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Post by murph on Jul 15, 2012 20:25:49 GMT -6
Rich is there. I got 2.89964. I won't squabble over 1/10,000th of a second. Pure Stock or limited? Any special requests? Send me PM. Murph
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Post by CycRunner on Jul 15, 2012 20:58:28 GMT -6
My calculations show 2.89903
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Post by murph on Jul 15, 2012 21:26:37 GMT -6
12.3 *10^(-6) in/in*F X 10F = 1.23*10^(-4) dimensionless 480 inches X 1.23*10^(-4) = 0.05904 inches 2.9s/480 inches = 0.006s/in 0.006s/in * 0.05904 in = 0.00035424 seconds 2.9 seconds - 0.00035424 seconds = 2.89965 seconds
Right me if I am wrong! There are a couple of different ways you can get from start to finish. This is the way I took. Murph
I will do more contests from time to time. Use to do them all the time!
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Post by CycRunner on Jul 15, 2012 22:00:13 GMT -6
12.3 *10^(-6) in/in*F X 10F = 1.23*10^(-4) dimensionless 480 inches X 1.23*10^(-4) = 0.05904 inches 2.9s/480 inches = 0.006s/in 0.006s/in * 0.05904 in = 0.00035424 seconds 2.9 seconds - 0.00035424 seconds = 2.89965 seconds Right me if I am wrong! There are a couple of different ways you can get from start to finish. This is the way I took. Here is how I got 2.89903 sec: Change in inches for 10 deg.F = .000123 inches (same as Murph) Speed = 2.900 sec. So time in inches/sec= 40X12/2.900=165.57241inches per sec. 70deg track length = 480"-0.000123= 479.999877" New track length inches/speed in inches /sec= 479.999877/165.5724= 2.89903 sec cycrunner Murph
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Post by murph on Jul 15, 2012 22:12:57 GMT -6
cycrunner's calculations:
Here is how I got 2.89903 sec: Change in inches for 10 deg.F = .000123 inches (same as Murph)This number is dimensionless, not in inches Speed = 2.900 sec. So time in inches/sec= 40X12/2.900=165.57241inches per sec.Cool 70deg track length = 480"-0.000123= 479.999877"Cyc: you are subtracting a dimensionless number from a dimensional number > that does not work. You need to multiply the dimensionless number by the track length in inches to get the delta length New track length inches/speed in inches /sec= 479.999877/165.5724= 2.89903 secClose, but one manipulation error cycrunner
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Post by Mr. Slick on Jul 15, 2012 23:53:25 GMT -6
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Post by murph on Jul 16, 2012 1:41:29 GMT -6
Slip Evo a $20 and he will sent the temp for you. (Just kidding). With Krytox tune at colder temperatures based on my observations. My oil is pretty resilient to this issue.
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Post by Mr. Slick on Jul 16, 2012 10:17:23 GMT -6
Are we really talking about 354 microseconds? We could probably account for that from the "shipping randomization" process.
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Post by CycRunner on Jul 16, 2012 11:03:44 GMT -6
cycrunner's calculations: Here is how I got 2.89903 sec: Change in inches for 10 deg.F = .000123 inches (same as Murph) This number is dimensionless, not in inchesSpeed = 2.900 sec. So time in inches/sec= 40X12/2.900=165.57241inches per sec. Cool70deg track length = 480"-0.000123= 479.999877" Cyc: you are subtracting a dimensionless number from a dimensional number > that does not work. You need to multiply the dimensionless number by the track length in inches to get the delta lengthNew track length inches/speed in inches /sec= 479.999877/165.5724= 2.89903 sec Close, but one manipulation errorcycrunner Oops! Murph, In my above calculations I neglected to account for the 480" of track. My new calculations are: For the aluminum expansion calculation I see that there is a 10 degree F temp delta. So for every inch of track length for the 10 Deg delta the change is 0.0000123 inches X 10, or 0.000123 inches. For 40' (480") this is 480 x 0.000123 = 0.05904". For a 2.9000 second time on the 40 foot track the speed is 480/2.900 or 165.57241 inches/sec. The new track length at 70 deg is 480."- 0.05904" = 479.94096". So, the new speed is the new track length in inches divided by the speed in inches/sec or 479.94096/165.57241 = 2.89867714 sec. Rounding to 5 places the new track time in November should be 2.89868 sec.By the way, I ran this by my brother in law who is a math prof at Clemson University and he agrees with this.
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Post by Mr. Slick on Jul 16, 2012 12:01:09 GMT -6
I would disagree with all of the calculations since they assume a constant AVERAGE speed. The end of the track right before the timer detection holes is where the length would be effectively "lost". The last part of the track is also where the car would be traveling the slowest.
So, what is the inverse of the speed of the car at the end?
EndOfRunRateofTravel (Seconds per Inch) * TrackLengthDelta(0.05904" Inches)
Also, do we probably should say that each track section actually contracts/expands independent of the other sections. If so, then we get to use the inverse speed of the car at each track section end times the distance change for that section. Taking into account that the transition/curve is shorter.
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Post by B.S.B. Racing on Jul 16, 2012 12:07:53 GMT -6
I say who cares! I want to get to the end......first! ;D
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Post by *5 J's* on Jul 16, 2012 12:24:33 GMT -6
Guess being left handed I do my math a bit different - but regardless I get 2.8996433 with no rounding, or 2.89964 if rounded as specified.
I do mine as a ratio such that:
2.9000s x -------- = ------------------------------- 40ft (40ft - (12.3x10^-6)*10*40ft)
40x = 2.9(40ft - (12.3x10^-6)*10*40ft)
x = 2.8996433s
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Post by *5 J's* on Jul 16, 2012 12:26:24 GMT -6
oops - formatting kind of messes up the ratio - oh well
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Post by ZZ Racing on Jul 16, 2012 13:04:19 GMT -6
Not being a smart A but 480/2.900=165.51724 so 479.94096/165.51724=2.899644.
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Post by Evolution on Jul 16, 2012 13:38:39 GMT -6
Ok I am going to stop this right now. The temp in the shop is as constant to 70 as it can be, so I win.
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