So, basicly if I just want to have better acceleration...... instead of going through the trouble of changing out final gear, just go buy some really wide 15" steel rims with low profile drag tires and spacer to clear the Brembo?

(Oh yeah, just change out the rear tires and leave the front alone. The new 'stagger' will give the car more rear weight bias, thus giving it more grip in the rear)

 

gdup

You are absolutely right. I misunderstood you. I thought that by 'lower gear ratios' you meant lower number, because "higher overall numberical" value is equal to a shorter gear ratio not 'lower'.

G

 

meta, it is definitely a little too conservative. I agree with q45 tech

Also, your trap speed would not go down, it will improve, just like your time. How can you time improve by you going slower? What will be hurt is your ABSOLUTE top speed that can be achived in that car, which is not much of a concern with many, as going from a ~170 possible top mph to ~15x is not that big of a deal.

 

Q, gurgen,

The rule may seem conservative but I think it's correct.
Still, I'll see if I can dig up a reference.

Gurgen,
You're incorrect.
I know it may seem counter-intuitive, but as I said originally, your trap speed will go down even though your overall time is faster. Your ET will improve due to improved acceleration, despite the lower trap speed. In other words, you will accelerate faster to the lower trap speed, and that faster acceleration will allow you to cover the quarter mile distance in less time.

In the G, for example, our final drive is 3.35 (to be exact it's 3.358) which means the driveshaft will turn 3.358 times each time the rear wheel rotates once. Since the engine is connected to the driveshaft, a higher final drive means the engine is turning faster (revving at higher rpm) per single revolution of the rear wheel -- there will be a faster increase in the number of revolutions per given amount of time (greater acceleration) BUT less overall number revolutions per given amount of time (lower speed).

The end result is that at the car's maximum performance limit, the fastest speed achieved will be lower, but the acceleration TO that speed will be higher. The higher acceleration is what lowers the ET.

Regarding absolute top speed, like I said before it will also decrease (...but unless you lift the speed governor that absolute top speed is limited anyway).

 

We have dealt with half a dozen customers changing the 3.538 [stock old Q diff ratio] to 4.083. Plus [4] - Lexus SC400 V8 AT substituting the SC300 manual diff ratio.
Both increased rpm by ~~15%.

Study the effects carefully before changing ratios especially with automatics.

 

Quote: ***meta, it is definitely a little too conservative. I agree with q45 tech ***

Are you sure about using the word "definitely" here? ;-)
I checked a few references which all agree exactly with what I've previously stated.

As we speak, my car is being worked on by some very competent race-engine builders (I'm getting several things done, including an Exedy carbon racing clutch and Exedy flywheel, among other things -- pics will be posted soon). Anyway, I've had some long discussions with these guys regarding various issues -- final drive ratios being one of them.

I won't post them all, but here's one of several references that say the exact same thing I've been saying.

<A HREF="http://www.intense-racing.com/Merchant2/merchant.mvc?Screen=PROD&Product_Code=GCK-3xx&Category_Code=T">How increasing the final drive affects quarter mile results</A>

 

Meta

It is hard to follow your logic... You say: <blockquote><font class="small">In reply to:</font><hr>

you will accelerate faster to the lower trap speed, and that faster acceleration will allow you to cover the quarter mile distance in less time

<hr></blockquote> . First of all, "accelerate faster" means, by definition of the word acceleration, that your speed will increase at a higher rate. How else can you get from point A to point B quicker? To cover a set distance (1/4mile) in a shorter time, your average speed must be higher - this is fundamental kinetics. Saying that the time will decrease while the top speed will decrease (worsen) goes very much against basic physics laws. What you say will be possible only if this mod can somehow decrease/increase the magnitude of acceleration along the 1/4 mile run; e.g. increase acceleration on the low end to get a "jump start" and then decrease the magnitude of acceleration so as to cover the 1/4 mile mark in a quicker ET but lower speed. Obviously, that's ludicrous for this mods, as it just replaces one constant ratio for another constant ratio.

This, <blockquote><font class="small">In reply to:</font><hr>

Since the engine is connected to the driveshaft, a higher final drive means the engine is turning faster (revving at higher rpm) per single revolution of the rear wheel -- there will be a faster increase in the number of revolutions per given amount of time (greater acceleration) BUT less overall number revolutions per given amount of time (lower speed).

<hr></blockquote> makes no sense to me at all. It is possible that I am wrong here, but I just don't think so. "Less overall number of revolutions per given amount of time (lower speed)" does not make sense. Yes, in a given amount of time, the number of revolutions will be less with a shorter gear ratio such as this one, but you are forgetting that it is now EASIER for the engine to turn faster because there is "less work" to do per engine rotation than there was before.

G

 

You can have a quicker time with a lower trap speed.
You accelerate faster earlier and not as fast down the track.

Current NHRA records:

1/4 mile time: 4.441 sec 10/04/03 Anthony Schumacher - Long Grove, IL '03 McKinney R.E.D Reading, PA

Top speed: 333.41mph 05/22/04 Brandon Bernstein - Lake Forest, CA

The difference is "average speed": you're right, the average speed will be higher. But the faster car was slower at the front of the track and higher at the end.

