Shift Anticipation Max Acceleration

This is set to 35 in my 2016 GT. As far as I can gather this is 35 Ft / Sec....

If I run the 1/4 mile in, let's say, 12 seconds, that is 1320 ft in 12 seconds, which is 110 ft per second, 99% of which is at WOT (except for the first fraction of a second at launch).

Why are these two numbers so far apart? Is anticipated shifting only for part throttle?

35 ft / sec is 23.86 MPH.....but that is steady state MPH and I guess we are dealing with Acceleration, or a rate of change in a change in distance.

a 60 ft time of 1.60 is typical for a lightly modified drag radial car, that is 37.5 ft per second so that is close....a 1.4 60 ft is pretty fast and that is 42.85.

but what about 60 - 330 ft? A 1.6 60 ft might give you a 4.6 330 ft which is 3 seconds over 270 ft which is 90 ft per second and so on.

60 Foot 1.6 ------ 37.5
330 Foot 4.62 ----- 90
660 Foot 7.26 ------ 125
1000 Foot 9.55 ------ 148
1/4 Mile ET 11.51 ----- 163


So how do you determine how many feet per second is correct for Anticipated Shifting?

35 ft per/second squared. so the fastest acceleration it can anticipate is 35 ft/second per second. So in three seconds it can anticipate 105ft/second(change in speed) stock. and in that same three seconds it can anticipate 315 ft(change in distance) in 3 seconds.

acceleration = distance divided by (seconds times seconds)

so 60 ft in 1.6 seconds = 60ft/(1.6s * 1.6s) = 60/2.56= 23.4ft/s*s
330 in 4.62 = 330ft/(4.62sx4.62s) = 15.5ft/s*s

we know the fastest time our cars accelerate is at the start so acceleration will decrease as we speed up.

35 ft/s/s is quite adequate and if you are having problems with your transmission shifting before your shift point due to the tires spinning or you're pulling faster than 1 second 35ft times revising this number can help. If you go to 2 seconds 60ft times its 60 divided by 4 and you only need 20 ft/s*s.

Hope this clears it up.

60ft/x squared= 35ft/s*s

60ft=35(ft/s*s) * (x*x)

60ft/ 35ft/s*s=x*x

squareroot(60ft/35ft/s*s)= x

x=1.3

so if you are doing faster than 1.3 60ft you need to raise the anticipation value.

Just remember ecu assumes no tire spin. Thats why you will see your transmission shift at something like 5k while doing a burn out.

Quote:

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Originally Posted by murfie

35 ft per/second squared. so the fastest acceleration it can anticipate is 35 ft/second per second. So in three seconds it can anticipate 105ft/second(change in speed) stock. and in that same three seconds it can anticipate 315 ft(change in distance) in 3 seconds.

acceleration = distance divided by (seconds times seconds)

so 60 ft in 1.6 seconds = 60ft/(1.6s * 1.6s) = 60/2.56= 23.4ft/s*s
330 in 4.62 = 330ft/(4.62sx4.62s) = 15.5ft/s*s

we know the fastest time our cars accelerate is at the start so acceleration will decrease as we speed up.

35 ft/s/s is quite adequate and if you are having problems with your transmission shifting before your shift point due to the tires spinning or you're pulling faster than 1 second 35ft times revising this number can help. If you go to 2 seconds 60ft times its 60 divided by 4 and you only need 20 ft/s*s.

Hope this clears it up.

60ft/x squared= 35ft/s*s

60ft=35(ft/s*s) * (x*x)

60ft/ 35ft/s*s=x*x

squareroot(60ft/35ft/s*s)= x

x=1.3

so if you are doing faster than 1.3 60ft you need to raise the anticipation value.

Just remember ecu assumes no tire spin. Thats why you will see your transmission shift at something like 5k while doing a burn out.

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yep, didn't know it was squared, thanks for the info!

anything to gain by lowering it to suit your needs? 1.6 60 ft and change the number to 24 ft^2 will it shift better at part throttle or any other gain/change?

I wouldnt lower it. You want to leave room for tire spin. You also have to remember these are average acceleration values. If you want to get the exact value you would need to use a calculus equation. Derivatives and limits to be exact. I can go through this if you want me to, but most people's eyes glaze over.

Quote:

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Originally Posted by murfie

I wouldnt lower it. You want to leave room for tire spin. You also have to remember these are average acceleration values. If you want to get the exact value you would need to use a calculus equation. Derivatives and limits to be exact. I can go through this if you want me to, but most people's eyes glaze over.

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yeah I leave the calculus to the calculator....no worries. ;-)

After learning calculus I don't think of it as harder than algebra but rather the short cuts to algebra.

I took business calculus in college and learned what I needed to and went straight back to addition and subtraction for daily use. ;-)

It could be worst. Common core.