Errr, so (Quote): ...'what sort of cost would be involved in raising ratio on 5th or 6th or overall gearing?!'

Thread author

If you can get the parts second hand then labour in the box build will be the main thing. But you would need either ratios which are higher and a matched pair to replace the current gears or a final drive and crown wheel with a lower numerical ratio both of which I'm not sure you'll get, But try finding out the ratios & FD for the alfa 3.2gta, 2.4jtd boxes as a start.

Oh about 8 hours labour to build.

i am using the final drive from the 3.0gtv and that is pretty much a straight swap, that will lower all the gear ratios and as sedicirich says the parts for that second hand aren't that expensive, labour will be the big cost as you will have the labour for the build of the box, the labour to take the box in and out and you might as well put a new clutch in while you are at it as well so unless you are very handy with a spanner you will be looking at least 12 to 14 hours labour charges

to shorten the intermediate ratios and have a higher 6th gear which i chose to do is a different ball game as the only way i could find to do that was to use the whole main shaft from a lancia thesis 20vt and getting one of those second hand proved to be impossible to I bought a new one along with the final drive from the 3.0gtv and a host of new parts and the cost for all that is not something i am going to share on an open forum, if you want further details about that then i would be happy to discuss via pm with you

Right, I've done some digging and found the formula explained (very nicely indeed ) on this website
Horsepower and Torque

Please have a read through and, in particular, look for the power method and the formula I have used in my calculator.



Force= 375*(HP/Speed)

I have the powerband of the car at the wheels, the weight of the car, the gearing of the car and the aerodynamic drag of the car. Throw in some formulas to show clutch slip off the line and maximum grip to represent tyre adhesion and you can see why my spreadsheet works.

Oi Trappy that's the link I posted earlier on in this discussion!

Works for me - must be your spreadsheet. See my next post:

+ speedo recalibration if the feed is from the box

Damnit, the work filter blocked it at the time! I wish I'd seen it then now...

RPM = 5750
BHP = 216

Speed 1st gear, 3.11 final drive: 40.52 mph (my tyre spec)

Force = 375 * (216/40.52) = 1998.773

Acceleration = 6.42 m/s^2

Speed 1st gear, 4.0 final drive: 31.50 mph

Force = 375 * (216/31.5) = 2570.769

Acceleration = 8.25 m/s^2


QED

Are you answering a question Griffster previously asked or has a point been made there?

Read my post above - #1316861

Your spreadsheet is obviously wrong. You can see from my latest post that, using your formula, acceleration is greater with the 4.00:1 final drive than the 3.11:1 final drive.

The same would be true if:

a) the tyres were smaller in diameter
b) the gear ratios were changed rather than the final drive

If you can't see that, then I don't know what more to say.

You're right acceleration is greater in your example... because you used a lower speed!

To make any comparison fair, you NEED to use the same speed. To work out which car is faster, you must find out what the power output is at the wheels at that speed. The fun bit is using the gear ratios, tyre size etc, to work out what power the car is making at any given point.

For the sake of argument, I'll stick with the 3.11 and 4.00 final drives. Let's not use first gear though, as the FIAT Coupé 20vT can only put down a force of 1611 on road tyres which equates to 5.13m/s^2 at these speeds...

Let's do a few 'rolling starts' and what out what would happen if both cars were in the optimal gear and booted it from a few different speeds.

Here is the approximate power at the wheels data for a standard 20vT;
2000: 34.2
3000: 98.4
4000: 157.0
5000: 175.1
6000: 176.8
6800: 148.9

So, from a;
50mph roll
3.11
2nd gear @ 4,856.4rpm
166bhp@ wheels- wind resistance
Force=1241
M/S^2= 3.95

4.00
2nd gear @ 6246.2rpm
165bhp@wheels- wind resitance
Force= 1221
M/S^2= 3.89

60mph roll
3.11
2nd gear @ 5827.7rpm
169bhp@wheels- wind resistance
Force= 1055
M/S^2= 3.36

4.00
3rd gear @ 5097.6rpm
165bhp@wheels- wind resistance
Force= 1033
M/S^2= 3.29

80mph roll
3.11
3rd gear @ 5284.5rpm
154bhp@wheels- wind resistance
Force= 719
M/S^2= 2.29

4.00
4th gear @ 5169.1rpm
152bhp@wheels- wind resistance
Force= 710
M/S^2= 2.26

100mph roll
3.11
4th gear @ 2023.7rpm
126bhp@wheels- wind resistance
Force= 470
M/S^2= 1.5

4.00
5th gear @ 5108.8rpm
127bhp@wheels- wind resistance
Force= 475
M/S^2= 1.51

The simple fact is this. The current power developed determines the current rate of acceleration. Gearing can be used to keep an engine in the power band more of the time. It can't be used to increase power; so it makes no difference to acceleration. It makes it easier for the engine to turn, but the work done always remains the same.