2004 G35C 6MT Black. Killer.

 

yes

and what i am saying is that this mod, a final drive ratio, CANNOT cause such a behavior. It replaces one constant ratio for anothe rrconstant raio,period.

 

Gurgen,
Just because you can't follow my logic doesn't mean my logic is wrong ;-)
But I'll admit I did not explain it as clearly as I could have.

Okay, here goes:
Say you are racing someone to a marker 440 yards away. You have the higher final drive, but the two cars are otherwise exactly the same.

You both launch simultaneously but you pull ahead much faster, by the end of second gear you are a car length ahead. You are first to pass the marker, at which point 14 seconds have passed, and your top speed at the marker point was 100 mph.
The other car takes 14.2 seconds to make the marker, but his top speed was 102 mph. He reached a higher top speed at the marker but it took him longer to accelerate TO that top speed.

Like I said, your car with the steeper final drive will have a lower ET, despite the fact that you had a lower trap speed. The greater acceleration took you to that lower trap speed in a shorter amount of time.

Now in response to what else you said, IT HAS NOTHING TO DO WITH AVERAGE SPEED. Trap speed and average speed are two totally different measurements. Maybe that's where you are getting confused?

You said, *** "Less overall number of revolutions per given amount of time (lower speed)" does not make sense. ***

Oh yes it does. Speed is distance over time. Distance can be thought of in terms of the number of revolutions your wheel makes... an 18 inch Pilot Sport covers about 83 inches in one revolution, let's say 7 feet. So in 1320 feet (440 yards or one quarter mile) your wheel makes about 189 revolutions. If you need 14 seconds to transverse that distance, or 16 seconds to transverse that distance, your wheel still makes 189 revolutions to cover the quarter mile.

With me so far?

To determine the ET (elapsed time), you could ask: How much time is required to make 189 revolutions of my wheel?

To determine trap speed, you could ask: What was the maximum number of revolutions my wheel made (what distance did it cover) in the interval of time, delta t?
Note: you want v (max) not v (average)... i.e. maximum speed, NOT AVERAGE SPEED!!

To determine acceleration, you could ask: What was the rate of increase of the number of revolutions per given interval of time (i.e. what was the rate of increase of "speed"?)

Get it now?

 

DED,
Exactly.
You said it much more succinctly than I did.

 

OK, Now I see what you are saying... thanks... I'll admit my thinking was a little off given that this concept was a little intricate and counterintuitive. Thanks for the explanation.

Gurgen

 

Think about how a dyno measures torque, it divides the drum acceleration/speed [time into 1/1000 of seconds].

It takes x amount of torque to change the drum speed 1.0 mph in y fractions of a second.........differential calculus was created to solve such problems.

Unfortunately the mass of the drum is not the same as the mass of the car so you must use a CORRECTION FACTOR.

Measure the drums acceleration in seconds and you will see it accelerates faster than the car ever can. [sources of errors]

Anyway the point was that if you know the torque curve and the gear multiplication and the correction factor for loaded/effective tires not being 12" in radius [the loaded radius]. Plus the accurate weight plus the weight shift. YOU can accurately approximate the real acceleration from any speed to any speed.

Easy to estimate the effects of a diff gear change with EXCEL.
Measuring the changes if any on a dyno will just confuse the issue.......again because of lighter drum interia.

http://www.hondata.com/gsensor.html
http://www.tomstickland.co.uk/astra/...del/model1.htm
http://phors.locost7.info/phors03.htm
http://phors.locost7.info/contents.htm

 

I think that the thing with trap speeds people dont understand is that acceleration is not linear. I'm not sure what a graph of the acceleration would look like EXACTLY, but im sure that its a curved graph with a really agressive slope that eventually flattens out at the top... i included a VERY crude drawing of what im talking about (just for illistrative purposes, not for anything else like accuracy) I just kinda did it to show graphically what he's talking about about having similar trap speeds at the same time but acceleration be different. This will (not shown by CRUDE graph) mean that the distance traveled in that amount of time will be different for each because they were not traveling at the same speeds at the same times. a very aggresive curve up to 100mph and slower acceleration between say... 80-100mph will yeild a quicker time than someone who has a little slower acceleration 0-80 but quicker acceleration 80-100. I'm really bad at explaining math..... its just somehting i understand. If u didnt follow... ur probably not alone. This is just how i understand it (took AP algebra 2 this year and did a lot of stuff like this involving speeds distance and time when accleration (or a rate) is going through exponential growth rather than linear. I realize also that acceleration is not a perfect exponential growth, but i also would be willing to bet that if there was an x,y table then a calculator could come up wiht a pretty accurate curve of best fit that would be close enough to show approximately what speeds would be at different times. It could also be done for distance/time or speed/distance.

2004 Black on Black g coupe 6spd.... aluminum pedals, clear corners, z tube, and pop charger. soon to have 19" volk gt-c's and PIAA xtreme whites in the clearance and corners
Houston Texas

 

If maximumum acceleration occurs at torque peak rpm, consider that the acceleration at the HP peak rpm will be 15-20% less [few lb/ft] and at redline may be 25% less.

If the next gear multiplication factor is less than the previous gear times the torque you are better off staying in that previous gear till redline.

The span between gears vs the torque fall off curve.