Need a little Math Help

[Mike Pearson]

I came up with .054 inches. would not be visible by the naked eye with the best cameras. I had a round where both of the cars had exactly the same RT to the thousandth and ran exactly the same over our dials to the thousandth. I won the round. not sure how the computer picked the winner. We both just shook our heads on that run.

[HawkBrosMav]

I also got .547 inches..


Not to completely over complicate things but I've always wanted to know what others think about how the difference in speeds of the 2 cars would change this calculation? Or if we know one or both guy(s) was/were on the brakes does the computer system give us enough info to accurately calculate this distance? Would we not actually want to know the rate of acceleration not the measured MPH to get this distance totally correct? or am I just over thinking it?


the .547 in a very good estimate to me but is it truly the distance between the 2 cars when the first one crosses the finish line?


Brad

[Troy Henderson]

Without doing all the conversions to complicate it..

1 mph = 17.6 inches/second

So: 141.5*17.6 = 2,490.4 inches/second

2,490*(2/10000) = 0.498 inches

My guess anyway.

[Spyphish]

Jeff, you need Jethro!!! Remember when Jed asked him to cipher since he was in Beverly Hills school.


Naught plus naught = naught
Naught minus naught = naught
Naught times naught = naught


Naught divided by naught = naught??????


Not really, that's undefined. But it made a great show.


See you in two weeks at BR?

[Larry Sullivan]

Originally Posted by HawkBrosMav

I also got .547 inches..


Not to completely over complicate things but I've always wanted to know what others think about how the difference in speeds of the 2 cars would change this calculation? Or if we know one or both guy(s) was/were on the brakes does the computer system give us enough info to accurately calculate this distance? Would we not actually want to know the rate of acceleration not the measured MPH to get this distance totally correct? or am I just over thinking it?


the .547 in a very good estimate to me but is it truly the distance between the 2 cars when the first one crosses the finish line?


Brad

All you need is the speed of the losing car since that''s how far he was behind when the winner broke the beam. Assuming the speed isn't changing very much in the last 66 feet, of course.

[Greenlight]

Remember, that the speed is now calculated as an average speed using timers at a distance 66 feet before the finish line, and the timer at the finish line.

So, he was probably actually going quicker than 141.5 at the finish line.

But, using 141.5 mph = 207.5383 ft/sec.

2/10000 seconds times 207.5383 ft/sec = 0.041508 ft.

0.041508 ft times 12 inches/ft = 0.498 inches.

At 143 MPH it is 0.503 inches.


*In the old days there was a timer 66 ft. before the finish line and one 66 ft. after the finish line. The average time was a lot closer to the actual finish line MPH back then.

[SSDiv6]

Taking into consideration the car is accelerating, the result is 0.4980 inches.

[Bobby Brannon]

Just saying. 3856 MOV. 3/4" of inch
but I am old

[HR9121]

A little more than a split frog's hair would be a guesstimate.