As an aside, you'll note that the gear each car is in in my examples above will sometimes be one higher than the speed the car can reach in a lower gear. That's the short-shift function kicking in at the optimal change-up point

You can lead a horse to water......

true


true, kinda, but a bit of a moot point. cam profiles usually establish the power band and gearing is matched to the engine spec, not the other way round.



true in its strict sense, ie the engine will deliver certain power and this wont change
BUT
it is not true in relation to the torque applied at the tyre or box


wrong, it has a significant effect on acceleration - and top end



????

I dont think you get the maths bud, relying on a spreadsheet incorporating all the mitigating factors such as shift time, mechanical grip, wind resistance only confuses the matter of the extent to which gearing & 'torque' affects acceleration

Trappy

From a quick look, the constant in that formula makes the formula work for imperial units yet you give acceleration in metric?

Without checking your maths I believe in the science but not your analysis. Most people understand torque but they don’t really understand power. Since you are taking a power approach, this is quite interesting. You have really had me thinking over the last day.

In your examples, you match the RPM by selecting a different gear, this is a scientific error, as it result in an unknown ratio, maybe even a similar overall ratio. It is at least going to be different by more than the FD. If the ratio is similar, then sure, it will accelerate the same. However if you use the same gear, RPM and therefore power will be different, which is the point. Also, in your example it does not take into account the change in acceleration throughout the gear. If you have truly shorter gearing it will rev out faster so you can achieve more power sooner.

Shorter gears will allow you to operate the engine around its peak power output more of the time. You may drop 2500rpm per shift rather than 4000rpm. Hence if you plot power output as a function of time or distance, it's going to be higher integral (more area under the curve) for a shorter ratio box than a longer ratio box. I.e. you are extracting more power from the engine over the same distance.

in very simple terms, take a modern diesel car. 140mph+ top speeds are the norm, but they still take ages to get there because they are so high geared

Partly...

Modern diesels are high geared because their range of useful revs is so narrow, but the torque is high. This is why there are many instances of six-speed diesels and (7+ speed when an auto-box is used)

why you led this disscussion - the cheapiest way how to check if it is faster or not is to fit wheels with various circumferences (as I did it and found that shorter final gera means faster almost by 0,4s..)so set of some standart 205/50/16, some set of 17/18or even 19" wheels will be necessary (also 15" wheels fit, but very limited amount of rim type -ATS DTC for example..) a do some ride with some Gtec, racetech and simmilar clever boxes..

and as was already written - the final ratio cost few $.. I get my final gear from AR 1471.6 88kW for only 40euros... the swap is about a working day...so te cost depends on hour taxes of garage... I have my own record in swapping final gears and some sincro rings on my mates skoda felicia kit car in 3ours...

375*(bhp/speed)= force-lbs

9.80665*(Force-lbs/Weight-lbs) gives me the m/s^2

In order to understand how this works, you need to understand and accept the following statements;

-If the vehicle's speed and the power of its engine are known at a given instant, the force of acceleration can be calculated without knowing anything about the drivetrain gearing, tire diameter, or even the engine torque.

-Power is what gives acceleration, not torque.

-The only reaon cars have gears, is because the power is very low at low rpms. If a car had a completely flat power curve (note not torque curve) then it would accelerate at the same rate in ay gear according to Force= 375*(HP/Speed) (let's discount aerodynamic drag for now). This is fundamental. There, with a flat power curve from 0rpms, a car with a single gear capable of taking it to 150mph would accelerate at the same rate as another car with 5 gears to do the job if the powercurve was flat.



I can't understand your point here. In my examples I'm using a fixed speed, which I believe is the whole point of this discussion. We want to establish if a car is faster with a closer ratio'd gearbox at the same speed. With a different FD, both 'cars' will have a different power output at the same speed because they will be using a different part of the powerband. The gear is irrelevant.



The change in the rate of acceleration throughout the gear is determined by bhp, nothing else. This can easily be determind by using a simple vlookup function of rpm vs bhp.



