Top speed on a windy conditions.

[CHICAGO]

I own an old and small car with only 30HP and a very poor aerodynamic coefficient, Cx=0.51.

Based on this and on the rolling coefficient, I just solved out what would be the maximum speed when driving on a leveled road and there is no wind at all. It came out to be 120Kms/hr which can be actually checked-out.

My question is: How can a, say, 50kms/hr frontal backward wind, affect this top speed?.

The engine power, as I ponted out, is 30HP when driving at forth gear-6000 RPMs- at 120kms/hr.

In this conditions, rolling resistance absorbes 4.3HP and drag resistance requires 25,7HP from the engine power. So 30HP in total.

Now let's imagine that the engine power does not change in a range from 3000 to those 6000 RPMs.

If we had a backward wind of 50 Kms/hr, do I need the same 25,7HP to keep (120-50) 70kms/hr, and would this be my top speed (with a slight increasing due to a less rolling resistance? Assumed we drive in the same forth gear.

Thanks a lot in advance a have you all a happy new year.

.

 

[mgb_phys]

You can't just combine win speeds like that, (because it doesn't take into account the cars shape)
The force from the wind is = density of air (1.2kg/m^2) * Cd * front area * velocity^2

So you would have to work out the drag force at a certain speed, then use power = force * speed to work out how much power you are using to beat this

At freeway speeds almost all of the engine power is to overcome aero rag, so when this power = engine power that's pretty much your top speed (assuming the gears could reach this)

 

[SystemTheory]

Convert peak horse power to watts and measure relative velocity in meters per second. The relative velocity into a head wind is the speed of the car plus the speed of the head wind.

The power caused by drag is:

$$P = \frac{1}{2}\rho C_{D}Av^{3}$$

where rho is air density, Cd is drag coefficient, A is frontal area of car, and power is proportional to relative velocity cubed. Plug in peak engine power and the other factors and solve for top speed estimate (assumes a properly designed gearbox for top speed).

 

[HallsofIvy]

Note that this question is very different from finding the speed of an airplane with a wind or a boat on moving water. The airplane is being supported by the moving air and the boat by the moving water. They are equivalent to a toy car moving across a table while the table is carried in some direction.

The only effect of wind on a car is the resistance force the car offers in the wind.

 

[CHICAGO]

The force from the wind is = density of air (1.2kg/m^2) * Cd * front area * velocity^2

This is the formula I used to learn what would be my old car top speed.
And this:

density of air (1.2kg/m^3) * Cd * front area * velocity^3
To solve the power required.

Now, I see that SystemTheory writes:

The power caused by drag is:

$$P = \frac{1}{2}\rho C_{D}Av^{3}$$

.

In the case our car speed is exactly the same as the head wind:
Does this mean that the power to beat a head wind is half the power required to move the car in a no wind conditions?.

Sorry, gents, I still do not have it clear enough.

In my example: my car top speed is 120 Km/hr in a calmed wind.

In case of a head wind of 5 Km/hr, will my top speed be 115 Kms/hr? (assuming that rolling resistance is the same and that the engine has the same power at both 120 and 115Kms/hr) .

Thank you for clarifying my confusion.

 

[mgb_phys]

In no wind conditions, and assuming that all the friction is due to aero drag (rather than the mechanical friction) then the power equation given by SystemTheory is the limiting top speed (terminal velocity)
The velocity here is relative to the air flow, it doens't matter if you or the wind is stationary relative to the road.
If you had a head wind this is added to the speed of the car, so in theory if the top speed is 120km/h then you would just fail to go forward into a 120km/h head wind
Alternatively you could go 60km/h (road speed) into a 60km/h head wind an so on.

 

[mugaliens]

I'm not sure why LaTex is displaying things properly in preview mode, but it wasn't displaying things properly here, so please forgive the image vs actual LaTex:

 

[SystemTheory]

Does this mean that the power to beat a head wind is half the power required to move the car in a no wind conditions?

No. The factor 1/2 is part of the formula for quadratic drag force:

http://hyperphysics.phy-astr.gsu.edu/HBASE/airfri2.html#c2

and to get power (in coherent unit system) multiply by velocity.

I'm not sure why LaTex is displaying things properly in preview mode, but it wasn't displaying things properly here

When this happens to me, I copy the text/code/message to clipboard, delete the "glitchy" post, and paste the message into a whole New Reply. Then it shows up properly for some reason.

I'm not sure the OP is applying k*v for rolling friction. The main point is that a headwind will slow down the car's road speed by direct subtraction.

