Acceleration

Given speed and distance figures comming in at the rate of fif*** samplings per second, how do I compute the acceleration? -- Chuck Stuart - Mesquite TX USA

>Given speed and distance figures comming in at the rate of fif*** >samplings per second, how do I compute the acceleration? >-- >Chuck Stuart - Mesquite TX USA Try this: accel (fps) = (S - S ) * 15 n n-1 Where S is your speed samples. You don't need the distance samples if your speed samples are accurate. Dave (davids) Sparks Late Night League http://www.sequoia-dev.com/Hawaii/latenite.html

The calculation for acceleration between two consecutive velocity points, v1 and v2, is as follows: a = (v2 - v1) / (1/15) You don't need the distance figures because acceleration is simply a derivative of velocity over time. Lawrence Lo > Given speed and distance figures comming in at the rate of fif*** > samplings per second, how do I compute the acceleration? > -- > Chuck Stuart - Mesquite TX USA

last I knew, aceleration (medium) was a= (v2-v1) / (t2-t1) . The aceleration (instantaneous) is a= lim (v2-v1) / (t2-t1) when (t2-t1) tends to 0. I don't know how the written simbol is but normally (v2-v1) is represented by /\t (with a base line) I don't know a lot of physics so I maybe wrong but I doubt it, since it's was it taught everywhere else in the world. ...:-=>> The freaking Mail Band <<=-:... http://www.geocities.com/MotorCity/3039

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>>Given speed and distance figures comming in at the rate of fif*** >>samplings per second, how do I compute the acceleration? >>-- >>Chuck Stuart - Mesquite TX USA >Try this: >accel (fps) = (S - S ) * 15 > n n-1 >Where S is your speed samples. You don't need the distance samples if your >speed samples are accurate.

Just in case anyone else was eavesdropping, this equation didn't come out very clear in ASCII, and there was an error in the units. Here's another attempt: accel = (S[n] - S[n-1]) * 15 ft / (sec**2) Where S is the vector of speed samples and n is the index. Dave (davids) Sparks Late Night League http://www.sequoia-dev.com/Hawaii/latenite.html

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>>The calculation for acceleration between two consecutive velocity points, >>v1 and v2, is as follows: >>a = (v2 - v1) / (1/15) >>You don't need the distance figures because acceleration is simply a >>derivative of velocity over time. >last I knew, aceleration (medium) was a= (v2-v1) / (t2-t1) . >The aceleration (instantaneous) is a= lim (v2-v1) / (t2-t1) when >(t2-t1) tends to 0. I don't know how the written simbol is but >normally (v2-v1) is represented by /\t (with a base line)

Yes, the symbol you are refering to is the greek letter delta, looking like a triangle. Everything posted to this point has been correct, just different interpretations of the same equation. 1/15 is the same as delta t in this case because the data is given 15 times/s, so each set of data is 1/15th of a second apart. Gregory Fung Vancouver, B.C., Canada IICC3 Rebel Alliance Lola/Ford/Firestone ITCC Player's Ltd. Audi A4

why do you multiply it by 15 ?? i don't know that formula. ...:-=>> The freaking Mail Band <<=-:... http://www.geocities.com/MotorCity/3039

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>>accel (fps) = (S - S ) * 15 >> n n-1 >>Where S is your speed samples. You don't need the distance samples if your >>speed samples are accurate. >why do you multiply it by 15 ?? i don't know that formula.

It's actually divided by one-fif***th, which is the same as multiplying by 15. you know a=v/t, there you go. Gregory Fung Vancouver, B.C., Canada IICC3 Rebel Alliance Lola/Ford/Firestone ITCC Player's Ltd. Audi A4