[mistermojorizin]

Originally Posted by hotrod182

My trap speed went up with DRs. when the 60' time went down.

On DRs

12.133 1.729 113.49mph
12.261 1.80 112.40mph

On Stock RFT

12.633 2.00 111.22 mph

right, that's why i said "ceterus parabus" - holding everything else equal. with drag dites, you can get better 60 foot and put all the power down. but if the same unchanged car runs a quicker 60 foot, it's not putting down the same amount of power, it reaches that 60 foot quicker but at a slower speed, and it reaches the gates quicker but with less time to accelerate.

[Mr. 5]

I'm actually a statistical analyst for the Department of Defense (have a degree in Pure mathematics) and when I first saw this with my runs a long time ago, I was a little stumped.

I've tried to come up with some proofs with why but I always seem to prove the opopsite.

[trueblue]

Physics 101:
Vx=Vxo+aT
For your 60' Vxo = 0 (standing start), acceleration is constant (due to same car, same power unless wheel spin is experienced), Vx at 60' is lower because the time has been reduced (faster 60' time).
Now re-apply the same equation but plug in Vx at 60' for Vxo for your starting velocity you will find that a being constant again with lower Vxo and T (1/4 mile ET) it would yield a lower Vx which is your 1/4 mile mph.
There will be a quiz on monday. :-)

[e90AW335i]

Originally Posted by trueblue

Physics 101:
Vx=Vxo+aT
For your 60' Vxo = 0 (standing start), acceleration is constant (due to same car, same power unless wheel spin is experienced), Vx at 60' is lower because the time has been reduced (faster 60' time).
Now re-apply the same equation but plug in Vx at 60' for Vxo for your starting velocity you will find that a being constant again with lower Vxo and T (1/4 mile ET) it would yield a lower Vx which is your 1/4 mile mph.
There will be a quiz on monday. :-)

Originally Posted by MarioKart

I love going to the track but I'm horrible in math, can someone explain how a faster 60' time results in a slower MPH at the end of the 1/4 mile?
Here are my results at my last track event
1st run 2.02 60' @12.52 @ 113mph
2nd run 1.7 60' @ 12.55 @ 111mph

How does this work?
TIA
Mario

60' is lower but ET is higher. Please explain your equation again....

[MarioKart]

Originally Posted by trueblue

Physics 101:
Vx=Vxo+aT
For your 60' Vxo = 0 (standing start), acceleration is constant (due to same car, same power unless wheel spin is experienced), Vx at 60' is lower because the time has been reduced (faster 60' time).
Now re-apply the same equation but plug in Vx at 60' for Vxo for your starting velocity you will find that a being constant again with lower Vxo and T (1/4 mile ET) it would yield a lower Vx which is your 1/4 mile mph.
There will be a quiz on monday. :-)

Yes, but what color are my eyes? I'm impressed!

[E82tt6]

Generally speaking, it's because you spend less time accelerating. On a car running low 12s at 115, .2 seconds are equal to almost 2mph of accleration.

[trueblue]

Originally Posted by e90AW335i

60' is lower but ET is higher. Please explain your equation again....

You happened to have highlighted the answer to your own question. Acceleration in this case is no longer constant, it can be and very likely is caused by human variations.

[trueblue]

Originally Posted by E82tt6

Generally speaking, it's because you spend less time accelerating. On a car running low 12s at 115, .2 seconds are equal to almost 2mph of accleration.

This is probably the best way to explain the equation verbally without the math.

[lauer87]

Its very hard to explain why better 60' times can lead to lower trap speeds, and can be hard to comprehend at first. There are some good explanations posted up already, just take some time and think it out for yourself.

[Driver72]

I've (along with Hotrod) and others have drag raced many many times.
I will tell you this.
A lower 60' time does not = slower trap speeds.

Slower ET often results in Faster trap speeds and vice versa.
But that's not always the case.

Some may then say, well if a faster ET results in a slower trap speed and a faster 60' time means a faster ET therefore that means a faster 60' time means a slower trap speed.

Well, saying that means you are DISCOUNTING everything that happens between the start of the run (first 60 feet) and the end.

