What Exactly happens After I change Gears in a Q?
What Exactly happens After I change Gears in a Q?
If I change from the 3.7? to 4.08? gears, what does this exactly do in terms of drivability. I am not very familar with how gearing works. Does this change each of the gear ratios? For instance, If I did the change in a 94, what would 1st/2nd/3rd/4th redline at? Does it reduce the gearing in each gear? Would this effect my top end performance? Anyone have this done on a 94, and would like to comment on it? Is it worth doing? I don't want to go level 10, I guess I'm wondering with an upgraded tcu, would it drive normally? I'm considering the ecu/tcu and gears. I think I can get all the parts, and labor for around 2,000-2,500. Not exactly a huge sum of money, not exactly cheap either. just from reading the past posts, it seem slike these three mods would make a big difference in my Q, and probably reduce my 1/4 time down to low to mid 14's, and high 90s for traps?
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Q45tech
- Moderator
- Posts: 14296
- Joined: Tue Apr 30, 2002 3:19 am
- Car: 1990 Q45 342,400 miles 22 years ownership with original engine
1995 G20t 5 speed 334,000 miles 16" 2002 wheels - 205/50/16 Sr20ve vvl
Since the effective delivered torque to the wheels is increase by 15% you realize the cube root or 2.5% faster acceleration while in each gear. Unfortunately instead of using 2 gears to go through the quarter mile with a 3.538 diff gear a 4.083 requires that you shift from 2nd to third at around 75-80 mph this extra shift requires a second so the 0.3-0.4 of a second you gained by gears up to 80 mph plus the 0.1 from 80 to 95 means you may actually be slower in quarter mile in a 1994 and above Q with the diff change.
In the case of the 90-93 the accelleration is faster due to the more aggressive 1st and 2nd gear........the JWT ecu allows to to rev to 7300 vs 6900 stock......... and reach 97-98 mph in 2nd as you hit the new revlimiter crosssing the quarter.
The problem is that the Q was designed as a highway cruiser producing maximum acceleration from 55-80 mph not a quarter mile machine nor a zero to 30-60 machine!
Trying to optimize the above will require much more than a diff change as the 4 speed transmission ratios are not ideal.
In the case of the 90-93 the accelleration is faster due to the more aggressive 1st and 2nd gear........the JWT ecu allows to to rev to 7300 vs 6900 stock......... and reach 97-98 mph in 2nd as you hit the new revlimiter crosssing the quarter.
The problem is that the Q was designed as a highway cruiser producing maximum acceleration from 55-80 mph not a quarter mile machine nor a zero to 30-60 machine!
Trying to optimize the above will require much more than a diff change as the 4 speed transmission ratios are not ideal.
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maxnix
- Posts: 22627
- Joined: Mon Jul 22, 2002 8:11 pm
- Car: 1995 Infiniti Q45
1995 Infiniti Q45t
2000 Infiniti Q45
Install a five speed
Quote »Trying to optimize (0-60 and 1/4 mile acceleration elapsed times) will require much more than a diff change as the 4 speed transmission ratios are not ideal.[/quote] So how do we get the 5 speed from the 2002Q45 into a generation 1? I assume it is in no way swap.
I went with shorter tires which gives much the same effect as a short rearend. My 94Q has a noticeable improvement in low end and mid range acceleration. Now jumping into a tiny opening in traffic is nothing more than a pedal stroke away. I like it.
The fuel mileage penalty was bigger than I expected. I went from 21 - 23 mpg to 16 – 17 mpg. I still like it.
The speedo runs about 5-6% optimistic. An indicated 80 mph is about 75 mph. I really, really do like it.
If you are looking at new tires soon, consider this approach. Much cheaper. Then, if you don't like it, you are in the wrong forum...wait, no. I mean, if you don't like it, just change back to standard tire sizes.
The fuel mileage penalty was bigger than I expected. I went from 21 - 23 mpg to 16 – 17 mpg. I still like it.
The speedo runs about 5-6% optimistic. An indicated 80 mph is about 75 mph. I really, really do like it.
If you are looking at new tires soon, consider this approach. Much cheaper. Then, if you don't like it, you are in the wrong forum...wait, no. I mean, if you don't like it, just change back to standard tire sizes.
I went to 245/45-16s.
If you think about it, my odometer should show more miles than actually traveled, so the milage is even worse.
My cruise RPM went from 2200 to 2500 RPM with an indicated 80 MPH. Typical New Jersey traffic. I must be out of the FI cruise range.
