better brakes

Ilya A. Kriveshko ilya.kriveshko at verizon.net
Thu Sep 25 14:44:54 PDT 2008


Jo Rhett wrote:
> 100 more horsepower gets you 15mph?   That's hard to believe.
> (not saying you are wrong, just doesn't make any sense to me)
>
> On Sep 25, 2008, at 1:20 PM, zoran wrote:
>> it is about 15-20 miles per hour difference between two bikes.
>> it is acceleration of gsxr that makes you think it is much faster
>> than that.
>>
>>
>> Zoran Vujasinovic
>> Twin Works Factory
>> 775-786-4881
>> www.twfracing.com


If you have a very long (e.g. infinite) straight to accelerate, it's the
force of air resistance, not mass, that determines your terminal speed.
Here is the formula for air resistance:

The image
“http://upload.wikimedia.org/math/7/9/c/79c64614a64fc3fffe0dae37cc9d4d7e.png”
cannot be displayed, because it contains errors.

[http://upload.wikimedia.org/math/7/9/c/79c64614a64fc3fffe0dae37cc9d4d7e.png]

Cd is the coefficient of drag
A is the frontal area

Both of the above are different for different bike/rider combo. Both are
likely larger for a large-displacement 4-sylinder bike than for a small
displacement twin. But for the sake of argument, let's suppose we're
using the same frame/geometry/rider/bodywork on both bikes.

The important thing to note in that equation is that v term (the speed)
is squared. That is, if you are going twice as fast, you are getting
four times the resistance. If you're going four times as fast, you are
getting sixteen times the resistance.

So, if an SV with 75 rwhp can develop enough propelling force to reach
the terminal speed of 130 mph (for example), then in order to reach 150
mph (1.154 times faster) that same bike would need to develop a force
that is 1.331 (i.e. 1.154 squared) times larger. So, how much power
would we need?

This is where I'm dropping the ball and need to grab a napkin... How are
propelling force and power related?

Power is the ability to perform work per unit of time:
P = W/t

Work is force multiplied by the distance along the force:
W = F*d*cos(a) = F*d*1 = F*d

Distance is speed multiplied by time:
d = v*t

Put the three together:
P = W/t = F*d/t = F*v*t/t = F*v

Therefore, force is inversely proportional to speed:
F = P/v

So, it looks like given constant (maximum) power, available propelling
force decreases linearly as speed increases.

So, if speed increased by 1.154, the required force increased by
1.154*1.154=1.331, and the required power increased by 1.154*1.154*1.154
= 1.536. So, our 75 rwhp SV would need to develop 75*1.536 = 115 rwhp,
or 40 rwhp more than stock.

If our theoretical SV had 100 more horsepower, as measured at the rear
wheel, and was geared such that it would develop that peak power exactly
at the maximum speed that we're about to compute, then this maximum
speed would be... Anyone? Anyone?

130 * ((175/75)^(1/3)) = 172.38mph

So, Jo, if you can squeeze your GSXR1000 in stock SV bodywork, you might
just be going 170 mph, given a really long straight... The 100 extra
ponies would buy you an extra 42 mph.

Did I get anything wrong here?
--
Ilya

PS: Disclaimer. I pulled the HP and top speed numbers for this
theoretical SV out of my a$$. YMMV.




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