As soon as I read the title I knew someone would drag me into this.
Let's use Newton's F=ma to answer the question.
First we need to ensure we all understand what Force (F) and mass (m) are.
Horsepower is not force. Horsepower is a rate at which force can be applied. Torque is a measure of rotational force. Torque is the work an enging/gearing system can do, horsepower is how fast the system can deliver that work.
A pound is a measure of force, not mass. A pound is calculated by multiplying the mass of an object by the local force of gravity, which on Earth is 32.2 feet per second squared. What this means to us is that if we have a hypothetical 700 lb. motorcycle/rider combination, the mass is actually a number far smaller...after all, if we put our motorcycle/rider combination into space it wouldn't weigh 700 lbs, but it would still have the same mass.
Dividing 700 lbs. by 32.2 ft/second squared equals 21.7. To keep things simple with unit conversions I'm going to skip over all that, but you'll have to take my word that our 21.7 value is actually 21.7 slugs, which is the British unit of mass.
The Gen 2 torque at the crankshaft is claimed to be 115 lb*ft. Gearing multiplies torque, however, so we have to account for both the gearbox ratio and the final drive ratio.
Second gear ratio, or torque multiplier is 1.937. Final drive ratio (stock) I believe (correct me if I'm wrong) is 2.35...based on a 17-tooth sprocket and a 40 tooth sprocket. There's a reson I've selected second gear which I'll explain later.
So our "F" for the equation is 115 lb*ft*1.937*2.35. For simplicity's sake I'm only using second gear.
F=523.5 lb*ft
m=21.7 slugs
Now for the fun part...
F=ma
523.5 lb*ft = 21.7 slugs(a)
523.5 lb*ft/21.7 slugs = a
24.1 = a
Again for simplicity's sake I've ignored unit conversion. Again I ask your understanding as you read that I'm just going to add a feet/second squared to our answer.
a= 24.1 ft/s*s.
How does our result stack up against real world numbers?
From wikipeda, Cycle World is quoted as having run a 9.75 @142 m.p.h. quarter mile. We don't know all the details behind that run (did their bike/rider combo weigh 700 lbs.? How much torque did they lose due to friction deficiencies?) We also know the torque multiplication of the gearbox isn't a constant rate during a quarter mile run. Our formula also doesn't account for parasite drag because we're keeping things simple here.
Taking these things into consideration, let's see what speed our math says the Hayabusa should be able to accelerate to in 9.75 seconds.
The final velocity of an object can be found by taking the initial velocity and simply adding the rate of acceleration multiplied by the time available for acceleration. In formula terms it looks like this:
V(f) = V(i)+at
so
V(f) = 0+24.1 ft/s*s*9.75s
V(f) = 235 ft/s
To convert that into miles per hour, multiply by 0.681818182.
V(f) = 160.22 miles per hour
160 m.p.h. is what our math says the Hayabusa should be able to achieve in 9.75 seconds...assuming zero wheelspin, zero aerodynamic drag, a 700 lb total weight, and that we never had to shift out of second gear. The last one is tricky because obviously that's not what is going to happen in the real world. I selected the second gear torque multiplication ratio in an attempt to provide a reasonable average torque multiplication value for the gearbox. With more time a better number could be calculated.
I think given the limitations we've placed on our calculation for time's sake that 160 m.p.h. in 9.75 seconds is close enough to 147 m.p.h. in the same time period to say our math is correct given the limitations we're working with.
So...now we have all the tools we need to calculate how much more "a" we can expect given whatever "m" reduction we want to use.
And that's really what we wanted to know...isn't it?