Don't get me wrong, I will pay for the performance boost. First gear is almost entertaining.
If you think about it, my odometer should show more miles than actually traveled, so the milage is even worse.
My cruise RPM went from 2200 to 2500 RPM with an indicated 80 MPH. Typical New Jersey traffic. I must be out of the FI cruise range.
Don't get me wrong, I will pay for the performance boost. First gear is almost entertaining.
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maxnix
- Posts: 22627
- Joined: Mon Jul 22, 2002 8:11 pm
- Car: 1995 Infiniti Q45
1995 Infiniti Q45t
2000 Infiniti Q45
I'm with Fred
Quote »If you think about it, my odometer should show more miles than actually traveled, so the milage is even worse.[/quote]
Since your indicated mileage will be greater than your actual mileage, the mpg will appear to increase since the numerator in the ratio is larger.
Similarly, AZhitman will be reducing his warranty by a like percentage because the indicated miles will be greater than the actual miles, unless the time limit occurs first.
Of course, in 16" diameter, usually 255/50 yields the correct 26" unloaded diameter. Different manufacturers on Tirerack will show their 245/50 the same or smaller diameter. So measure (twice) first before you buy.
Since your indicated mileage will be greater than your actual mileage, the mpg will appear to increase since the numerator in the ratio is larger.
Similarly, AZhitman will be reducing his warranty by a like percentage because the indicated miles will be greater than the actual miles, unless the time limit occurs first.
Of course, in 16" diameter, usually 255/50 yields the correct 26" unloaded diameter. Different manufacturers on Tirerack will show their 245/50 the same or smaller diameter. So measure (twice) first before you buy.
Originally posted by Q45tech "]Since the effective delivered torque to the wheels is increase by 15% you realize the cube root or 2.5% faster acceleration while in each gear.
I think that is not correct:
Torque is directly proportional to force (thrust) from the drive wheels. The relationship between force, mass, and acceleration is:
F = ma
So acceleration, a, is directly proportional to force, given constant mass. Therefore, the acceleration increases by 15%, all other things equal.
Of course, when aerodynamic drag is the principal retarding force, then power is related to speed by:
P = k* v^3
Where P is power, k is a constant, and v is the speed. So top speed increases as the cube root of power.
Incidentally, power is force times speed, and aerodynamic force is equal to
Cd*A*1/2 rho * v^2
Where Cd is drag coefficient, A is frontal area, rho is air density, and v is speed.
Please consider this correction. The writer Qtech has had very impressive and interesting posts.
I think that is not correct:
Torque is directly proportional to force (thrust) from the drive wheels. The relationship between force, mass, and acceleration is:
F = ma
So acceleration, a, is directly proportional to force, given constant mass. Therefore, the acceleration increases by 15%, all other things equal.
Of course, when aerodynamic drag is the principal retarding force, then power is related to speed by:
P = k* v^3
Where P is power, k is a constant, and v is the speed. So top speed increases as the cube root of power.
Incidentally, power is force times speed, and aerodynamic force is equal to
Cd*A*1/2 rho * v^2
Where Cd is drag coefficient, A is frontal area, rho is air density, and v is speed.
Please consider this correction. The writer Qtech has had very impressive and interesting posts.
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Q45tech
- Moderator
- Posts: 14296
- Joined: Tue Apr 30, 2002 3:19 am
- Car: 1990 Q45 342,400 miles 22 years ownership with original engine
1995 G20t 5 speed 334,000 miles 16" 2002 wheels - 205/50/16 Sr20ve vvl
The problem is friction and the tire /road interface.Just because you change the rear diff by 15.0% you don't get 15.0% more torque to the tire road interface.
Testing on Q proves the cube root formula for quarter mile time since above 60 mph aerodynamic forces come into play.
However on zero to 30 mph the stock tires are pretty traction limited [@ 30 mph= 4,000 rpm .....307 x 0.787= 241 lb/ft x 9.85= 2380/2 tires less LSD imbalance means ~~~~1190 x [24/25.5=0.94] or 1120 lb/ft at each tire road interface yet the weight on each tire is roughly 989 pounds = tires slipping at a maximum peak acceleration of 0.553 G in 1st gear at 30 mph.
Testing on Q proves the cube root formula for quarter mile time since above 60 mph aerodynamic forces come into play.
