First Order Linear
Hi Mike,
This topic has finally arrived and I welcome it, but I gotta warn ya, at some point in the course of this discussion I will probably state some things that while perfectly valid in theory, may be hard to substantiate empirically. This is where experienced racers like yourself and Jon come in. I can theorize all day, but you guys are where the chain hits the wood and I refuse to step into the ring with you there because I'm simply not qualified to do so. The bottom line is that while I consider it acceptable that AS history may paint me as an out of control lunatic, I don't want to be remembered as someone who claims he can file a world class racing chain in an hour and a half.
First off, I think that in order to talk about this kind of stuff, we have to keep reminding ourselves that there are many variables to contend with. What has to be done is keep as many things constant as possible otherwise we'll never get anywhere. This being the case, I believe that your 1/4 mile draq racing analogy is not the best one to correlate with HP and how it relates to cutting times. In a drag race, the car is accelerating during the run. In a saw race the engine speed is constant and this makes things a lot easier to think about. Horsepower and torque are linked at the hip. Torque is what is actually measured and horsepower is derived from torque and RPM. Horsepower is what counts when you want to see how fast a car goes. Torque is what determines how fast you get there. If you gear a car too high, you won't have enough torque to get to the RPM point where you have enough HP to achieve it's maximum speed. If the car is geared too low, you'll go past the peak HP point in RPM and again, you won't have enough HP to achieve maximum speed. In any type of engine, HP=(Torque)(RPM)/5252. What this says is that on any engine, torque (in ft.lb.) will always be greater than HP below 5252 RPM and less than HP above this engine speed.
Drag and frictional losses don't increase linearly with speed. As you point out in your drag race scenario, it takes a lot more than double the HP to get to a 10 second ET from a 20 second one. If we get away from the acceleration case and revert to merely a constant top speed condition, we can say that in order to double the speed of something, with everything else being equal, it will take approximately 4 times the HP. This is what is referred to as a "square law" and merely states that HP increases as the square of the speed. Three times faster would require 8 times the HP, 4 times faster would be 16 times the HP.
Just to make things simple, let's leave sprocket size and modified engines' RPM increases (chain speed variables) out of the equation for a moment and assume, for the sake of argument, that the only thing we will change after a saw engine is modified is to increase the chain raker depth. What this says is that if we have the power to support the same engine speed as the stock engine with it's taller rakers and the pressure we put on the bar is also the same, then the cut will be faster. Yes, yes, I know that the RPM will increase and you'll put on a bigger sprocket and maybe leave the rakers the same and....whatever, but I have to do this in stages as what's left of my brain will explode if I don't. By taking bigger bites out of the wood with the increased raker depth of the modified saw, we have left the chain speed, hence the drag and friction of the chain in the wood, around the bar groove, etc., constant. What this does is allows us to further simplify things by eliminating the square law factor, thus making the equation what is called "first order" or "linear"; namely no exponents (squared, cubed, etc).
In this simplistic case, figuring the cutting time through the log as it relates to saw HP becomes the simple inverse function:
Time=1/HP. As an example; if a stock saw with 5 HP does a cut in 20 seconds, it would take 10 HP to do the same cut in 10 seconds, 20 HP to do it in 5 seconds, 40 HP to do it in 2.5 seconds. This is why Robert's V8 Predator goes through a 30 inch log in 0.9 seconds, but it takes 300 or whatever HP to do it. Note that as we keep doubling the HP, the cutting times are reduced by 50%, and as such it would take an infinite amount of HP to do a zero second cut. An inverse function such as this will never reach zero, as with each doubling of the HP, the cutting time will be reduced by half of it's value. I think this is where the confusion arises about the 100%/50% stuff. A 100% increase in HP will result in a 50% reduction in cutting time. Of course this is just a very simplified first order linear approximation.