I have also gotten the highest trap speeds with lower 60' times.
But I've also gotten worse.

There are many variables in between runs and what happens between the 60' time and the 1/4 mile time.
As someone said, you HAVE to look at the times and speeds at the marks in between to see where you slowed down or sped up.
A shift 100 rpms different in each gear could make the difference.
The temp of your engine could make the difference.
A gust of head wind in the middle of the run could make the difference.
Not keeping as straight of a wheel going down the track can make the difference. (this is one point many often overlook. It's a 1/4 mile IF you go perfectly straight down the run. If you need to make a correction or go "out of the groove" you are actually traveling more than a 1/4 mile in that particular run. It may only be a couple feet further, but that can make the difference.

As Mr. 5 pointed out, it's kinda impossible to know what EXACTLY will give you the best ET AND the fastest trap speed on the same run.
But it's safe to say you need to do everything right AND your CAR needs to be running it's best to get the best of both.

[Mr. 5]

Originally Posted by trueblue

Physics 101:
Vx=Vxo+aT
For your 60' Vxo = 0 (standing start), acceleration is constant (due to same car, same power unless wheel spin is experienced), Vx at 60' is lower because the time has been reduced (faster 60' time).
Now re-apply the same equation but plug in Vx at 60' for Vxo for your starting velocity you will find that a being constant again with lower Vxo and T (1/4 mile ET) it would yield a lower Vx which is your 1/4 mile mph.
There will be a quiz on monday. :-)

This actually proves the opopsite if you are saying that acceleration is constant, but acceleration is not constant. It doesn't matter if you are using the same car with the same driver.
It will never be constant unless you set it up roboticly.
Just look at the 60 foots.

We have:
D=VT
A = (V1-V0)/(T1-T0)

Car A gets to 60 foot at 2.0
Car B gets to 60 foot at 1.8

Car A
So 60 feet = 2.0V
T = 30 ft/sec = 44 mph

Car B

So 60 feet = 1.8V
T = 33.33 ft/sec = 48.8 mph

By your thinking, Car A would have to have a faster average velocity between 60ft and 1320ft to keep up. But this would never happen since acceleration is constant.

[mistermojorizin]

Originally Posted by trueblue

Physics 101:
Vx=Vxo+aT
For your 60' Vxo = 0 (standing start), acceleration is constant (due to same car, same power unless wheel spin is experienced), Vx at 60' is lower because the time has been reduced (faster 60' time).
Now re-apply the same equation but plug in Vx at 60' for Vxo for your starting velocity you will find that a being constant again with lower Vxo and T (1/4 mile ET) it would yield a lower Vx which is your 1/4 mile mph.
There will be a quiz on monday. :-)

this is basically what i said but in laymans terms. you cover the same ground in less time, giving you less time to accelerate, yielding a lower speed, assuming the rate of acceleration for the car has not changed (e.g, you're not using different tires, or missing shifts, or the car has gotten hotter, etc. etc.)

Originally Posted by e90AW335i

60' is lower but ET is higher. Please explain your equation again....

because the assumption didn't hold true. either the driver or the car did something to accelerate slower than on the first run. maybe it was hotter. this more than offset the lower 60' advantage.

[E92Pilot]

Originally Posted by MarioKart

I love going to the track but I'm horrible in math, can someone explain how a faster 60' time results in a slower MPH at the end of the 1/4 mile?
Here are my results at my last track event
1st run 2.02 60' @12.52 @ 113mph
2nd run 1.7 60' @ 12.55 @ 111mph

How does this work?
TIA
Mario

If you picked up three tenths in the 60' on the second pass you should have ran .3-.4 quicker in the ET. You must have hit the limiter or shifted early or possibly lifted somwhere in that run. All things equal you should have run at least 12.1 or 12.2.

[Mr. 5]

There are so many variables when looking at this that it's hard to determine an exact equation because I've really tried to prove this one way or the other.
When you think of it simplistically, it really doesn't make sense, but in my experience, I've usually determined that the faster (lower) ET, the slower the trap.
Because simplistically, the quicker, I get from Point A to point B, the faster I need to go correct? You would think.

