Sorry to have been absent from the discussion. I thought my question was a dead horse.
LoneOak is right, I’m trying to find an equation that will show the effects of gearing a chainsaw. It is my understanding that with all things being equal, more sprocket teeth = higher chain speed / less cutting torque due to an increase in the sprocket diameter, and fewer teeth = slower chain speed / more cutting torque.
Knowing the powerhead’s RPM and HP and the chain pitch, can the benefits of a 6, 7, or 8 tooth drive sprocket be calculated for a given saw. Or at least the difference.
{If I recall correctly Hp was derived from noting the time taken for a (pit-pony) to pull a load of 1 ton over a distance of 1 foot in 1 minute but continuously for, again from memory, 4 hours. }
When the engine is not operating, torque is nil. When operating with no load other than the crank the turning power depends on the thrust of the pistons. The speed with which that torque is applied results in measurable 'horsepower'. The formula used as earlier gives the wrong impression....Torque produces Hp not the other way around and that Hp varies with the applied torque and the speed of rotation. Ingneral terms Torque = force applied x distance of the travel. Hp =torque x rpm.
Timing, cam profile, headwork, nature of aspiration and so on affect performance. In a chain saw fuels may alter the performance but with little adjustment to utilise it....the type of chain teeth used might give a sensory indicator.
Cars track raced at high revs commonly use light if any flyweels and may use lightened gears as not needing so much torque sustainability as might a road car with mechanically-unskilled drivers allowing the engine speed to fall below best torque production in any gear....as the driving wheels slow down Hp is reduced....Hp being the concept of how quickly work can be done. It's not so easy to hold a chainsaw at constant speed at which torque is highest so we run them 'flat out' and the conditions pull the Hp up and down.
With transmissions...and a chainsaw is a 'power/torque' transmitter come vectors which will reduce the 'crankshaft' Hp dependant on the torque sustainability or otherwise. To asses the torque difference between drive sockets use the factory figures and ratio the diameters, but also the mass. I suspect the usual 'square law' will come into it so I imagine the comparative formula might be, change in torque... at constant revs....= (Diameter -small) squared x D-small mass /(Diameter -large) squared x D-large mass. Unlike a gear box in a car changing teeth numbers when one gear drives another, changing teeth numbers on a chainsaw drive gear makes little difference. One can argue that a chain is effectively a gear but I'd say unless the bar is very small any change in normal drive gear teeth (one or two) would be marginal...
The torque at chain-drive gear centre will be the close to being the same not withstanding gear-teeth ; more torque at some marginal amount will be needed to turn the heavier gear. It is the mass, condition, suitability and condition of the chain which demands torque adequate to travel around the bar ....then the cutting itself requires torque increase over the mere turning of the chain.
Chainsaws are run at high rev's to maintain torque and thereby, horsepower. They are in that way interdependent. There is a stage at which the engine perfoms 'best'...as it looks at all factors of rotating mass, fuel, heat...Curves can illustrate that.