I once drove a 4 cylinder Nissan pickup with rear cap from New York to New Hampshire behind tropical storm "Bob." Due to cyclonic winds (I forget which way they rotate in Northern Hemisphere) I faced all three conditions, headwinds, tailwinds, and calm eye of storm. In the headwinds I had the throttle to the floor doing maybe 45-55 mph. In the tailwinds with little to no throttle doing 70 mph (rear cap surface broad and flat)! Eye of the storm normal response. Interesting trip.

 

[CHICAGO]

... The main point is that a headwind will slow down the car's road speed by direct subtraction...

Thanks a lot for your responses, that is what I expected. In case of a headwind I have to vectorially add both the car/road and the wind speeds to solve these two drag resistances and figure out my top speed.

But now, the main reason of my question:

Lets suppose the car limit speed is 120 km/hr (as it actually is) on a calmed wind conditions, with the engine developing its maximum power, 30HP at 6000 RPM rolling at 4th gear.

Now, we have a 60km/hr headwind. Even if my engine is capable of keeping its max power, I wouldn´t be able to go faster than 60 Kms/hr (more or less) in the 4th gear.

What is the explanation to the fact that in the 3rd gear and a 60 Kms/hr headwind I can go faster than 60 kms/hr, as I really can, up to 70 or even 75 Kms/hr? In such a way that it is moving at 135kms/hr "within" the air?.

The engine maximun power is when revs are 6000. In the 3rd gear is about 90 kms/hr.

That is my doubt. Why can my car go faster when the simple addition of both car/road and headwind speeds exceeds those 120 Kms/hr even when the engine is not developing its maximim power?

I hope to have exposed the issue clear enough to understand.

Many thanks in advance.

 

[Lsos]

I believe there's a complication here that we're all missing.

It was mentioned that if there was a 120km/h head wind, then the car would fail to go forward. But, surely if we had low enough gears, the car could make forward progress, even against 1000km/h wind.

So it's not as simple as saying that the car moving 120km/h is the same as the wind going 120km/h. Sure, the same power is required, but in one instance, the car needs to supply this power, where as in the other instance, it's the wind that's supplying it. It's not as simple as that either, but the reason why you can go 75km/h in a 60km/h headwind has something to do with the fact that some of the "extra" power required is supplied by the wind.

Perhaps the best and most accurate way to look at it is that even though the resulting airspeed is 135km/h, the car is still only moving 70km/h...after all, it's wheels are pushing it against the ground. They have to push harder as a result of increased air resistance...but the power required won't be nearly as high as if it really was going 135km/h.

 

[mgb_phys]

It was mentioned that if there was a 120km/h head wind, then the car would fail to go forward. But, surely if we had low enough gears, the car could make forward progress, even against 1000km/h wind.

No it would be blown backwards

So it's not as simple as saying that the car moving 120km/h is the same as the wind going 120km/h.

As far as the drag force on the car is concerned it is exactly the same.
Only considering the aero drag and not wheel friction if you can generate a certain force against the wind it doesn't matter what you speed is relative to the road - it's only the molecules of air hitting the car that gives you any force

 

[DannoXYZ]

But now, the main reason of my question:

Lets suppose the car limit speed is 120 km/hr (as it actually is) on a calmed wind conditions, with the engine developing its maximum power, 30HP at 6000 RPM rolling at 4th gear.

1. Now, we have a 60km/hr headwind. Even if my engine is capable of keeping its max power, I wouldn´t be able to go faster than 60 Kms/hr (more or less) in the 4th gear.

2. What is the explanation to the fact that in the 3rd gear and a 60 Kms/hr headwind I can go faster than 60 kms/hr, as I really can, up to 70 or even 75 Kms/hr? In such a way that it is moving at 135kms/hr "within" the air?.

The engine maximun power is when revs are 6000. In the 3rd gear is about 90 kms/hr.

That is my doubt. Why can my car go faster when the simple addition of both car/road and headwind speeds exceeds those 120 Kms/hr even when the engine is not developing its maximim power?

You need to calculate RPMs versus speed in 4th and 3rd gears. In case #1 with 60kph in 4th gear, what is the actual RPM of the engine at that speed? It certainly isn't over 6k-rpm. I bet it's closer to 3k-rpm and outputs a lot less than maximum power.

Compare to case #2 with 60kph in 3rd gear and I bet the engine is at a higher RPM and generates more power. It's probably closer to 4.5k-rpm and generates more power and allows you to go faster into the 60kph headwind.