However on zero to 30 mph the stock tires are pretty traction limited [@ 30 mph= 4,000 rpm .....307 x 0.787= 241 lb/ft x 9.85= 2380/2 tires less LSD imbalance means ~~~~1190 x [24/25.5=0.94] or 1120 lb/ft at each tire road interface yet the weight on each tire is roughly 989 pounds = tires slipping at a maximum peak acceleration of 0.553 G in 1st gear at 30 mph.
AZhitman wrote:What kind of odometer reduction can I expect if I go to a 245/45/16 tire?
I'm on 245/50/16 now, so I think my odo/speedo is right about even with stock.
If it's significant, I may swap (just so my warranty lasts longer...):D
I'd calculate that your mileage would be 11.8% less based on overall tire diameter alone then add the extra fuel consumption due to higher engine rpms to stay at typical speeds. Say 15% less overall?
(You're heading into SUV territory, buddy .
)Testing on Q proves the cube root formula for quarter mile time since above 60 mph aerodynamic forces come into play.
Yes, well.. to the extent that the time is governed by aero drag,then the V^3 terms are at work. BUt for two identical cars at the same speed, instantaneous acceleration would be proportional to torque at the rear wheel (subject to traction issues discussed below).
However on zero to 30 mph the stock tires are pretty traction limited [@ 30 mph= 4,000 rpm .....307 x 0.787= 241 lb/ft x 9.85= 2380/2 tires less LSD imbalance means ~~~~1190 x [24/25.5=0.94] or 1120 lb/ft at each tire road interface yet the weight on each tire is roughly 989 pounds = tires slipping at a maximum peak acceleration of 0.553 G in 1st gear at 30 mph. [/QUOTE]
I didn't get the last of that.
Here you have derived the torque at each rear wheel ( engine torque times two reduction gears, divided by 2) as 1120 lb/ft. But without knowing the radius I wouldn't know the tangential force on the road. The max tangential force is mu (.8 for dry pavement?) times weight on the wheel. In a car balanced 25/25/25/25, driving 2 wheels, the max accel is of course .5 x mu, ignoring weight transfer.
Please see a new thread, wherein I ask some questions about valve timing. I appreciate your thoughtful responses.
Yes, well.. to the extent that the time is governed by aero drag,then the V^3 terms are at work. BUt for two identical cars at the same speed, instantaneous acceleration would be proportional to torque at the rear wheel (subject to traction issues discussed below).
However on zero to 30 mph the stock tires are pretty traction limited [@ 30 mph= 4,000 rpm .....307 x 0.787= 241 lb/ft x 9.85= 2380/2 tires less LSD imbalance means ~~~~1190 x [24/25.5=0.94] or 1120 lb/ft at each tire road interface yet the weight on each tire is roughly 989 pounds = tires slipping at a maximum peak acceleration of 0.553 G in 1st gear at 30 mph. [/QUOTE]
I didn't get the last of that.
Here you have derived the torque at each rear wheel ( engine torque times two reduction gears, divided by 2) as 1120 lb/ft. But without knowing the radius I wouldn't know the tangential force on the road. The max tangential force is mu (.8 for dry pavement?) times weight on the wheel. In a car balanced 25/25/25/25, driving 2 wheels, the max accel is of course .5 x mu, ignoring weight transfer.
Please see a new thread, wherein I ask some questions about valve timing. I appreciate your thoughtful responses.
I disagree. The mileage loss will always be less than the percentage reduction in tire circumference. To understand this, assume car A makes 2200 RPM at 60 actual (GPS verified) mph, using 10 inches of manifold pressure (20" vacuum) and a fuel flow of 3 gallons per hour. Obviously it gets 20 mpg. Consider that similar car B changes its tires and it reads 2400 RPM at 60 mph. This is an increase in RPM of 200/2200 = 9%. Now slow car B down to 2200 RPM. IT now is going 55 mph. What is the fuel flow? Obviously it CANNOT be larger than 3 gph, because we know the engine requires 3 gph to go 2200 rpm and 10 inches mp. We might not know the mp required for 55 but it will be less than that req'd for 60, given equal RPM. If it is 3 gph, the maximum possible, then the fuel penalty is 9%. If it is less than 3 gph, and it must be, because the engine is under less load than it was at 60/2200, then the fuel mileage penalty is less than 9%.
He's still sneaking into SUV territory when he increases the rolling resistance with those fatter tires. I believe he described his "looks like a steam roller from the back". And I assume - a fairly safe assumption - that because he normally cruises outside of the optimum mileage range of the engine it, will be that much more innefficient. Don't worry, though. AZ's cool with that.