Cycle World article

[Wadejesu]

I need more practice

 

[Lamonster]

I met Rickey at a show years ago and even got a pic of him on one of our custom VTX's. I had no idea this bike would run 10's with the right rider.

 

[1olbull]

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.

The above is not accurate enough and incorrect.
Please see the following post #40 that has the correct calculations.

 

[Daytonastar]

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.

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.

 

[Boog]

@1olbull do you really think the writer would do all that math?

 

[Steel]

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.

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

 

[scot in exile]

come on now everyone knows that --do you never watch street outlaws

 

[FE_rex]

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

I'll vouch for that at those times. I've been there / done that bracket racing in cars; winning and losing by estimating opponent's reaction and et while trying to decide if I should take the stripe or not. You can shave a few hundredths by dragging the brake, coming off the loud pedal is worth a few 1/10s.

1/10 is about a car length. Also acceleration is not constant throughout the 1/4, and acceleration lessens significantly throughout the 1/4.

Assuming 850 lbs with rider. ET showing about 140 hp and mph is showing 135 hp

 

[1olbull]

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.

Run your own number of 1.39 second.
.5 x 13.22 x 1.39 sq = 12.77 feet
That would actually yield a distance closer to a "bike length".

WRONG.
Please following post #40 with the correct calculations.

 

[Steel]

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.

There's nothing constant in the acceleration of a vehicle. Here's a clip of a 8.8 vs 9.68 that's a .88 difference which by your calculations on the Rocket's time (I know different speed, but close enough for me) would be 9.71 inches. Looks a bit further to me.