For the Indianapolis 500, the starting order is set by average speed. Cars run four laps, and the four lap average is taken for the starting lineup. Each lap the car takes is averaged in as an average speed. My question is, how do you find that average speed, the average speed of a lap?
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I happen to be a fan of "The Greatest Spectacle In Racing".
so let's start with some basics:
88 ft/s = 60 mph
the oval race course itself is 2.5 miles which = 5280(2.5) = 13,200 ft
IF a car's average speed for the 4-lap qualifying = 227.375 mph
that's (227.375/60) x 88 = 333.483 ft/s
and it took an AVG of 13,200/333.483 = 39.582 s
for that car to qualify at 227.375 mph
so yes, the four(4) electronically obtained times per lap can be averaged
to find the avg speed => just take the average time for these laps and
divide that into the 13,200 ft length of a lap. This gives U the speed in ft/s.
Convert "ft/s" into "mph" by multiplying by 60/88 = 0.68182
so let's start with some basics:
88 ft/s = 60 mph
the oval race course itself is 2.5 miles which = 5280(2.5) = 13,200 ft
IF a car's average speed for the 4-lap qualifying = 227.375 mph
that's (227.375/60) x 88 = 333.483 ft/s
and it took an AVG of 13,200/333.483 = 39.582 s
for that car to qualify at 227.375 mph
so yes, the four(4) electronically obtained times per lap can be averaged
to find the avg speed => just take the average time for these laps and
divide that into the 13,200 ft length of a lap. This gives U the speed in ft/s.
Convert "ft/s" into "mph" by multiplying by 60/88 = 0.68182
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You take the speed of each lap for each car add them up, and divide that by the number of laps..
Ex.
One car takes 100 seconds for lap 1, 112 seconds for lap 2, 101 seconds for lap 3 and 144 for lap 4...
100 + 112 + 101 + 144 = 457 seconds
457 / 4 = 114.25 seconds on average
Ex.
One car takes 100 seconds for lap 1, 112 seconds for lap 2, 101 seconds for lap 3 and 144 for lap 4...
100 + 112 + 101 + 144 = 457 seconds
457 / 4 = 114.25 seconds on average
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By trying to figure out what "average speed" is.
Try Google.
Try Google.