[E92Pilot]

Originally Posted by Mr. 5

There are so many variables when looking at this that it's hard to determine an exact equation because I've really tried to prove this one way or the other.
When you think of it simplistically, it really doesn't make sense, but in my experience, I've usually determined that the faster (lower) ET, the slower the trap.
Because simplistically, the quicker, I get from Point A to point B, the faster I need to go correct? You would think.

Yes a car will always mph more when it spins in the 60'. But the two runs were not equal. I need to see the 1/8 mph and et from both runs to figure out what happened.

[AndreyATC]

Another weird ET vs MPH
This guy that we discussed in earler thread (few days ago) ran couple of times
For some reason he also got better MPH on worse 60'

http://www.dragtimes.com/BMW-335i-Timeslip-16534.html
http://www.dragtimes.com/BMW-335i-Timeslip-16337.html

I have thought of it (not mathematical) and came up with my own explanation about this
If you spin more on take off you probably in higher RPMs while still closer to start line
This is possibly gives you more power earlier in the run while increasing your run time?
I totally pulled this out of my ass, but does this make sense?

[MarioKart]

Originally Posted by E92Pilot

If you picked up three tenths in the 60' on the second pass you should have ran .3-.4 quicker in the ET. You must have hit the limiter or shifted early or possibly lifted somwhere in that run. All things equal you should have run at least 12.1 or 12.2.

I agree, when I pulled around to get my slip the first thing I noticed was my improved 60' time. Instantly I thought I must have gotten a 12.2 because the weather was a little colder than my first run and I have an automatic. I was a little shocked to see a slower MPH and longer ET (only by 3hunderth) Not sure why I didn't get a better time with a better 60'.

[trueblue]

There are way too many variables happening between you foot and the wheels to account for. To look at it in the most simplistic form, F=MA: In this case, none of the basic elements are constant. Even the mass in this case (change in quantity of fuel) is variable. The force exerted on the car changes due to frictional components of the tires are different each time due to things like temperature, pressure, track surface condition etc. All of these things affects how the car accelerate, and to top it off there is your reaction time added to the mess.

[Mr. 5]

Originally Posted by MarioKart

I agree, when I pulled around to get my slip the first thing I noticed was my improved 60' time. Instantly I thought I must have gotten a 12.2 because the weather was a little colder than my first run and I have an automatic. I was a little shocked to see a slower MPH and longer ET (only by 3hunderth) Not sure why I didn't get a better time with a better 60'.

Your time should have improved.
Something had to have happened differently when you had the better 60'.

What I thought you were initialy asking was why does it seem like if you have a better 60' the trap speed is most of the time less than a higher 60'?

If you did everything the same, meaning shifted the same place, and didn't hit the rev limiter, then your engine could have been pulling timing during the second run.

[MarioKart]

Originally Posted by Mr. 5

Your time should have improved.
Something had to have happened differently when you had the better 60'.

What I thought you were initialy asking was why does it seem like if you have a better 60' the trap speed is most of the time less than a higher 60'?

If you did everything the same, meaning shifted the same place, and didn't hit the rev limiter, then your engine could have been pulling timing during the second run.

Everything felt the same, the car had cooled down for an hour, the outside temp dropped 5 degrees and I had less fuel, but most of all I have an automatic that seems to shift indentical every time. I agree something did happen but I thought it would have happened for the better.

[Mr. 5]

This is what I'm talking about.
It has happened to me time and time again.
Based on my experience, the quicker the 60' the lower the trap.
People say that it's because you have less time to accelerate, but quicker acceleration is what got you down the line quicker so then why is the trap speed less?
hmmmmm.

Attached are examples of quicker 60' times with quicker 1/4 mile runs with lower trap speeds.

[sleepdoc]

Originally Posted by Mr. 5

This is what I'm talking about.
It has happened to me time and time again.
Based on my experience, the quicker the 60' the lower the trap.
People say that it's because you have less time to accelerate, but quicker acceleration is what got you down the line quicker so then why is the trap speed less?
hmmmmm.

Attached are examples of quicker 60' times with quicker 1/4 mile runs with lower trap speeds.

I too see this general trend in my timeslips. I do believe it is a real phenomenon.