So you need to obtain a gearing-table for your car and a dyno-chart to really accurately answer this question.

 

[Lsos]

No it would be blown backwards

Well...of course. But we're dealing with simple textbook physics, let's make some simple textbook assumptions...

As far as the drag force on the car is concerned it is exactly the same.
Only considering the aero drag and not wheel friction if you can generate a certain force against the wind it doesn't matter what you speed is relative to the road - it's only the molecules of air hitting the car that gives you any force

Right, the drag force is exactly the same. But power is this drag force x velocity, and the velocity isn't the same. Even if the wind is blowing at 100000km/h but the car is not moving, the car's velocity is 0.

0 x 100000km/h = 0

 

[mgb_phys]

The velocity is still the relative velocity between the wind and whatever the wind force is being applied to - otherwise an aircraft propeller would work but a wind turbine wouldn't

 

[Lsos]

The velocity is still the relative velocity between the wind and whatever the wind force is being applied to - otherwise an aircraft propeller would work but a wind turbine wouldn't

So how much power must the car be using to sit still in a 120km/h wind?

How much power must a building use to sit still in a hurricane?

The answer is none. They don't move. And yet, we have a velocity and we have a force.

Yes, power is expended, just not by the car, but by the wind. We must separate the two.

Think about the propeller and the turbine example. Yeah, in both instances you can use velocity x force and come up with a power, except in one instance the power is in, and in another instance the power is out. It's a very big difference, and it pertains to the OP's question because he's only interested in the power OUT. From his car.

Attach a turbine to the still car. Not only will it use no power in the windstorm, but it can actually generate power...

Edit: I just noticed HallsofIvy saw this before me:

The only effect of wind on a car is the resistance force the car offers in the wind.

Use the WIND speed/ coefficient of drag/ area to figure out a resistive force.

Then use the CAR speed, to figure out the power the car needs to put out to overcome that resistive force.

 

[SystemTheory]

So you need to obtain a gearing-table for your car and a dyno-chart to really accurately answer this question.

This is correct. If you're only making 30HP, and it is geared to push air, you can't push more air by changing gears. However, picking a gear ratio to make maximum use of available power at speed is a design problem which may not be optimized for your car. You probably have an overdrive 4th gear to get good speed and fuel economy at normal highway speeds with no headwinds.

The gear ratio acts like a transformer, and I suspect the wheel bearing damping is proportional to the gear ratio squared when reflected back to the engine. This generates a "sweet spot" gear ratio to match maximum engine power to reflected friction. For your car this is roughly located in the 3rd gear (you should never be making more than peak power of course, just converting it to velocity effectively using a gearbox ratio matched to the frictional load for velocity optimization).

A rough analogy is impedance bridging, where one wants to optimize voltage to a load from a source with internal resistance:

http://en.wikipedia.org/wiki/Impedance_bridging

in your case the problem is matching the source/load resistance ratio through the gearbox to optimize velocity.

 

[CHICAGO]

I can show you this car engine performance curve, torque vs. RPM, and everything about its Cd, Cr, and so on. But its engine max. power is 30HP at 6000 RPM. just the power to overcome a drag resistance (plus the rolling one) of 120 kmh.

But why can my car beat a headwind of 60kmh moving at 75kmh (135 kmh respect to wind) when rolling at 3rd gear (about 4500 RPMs/27 HP according to engine performance and gearbox design)?. I guess the maximum engine power (30HP@6000 RPM) will never be increased, disregarding what gear I am engaged.

According to the formulas the car would a 40 or 45 percent more power (about 42 or 43 HP) to be able to move forward at 135kmh.

Am I missing any parameter?

What I would like to clarify here is that if I had a, say, a 5 or 7% uncertainty in the calculations I would not ask you about this. The issue is a 40 or 45% deviation from the theory you are exposing here, and that is why I am asking around if we are missing something.

Thanks for all opinions.

 

[Lsos]

The parameter you are missing is the speed. The wind might be 135km.h, but the car is still only moving at 75km/h through it.

The force trying to slow the car might be the same, but the car is moving much slower against this force...hence why it has power to spare.

I calculated that the air drag against the car from a 135km/h wind is about 850N.

So, plugging this into the power formula...

p = f x v
p = 850N x 37.5m/s

You get 32000 Watts. 43 horsepower. As you said, about 40% more power.