True, but if you modify only the FD, then you will have EXACTLY the same power plot through each gear! That's why it makes no difference. In order to improve the acceleration using gearing, you would need to customize every ratio independently and even then, you must remember that ONLY the current bhp determines how fast it accelerates.

No no and no agian that exactly what we are NOT talking about.
You have fundemental gap in your knowledge between instantaneous acceleration and what the rest of us are talking about is average acceleration a=(v1-V0)/t.

To maintain a body at speed need an instantaneous acceleration as you say it doesn't mater what gear ratio you have at what speed as long as force can overcome the forces of resistance then the speed will be maintained, BUT that is NOT the main theme of recent discussion. The rest of us are talking about the time it takes to change the velocity of the body from one speed to another. The stuff you have typed at fixed speeds is true but that is not the acceleration that will be changed by FD ratio change while the time it takes to change one speed to aother will and thats what the rest of the world on here are talking about. Even the article you linked to explains this when looking at gearing, in fact I'm not sure you understood that article in its entirity.



oh and a vlookup means nothing to anyone without the value of the cell you are looking up and the table you are looking up against, obviously not some sort of empirical maths command.

I can only hope you start to see the differences in the 2 scenarios, otherwise I'm afraid you will not see what the rest of us are talking about. 2 identical cars one with a 3.5:1 will be out accelerated by the other with a 4:1 ratio where the test conditions are say idle velocity in second gear on the 3.5:1 to max rpm in second gear with max velocity being set by the red line speed of the 4:1 FD car. Here we set the v0 and v1 speeds to be the same, but t will be reduced increasing a.

What I find most astounding is that, although I'm the only person in this discussion that actually has a tool that accurately estimates the performance of a car with just a few attributes, you're all keen to dismiss anything I say.

The very fact that I do have this calculator proves that I know what needs to be known to make such a thing! And of course I have those formulas, they're what make it work.



I'm using a fixed speed because if we want to determine if one car is faster than another after the fitment of a shorter FD, then you would want to use the same speed to compare them at!? (I'm getting worried here fella ). This has been the whole point of this discussion from the beginning (except maybe the cost of it eh Griffster ).



I don't understand why you are suggesting I don't have a grasp of the time it takes to change the velocity of the body from one speed to another. The data below shows that at 60mph, the two cars would be in the gear listed, have the power available for acceleration and the rate of acceleration that would give. My spreadsheet records this at every 0.1mph increment.

60mph roll
3.11
2nd gear @ 5827.7rpm
169bhp@wheels- wind resistance
Force= 1055lbs
M/S^2= 3.36

4.00
3rd gear @ 5097.6rpm
165bhp@wheels- wind resistance
Force= 1033lbs
M/S^2= 3.29



This doesn't make sense. At a fixed speed I can use the gearing to calculate exactly how much power the car is developing at the wheels by running a vlookup over the rpm at that speed against the power at the wheels. This is reduced by drag at that speed, converted to a force and and then converted into acceleration. Subsequent columns within the spreadsheet then tally up how far the car travels during this speed increase increment (using the forumlas you posted) and give me the full picture. I can then interrorgate the columns to bring back any metric I want (distance traveled at x speed, speed at x distance, time to reach x speed and so on).


Let me ask 'the rest of the world on here' to provide one shred of evidence that shorter gearing actually provides faster acceleration.

Well the only evidence I can think of is to test acceleration from. 0-30mph

Try it using 1st gear
Then try it in 2nd gear
Now try it in 3rd gear

Now which one gave the quickest time? Or am I missing something?

Sigh... I'm gonna update my signature to this soon!

Accelerative Force(lbs)= 375*(bhp/speed)

If you were to plot power against speed in those three gears, what do you suppose you would see!??????

Here's a good website that explains what a gearbox is for http://craig.backfire.ca/pages/autos/transmissions

I thought you asked for evidence that:

"Let me ask 'the rest of the world on here' to provide one shred of evidence that shorter gearing actually provides faster acceleration."

You asked, I gave.

And isn't that the link that I provided at start, why do you keep referring to it...

You didn't, you stated something that clearly points out that you haven't been reading what I've previously posted.



It isn't, it's a different page from the same website talking about gear ratios and why they're selected.;

In short, it's to optimise the range of the powerband in play. This whole multiplying torque = more acceleration nonsense isn't mentioned.

