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Old 10-12-2008 | 12:42 AM
  #41  
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DarZeelon
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From: Rosh-HaAyin, ISRAEL
Default RE: Overloading an engine


ORIGINAL: Juice

Dar... One thing I'm wondering is if you are taking into account the 2:1 gear reduction in your calculations. The RPM numbers I'm posting are the prop RPM. The crankshaft RPM is 2x the prop RPM. Again, I have to read everything carefully still, so if you have already taken it into account, please forgive my ignorance.
Juice,


Horsepower calculations do not take into account reduction drive ratios, since work is not affected...

Applying a force over a distance per time unit is equivalent to applying twice the force over half the distance.


Even though the torque is roughly doubled (I take gears at 100% efficiency), the HP output produced is unaffected, so the reduction ratio does not need to be taken into account.


When the power absorption of the prop is used, the engine is regarded as a 'black box' which produces this output... The RPM is what the prop is spinning (crankshaft RPM cannot be measured in the engine...).

I only mentioned these facts in my posts, since Greg posted graphs that got to 10,000 RPM (which is the actual crankshaft speed) and in some graphs he posted torque figures, that are roughly double those of typical .91 four-stroke engines... I felt an urge to explain...


I did not actually state any torque values; just comparative numbers, to show if torque is higher at one point than it is at another.


We will be waiting for your results... Do the larger and smaller props too, please, so we can have all the results together, at the same measuring sequence.


No need to apologize... People don't know this until they learn it.