The issue, though, is that 37.5m/s. Remember, that's NOT how fast the car is moving. It's only moving at about 21m/s, because it's applying it's power against the stationary road, and not, like an airplane, against the air.

p = 850N x 21m/s

About 17850 Watts. 24 horsepower. 3 horsepower off from 27...just enough to account for road drag and whatever other uncertainty...

 

[SystemTheory]

The motor does not make more power or torque in a headwind. It develops power according to the torque speed curve. For simplicity, at constant speed, let there be 1 power source and 2 power sinks.

Pin = P1 + P2

where P1 is power dissipation due to friction in the driveline. If this depends on velocity (reasonable) then in a headwind in 3rd gear the driveline is turning only in proportion to 75km/h versus 120km/h in static air in 4th gear. The simple answer is P1 is lower in the headwind making more of the limited engine power available as P2 to push wind at 135km/h equivalent relative velocity.

In no case should the car be able to push air with more power than the engine makes, so if your aero model and engine curves show violation of conservation of power, one of them is wrong! But here I see no reason to doubt your models on the face of it.

 

[CHICAGO]

Thanks for all responses.

So, in the case of a 60kph headwind the car speed limit must be 60kph plus that small speed I can get out of the diference in the rolling resistances between 120 and 60 kph. But it seems that power residual, proportional to the speed I guess, is not large enough to explain how my car moves at 75kph when I shift to 3rd gear.

Could the explanation be that in this old car the rolling resistance is not a linear function of speed?. I mean, for example, the front wheels are not well aligned. Could a misalignment like this make the rolling resistance fuction more exponential ?.

Thanks again.

 

[Lsos]

where P1 is power dissipation due to friction in the driveline. If this depends on velocity (reasonable) then in a headwind in 3rd gear the driveline is turning only in proportion to 75km/h versus 120km/h in static air in 4th gear. The simple answer is P1 is lower in the headwind making more of the limited engine power available as P2 to push wind at 135km/h equivalent relative velocity.

OP is trying to explain away a difference of about 13 horsepower, where as P1 would only give you about 2-3hp extra at best...

 

[SystemTheory]

Chances are the driveline friction is a nonlinear function of velocity. But that may not be the primary cause of your ability to go faster in 3rd gear into a headwind.

Suppose the axle speed is omega in radian per second and the wheel damping is linear with a factor D in standard SI units. The load 1 counter-torque:

cT1 = D*omega = D*(v/r)

where v is velocity at the center axle and r is the tire radius. But the reflected counter-torque at the engine is inversely proportional to the speed reduction gearbox ratio squared:

cT1e = cT1/G3^2 versus cT1/G4^2

therefore the engine sees potentially much less driveline counter-torque in third gear at lower road speed than it sees in forth gear at much higher road speed. I would do a graphical analysis if you have good enough data to plot some curves.

I can show you this car engine performance curve, torque vs. RPM, and everything about its Cd, Cr, and so on. But its engine max. power is 30HP at 6000 RPM. just the power to overcome a drag resistance (plus the rolling one) of 120 kmh.

But why can my car beat a headwind of 60kmh moving at 75kmh (135 kmh respect to wind) when rolling at 3rd gear (about 4500 RPMs/27 HP according to engine performance and gearbox design)?. I guess the maximum engine power (30HP@6000 RPM) will never be increased, disregarding what gear I am engaged.

Sorry I kinda missed this comment that it would take 43HP according to your calculations. How much power do you estimate it takes just to push air at 135kmh relative velocity? If you wish to find an error in your calculations or data assumptions please post your work to the thread.

 

[Lsos]

If we had a backward wind of 50 Kms/hr, do I need the same 25,7HP to keep (120-50) 70kms/hr, and would this be my top speed (with a slight increasing due to a less rolling resistance? Assumed we drive in the same forth gear.

Since my explanations were ingnored, I'm just directly going to answer the above question: NO. You DO NOT need the same power.

... The main point is that a headwind will slow down the car's road speed by direct subtraction...

THIS is NOT TRUE

If you have a backwards wind of 50km/h, driving at 70km/h will only require 15hp. Therefore, that WILL NOT be your top speed.
If you have a backwards wind of 100km/h, driving at 20km/h will only require 4.3hp. Therefore, again, that WILL NOT be your top speed.

Let's take it all the way. If you have a backwards wind of 120km/h, you only need 0hp to drive at 0km/h. Therefore, that will also NOT be your top speed.

And it's very easy to see why, using simple high school mechanics formulas. You can do the calculations to check it out yourself, but you MUST use the correct valuves, which you have NOT been doing. I have already done the simple calculations in this thread, and they work out.