Trappy,

Sorry mate I thought we were simply discussing the following:

"If I put shorter gearing on my car, my car will have a lower top speed but should accelerate faster"?

Now while all this talk of spreadsheets and formulae is going right over my head I can see the practical side and the only way I can think of to prove or disprove your theory is this:

Shorter gearing (eg first gear), will get me to thirty MPH faster than second gear (taller gearing). Obviously my maximum speed in first will be much less than it would be in second.

I thought that was the crux of the argument. If it isn't then I apologise. Perhaps you could explain what we are actually discussing then? (Please avoid any talk of Excel spreadsheets or mathematical formula or physics... though, just in layman's terms.)

In fact according to your own figures acceleration is altered. (Page 1 of this thread).
Fiat Coupé 20vT (Standard final drive 3.110
0-30mph: 2.53
0-40mph: 3.41
0-50mph: 4.78
0-60mph: 5.98
0-70mph: 7.85

Fiat Coupé 20vT (with 4.000 final drive)
0-30mph: 2.53
0-40mph: 3.7
0-50mph: 4.67
0-60mph: 6.26
0-70mph: 7.72

You will have the same power curve as a function of RPM but you will have a different power curve as a function of speed. i.e. To achieve a higher speed, you will have operated the engine over a different power range. If power range were to be the same then the gear ratio must also be the same.

OK, from that very link



Now 3 gears or 6 gears should make no dfference to your arguement. The only thing that has changed is the gearing.

The 3 speed could theoretically have another 3 higher gears again only they were not used as the first 3 covered the same speed range as the other 6 speed gearbox. This is basically the same as changing the FD by a huge ammount.

Oh I do *love* an argument where people are so fiercely agreeing with each other!

Here's the short and nasty: if you increase the final drive ratio (i.e. more teeth on the pinion or fewer teeth on the crown wheel), without making *any other change*, your vehicle will accelerate more quickly in all gears but will have a lower top speed (limited by engine maximum revs). If you reduce it, the opposite effect will occur. This is exactly the same effect as putting smaller or larger tyres on, respectively.

What determines the instantaneous acceleration is the instantaneous power available - and note that the peak power output from *any* engine is always above the peak torque output.

The function of the gearbox is to match available power to desired speed. Ideally, for the best acceleration, you want the maximum power transferred to the wheels at all times, so you match the ratios in the box such that having passed the power peak, changing up will reduce the speed of the engine to achieve the same power on the 'uphill' side of the power curve.

The closer the gearbox ratios are, the nearer the peak power you will be when you change and the more power you will deliver to the wheels, and therefore the greater the acceleration (ignoring such delights as the absolute number of gears in the box, the time it takes to change between them, the physical momentum of the engine parts, friction losses in the box etc).

The manufacturer will play with the gears generally not to give the maximum acceleration but to meet a number of other criteria - acceleration at the bottom end (because 0-60 is *important*, right?) or midrange; quietness or fuel-efficiency at nominated cruising speeds; even avoiding bodywork resonances at local speed limits. But within those constraints, they will generally attempt to arrange things in the box such that if you change down at optimum revs, you'll get much the same power on the other side of the curve.

Example: 16vt (for which I have the Fiat curves to hand) has a power of about 180BHP at 6000, and the same power at about 4500; between those two points the power increases. So you'd expect the gear ratios to be about 4500/6000 (0.75) between changes:



Looking at the NA, we see essentially the same


Looking at the NA curve, we see that the power at peak revs is about 135BHP at 6500; the same point on the up-side is around 5400 rpm. That would suggest a better performance could be achieved with ratios of 0.83 - but you'd need a seven speed box to do that, and you'd spend so little time in the lower gears it would probably not be worth it. (And that's probably why the ratios in first and second in both boxes are below the expected values; the acceleration has to stop while you change gear so you don't want to do that too often...)

One final thought: it's quite possible to get the seemingly contradictory position that decreasing the final drive ratio will *reduce* the top speed... maximum speed when maximum power is transferred to the wheels. You'd expect this to be the engine's peak power, but consider this scenario: at any speed on the road, what is holding you back is the friction with the road surface and the aerodynamic friction (form drag). Form drag is proportional to the square of the wind-speed, so ignoring tyre friction you need four times the power to go twice the speed. It's quite possible to gear the engine such that at a given speed, the power required to accelerate is more than the engine is capable of supplying at that instant, even though the engine is capable of delivering more power. This is why some cars with a long top gear are faster in the gear below...