that rpm.
at 4000 rpm the engine produce about 100 nm?
100x30 = 3000 nm... That just seems stupidly high am i doing anything wrong?
at 4000 rpm the engine produce about 100 nm?
100x30 = 3000 nm... That just seems stupidly high am i doing anything wrong?
It's much less than that. The torque load obviously can't go beyond what the engine/driveline is capable of producing. So, look up torque curves of your snowmobile engine. Then multiply the clutch. I believe a typical sled clutch is about 3:1. My guess is an 800 probably produces about 50ft-lbs at 4000rpm, which would give you roughly an ideal max torque at the track of 150ft-lb.
yeah that makes way more sense. But should not the clutch ratio and the gear ratio and the final driver be in the equation too?
Yes, definitely. Brain wasn't firing on all cylinders so early in the morning. You can just look up the final drive ratios online.
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