Just two simple formulas:

force (air resistance) = ½pCdAv$$^{2}$$

Power = force x speed

The v and the "speed" are NOT the same. The v represents the wind speed, the "speed" represents the road speed.

If you solve these formulas, you will see that a 120km/h wind only gives a force on your car of like 575N. This means that my wife could push your car against 120km/h wind. The only question is how fast can she push it.

 

[SystemTheory]

This means that my wife could push your car against 120km/h wind. The only question is how fast can she push it.

RIGHT! But without stalling or seriously loading down the engine, she could not drive it in 4th gear! She probably could not drive it in 3rd gear! She maybe could drive it OK in 2nd gear and probably could crawl along against the 120 km/h headwind in 1st gear at the rpm in which the engine power is exactly matched to the reflected torque from the drag force and driveline friction! The tire radius and gearbox are a source/load transformer and if you don't properly model this function you won't be able to identify the torque/power equilibrium condition in the differential equation of motion.

 

[Lsos]

RIGHT! But without stalling or seriously loading down the engine, she could not drive it in 4th gear! She probably could not drive it in 3rd gear! She maybe could drive it OK in 2nd gear and probably could crawl along against the 120 km/h headwind in 1st gear at the rpm in which the engine power is exactly matched to the reflected torque from the drag force and driveline friction! The tire radius and gearbox are a source/load transformer and if you don't properly model this function you won't be able to identify the torque/power equilibrium condition in the differential equation of motion.

Right...with a 120km/h headwind, you're probably not going to go anywhere in 4th gear.

With 30 hp, this car could theoretically go against this wind at around 58km/h...although probably less due to gearing constraints.

 

[CHICAGO]

Just two simple formulas:

force (air resistance) = ½pCdAv$$^{2}$$

Power = force x speed

The v and the "speed" are NOT the same. The v represents the wind speed, the "speed" represents the road speed.

Great ! now numbers and tests results match together !

I still have some deviations but they are really small and probably due to my way of doing the tests.

Since my explanations were ingnored, I'm just directly going to answer the above question: NO. You DO NOT need the same power. ...

Sorry Lsos. I never ignored your arguments. The problem is that I did not understand the basic concept (to me, now is basic, yesterday it wasn´t) that the power needed was basically that, power to do a work in a time period.

Really interesting.

You guys can´t imagine how happy I feel now.

By the way the guilty car is a 1981 Citroen 2CV. Small, simple, basic and powerless but enough to learn about this kinda things.

Best regards.

 

[Lsos]

A couple things I noticed about your test and calculations though...

First, you said you were driving into a 60km/h headwind. There has to be some large inaccuracy there because I doubt you have a pitot tube in your car...

Second, a car usually loses power between the engine and the wheels. I keep hearing around 15%, which means your car would only have 25.5hp with which to propel itself and fight the wind and the rolling resistance. I'm not sure how that would affect the results.

And third, when you mentioned:

If we had a backward wind of 50 Kms/hr, do I need the same 25,7HP to keep (120-50) 70kms/hr, and would this be my top speed (with a slight increasing due to a less rolling resistance? Assumed we drive in the same forth gear.

That "assume we drive in the same 4th gear" part should be taken out...as it makes it impossible to assume the same power output, which is probably what you had in mind!

Thanks for finally listening to me :)

Best regards.

 

[CHICAGO]

No, I don´t have a pitot tube tube in my car, although now I am thinking in about installing one (I am kidding)

I work in a Space Telecommunication Complex where there are several large dish parabolic antennas. One of them is 70meter-diameter and for both signal receptions and antenna security we have to monitor the weather conditions very closely.

We have several weather stations outside and a good computer showing the parameters all the time. One of them, of course is the wind speed a its direction. In case of gusting winds the computer also shows the mean value, the “RMS” force among some other interesting values.

In the Facility I have also a long (1200 meter) road where I have done the tests with my car.

It is long to explain how I checked my car speeds against those parameter and I, of course, am aware of the mistakes I could have made. But I took a lot measurements taking the mean results and so……

About the engine power, I understand it is measured at the clutch flywheel and the gearbox losses are included in the rolling resistance, am I right?

You know, in order to minimize the errors I drove back and forward many many times to get a reliable mean figure.

And as I said, had I found a 5 or even a 10% deviation in my calculations I would have blamed me and my way of doing those tests. But the actual deviation was much larger and I was suspecting about the model I was assuming.

Thanks for your good help.

