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Re: maxing out
There's always something faster... The faired SV's a little overgeared IMO, you're not the first to find they can't rev it out. Mine has something like 15-20bhp more than stock once you get near the redline (it makes around 80 at peak compared to 68 or so standard, but also doesn't fall off after peak as hard as stock does), and it still takes a fair old time to reach the limiter with the standard gearing. The actual difference in speed isn't huge though.
Another thing to note is that hitting the limiter in top is quite alarming when it happens, I can't recommend it :smt036 I thought I'd blown it up :geek: |
Re: maxing out
I've got a K3 and hit the rev limiter at an indicated 133. I was so tucked in you couldn't imagine-almost part of the bike. Believe it or not it was the first time I'd hit the limiter! I don't make a habit of it.
I wouldn't worry about the 9000 rev thing-the wind was probably against you or something. Do you eat a lot of pies? As far as outgrowing the SV, well I doubt you have, especially considering you only passed your test IN APRIL! DOH! (See first post) :) |
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Re: maxing out
Thanks for all the comments...
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Re: maxing out
Gear it down... You're basically in "overdrive" when in top so will never hit the limiter. Some say they can go faster in 5th than 6th...
Up one on the front and/or 3 on the front, then you'll probably be able to get closer to it. . |
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Drag force = (d * V^2 * Cx * S) / 2 Where: d = air density V = velocity Cx = Drag coefficient S = Frontal surface area or http://en.wikipedia.org/wiki/Drag_%28physics%29 |
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Power The power required to overcome the aerodynamic drag is given by: http://upload.wikimedia.org/math/1/d...54d626ea78.png Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times a work in half the time requires eight times the power. It should be emphasized here that the drag equation is an approximation, and does not necessarily give a close approximation in every instance. Thus one should be careful when making assumptions using these equations. |
Re: maxing out
I don't think anything I said was actually wrong (though if I'd gone on to say more it probably would have been). So in the spirit of things...
I'll offer you a draw? :-) |
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