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GS1100E - STOCK Dyno numbers - wheel HP? or crank HP?

freakhousecustoms

Forum Newbie
I'm sorry if this is a redundant question, but I've been searching the threads and I can't find the answer I'm looking for. If someone can point me in the right direction, it would be greatly appreciated.

I was looking for bone stock HP numbers for a GS1100E. suzukicycles.org says 88hp for the 1980-1981 and then 108hp for the 1982-83. 1/4 mile times across the years don't reflect this jump in HP, so I'm wondering if the 88hp was maybe measured at the wheel? And the 108hp was at the crank? As pretty much everyone who does mods usually quotes wheel HP, I'd like to know what number we are all starting with & I'm honestly more concerned with the stock rear wheel HP number.

I've been racing out at Bonneville for the last few years and I'm trying to plan out my next build and stock rear wheel HP would help me in my calculations.

Thanks,

-Fran
 
THANK YOU, posplayr!!!

I knew I had read that thread ages ago, but just couldn't find it.

I've been reading a bunch of your threads recently, especially the one about low cost/high hp head work!
 
Obviously not gospel, but a neat little tool to use.

I went 130mph this year with a stock engine (well, no air filter at all & 4 straight pipes). Hoping to return to the Salt in 2018 with just a little bit more power.

HP_Estimates.jpg
 
Obviously not gospel, but a neat little tool to use.

I went 130mph this year with a stock engine (well, no air filter at all & 4 straight pipes). Hoping to return to the Salt in 2018 with just a little bit more power.

View attachment 52941

Hp requirement goes up as the cube of the velocity ratio:

Hp(V2) = (Velocity_2/Velocity_1)^3*Hp(V1)

Hp(120 mph) = (120/100)^3 * Hp(100 mph) = 1.728 * Hp(100 mph)
 
Hp requirement goes up as the cube of the velocity ratio:

Hp(V2) = (Velocity_2/Velocity_1)^3*Hp(V1)

Hp(120 mph) = (120/100)^3 * Hp(100 mph) = 1.728 * Hp(100 mph)

In case there are people wondering why it is a cube function and not a square function it is because there are two equations at play:

1) Aerodynamic drag goes as the square of the velocity.
2) Power required for any given speed is defined as thrust x velocity. Thrust overcomes drag and is what is actually pushing you forward. This means as you go faster the power required goes up linearly with the speed, independent of drag.

Combine the two functions and you end up with P2/P1 = (v2/v1)^2 x (v2/v1) = (v2/v1)^3


Mark
 
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