 

[SystemTheory]

Chicago,

I'd like to generate some drag and wind power estimate curves to post to the board if you provide these input values:

A = ?, car frontal area in square meters
Cd = 0.51 per above
rho = ?, air density in kilogram per cubic meter

this will give a comparison to your values in a graphical depiction of drag force and drag power versus velocity. It will only take me minutes to generate since I have done this for small electric RC cars.

 

[SystemTheory]

So my model confirms Lsos' basic analysis. I misread your post above since you said the engine would require 43 HP (my bad). Now I see the point you were making. The graphical comparison is attached.

Numerical Code

`Little French Car
rho = 1.229 `{kg/m^3}
Af = 1.8 `{m^2}
Cd = 0.51 `{#}
vH = 120/3.6 `{m/s}
vL = 75/3.6 `{m/s}
vW = 60/3.6 `{m/s}

`Plot Drag Load
v1 = 0 .. vH `{m/s}
D1 = 0.5*rho*Cd*Af*v1^2 `{N}
v2 = 0 .. vL `{m/s}
D2 = 0.5*rho*Cd*Af*(v2 + vW)^2 `{N}

`Plot Drag Power
P1 = D1*v1 `{W}
P2 = D2*v2 `{W}

`Maximum Engine Power
Pmax = 24600

Plots are converted to show Newtons and horsepower versus kilometers per hour.

 

[CHICAGO]

.
Hi System Theory ! That is great ! I really thank your plots.

Where did you get that car front area?, I coldn´t find any document with this parameter. I solved my myself by a home made method to get 1,67 m².

Where I live the mean air density is 1,15 kg/m³.

Thanks again for your good job.

.

 

[Lsos]

I guess you did do some proper tests then!

About the engine power, I understand it is measured at the clutch flywheel and the gearbox losses are included in the rolling resistance, am I right?

You are right that engine power is measured at the clutch. However, the drivetrain losses are NOT included in the rolling resistance.

Rolling resistance is generally constant, no matter how fast you're going. As a result, power lost due to rolling resistance rises linearly with speed. If it's 4.3hp at 120km/h, then it will be around 2.2hp at 60km/h.
Drivetrain losses, while being complicated (dependent on rpm, torque, engine type, drivetrain layout, transmission, etc), is generally said to be 15% of engine output for a front wheel drive, manual car. Notice that unlike rolling resistance, it's not dependent on speed. This means that for your car, the actual power getting to the wheels is only about 25.5hp. So you have 25.5hp to overcome air drag AND rolling resistance.

I hope this doesn't mess up your calculations too bad. Whatever calculations I did were also ignoring drivetrain losses. And, I was using 1.69m$$^{2}$$ as the car frontal area, which I deduced from your calculations...

 

[SystemTheory]

Frontal area Af = 1.8{m^2} was an estimate based on a modern (199?) CV2 design that I saw in a website or the Wikipedia. If you provide some more system parameters I'll revise the model to your values.

I can overlay the source transformer force-velocity curves on the drag loads in these plots if you provide a sketch of the torque-speed curve, the tire radius, and the speed reduction ratio in each gear. How did you estimate or measure the torque-speed curve?

 

[CHICAGO]

Hi System

These are the only two plots I have.

On those, the blue graphs show the original engine performance. The orange ones are the result of installing what we call "turbito" which has nothing to do with a real turbo. It is a simple air conection between the front fan and the air filter intake in order to have some more air flow to the carburator.

I have this "turbito" mounted, so my engine performance would be those shown on orange.

http://img145.imageshack.us/img145/4074/2cvtorqueplotenglish.jpg
By chicago49 at 2010-01-07

http://img145.imageshack.us/img145/3033/2cvpowerplotenglish.jpg
By chicago49 at 2010-01-07

And here, my gearbox data. I use 135 tyres.

http://img189.imageshack.us/img189/9260/gearboxdata1i.jpg
By chicago49 at 2010-01-07

I really appreciate your help.

Thanks a lot.

.

 

[SystemTheory]

Good data. I need to install and test a software application to convert your torque-speed plots to a lookup table, or else create a piece-wise linear approximation, so I'll get back to you as time permits with a graphical study.

Just to confirm my initial understanding I would multiply to get G3 = 4.125*1.786 = 7.367 in third gear, and G4 = 4.125*1.316 = 5.429 in fourth gear.

If you see an error in this assumption let me know since those are the values I will use. It is not clear to me how to find the tire radius for 135 tyres, so if you can clear that up it's one less thing to google.