Cycle World article

1olbull said:

The Cycle World Article reads:
"139 thousandths of a second from the Triumph's initial elapsed time?
That's a bike length at the end of a quarter-mile, absolute domination in an actual race."

I question this writer's comments. I believe the writer likely just used the simple formula D=VT.
D=186.81 fps (127.37 mph) times .137 second=25.5 feet.
1) The MC speed of 127.37 mph was not constant.
2) The acceleration rate was variable throughout the 1320 feet.
3) The average acceleration rate would be: the ending Velocity squared minus the initial Velocity squared divided by 2 times the distance.
With the initial Velocity being 0, the acceleration rate would then be 186.81 fps squared divided by (.139 sec times 1320 feet).
acceleration rate = 13.22 fps squared.
4) Distance = initial velocity times time plus one half the acceleration rate times the time squared.
With the initial Velocity being 0, the distance would then be .5 times 13.22 fps squared times .139 second squared.
The actual distance over .139 second would then be .128 foot or 1.5 inches.
Hardly a bike length.

I do not subscribe to Cycle World, but if you would be so kind to post the entire article, author and the mag editor's email, I'd be happy to write them.

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1olbull said:

The Cycle World Article reads:
"139 thousandths of a second from the Triumph's initial elapsed time?
That's a bike length at the end of a quarter-mile, absolute domination in an actual race."

I question this writer's comments. I believe the writer likely just used the simple formula D=VT.
D=186.81 fps (127.37 mph) times .137 second=25.5 feet.
1) The MC speed of 127.37 mph was not constant.
2) The acceleration rate was variable throughout the 1320 feet.
3) The average acceleration rate would be: the ending Velocity squared minus the initial Velocity squared divided by 2 times the distance.
With the initial Velocity being 0, the acceleration rate would then be 186.81 fps squared divided by (.139 sec times 1320 feet).
acceleration rate = 13.22 fps squared.
4) Distance = initial velocity times time plus one half the acceleration rate times the time squared.
With the initial Velocity being 0, the distance would then be .5 times 13.22 fps squared times .139 second squared.
The actual distance over .139 second would then be .128 foot or 1.5 inches.
Hardly a bike length.

I do not subscribe to Cycle World, but if you would be so kind to post the entire article, author and the mag editor's email, I'd be happy to write them.

Click to expand...

Steel said:

Every racer I've ever heard talk about shaving a tenth off a 1/4 mile says it's about a car length.

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Daytonastar said:

Not really able to follow all the maths, but if your final conclusion is correct, then if you were 1.39 seconds slower over a standing quarter, then you’d only be 15 inches behind, and my gut feeling is you’d be way further back than that.

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1olbull said:

At a constant speed of 100 mph over any distance you move 147 feet a second and in .1 second, 14.7 feet (about a car length).
At a constant speed of 200 mph, 293 feet in a second and 29 feet in .1 second.
That .1 second distance is dependant on the time, velocity & acceleration over a given distance, quarter mile or whatever.

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