# Top speed on a windy conditions.

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.

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:

CHICAGO said:
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.

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.

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.

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.

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• frenchcar.png
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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.

.

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...

Frontal area Af = 1.8{m^2} was an estimate based on a modern (199?) CV2 design that I saw in a web site 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?

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 [Broken]
By chicago49 at 2010-01-07

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

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

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

I really appreciate your help.

Thanks a lot.

.

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• 2cv_torque_plot_english.JPG
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• 2cv_power_plot_english.JPG
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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.

....I guess you did do some proper tests then!.............You are right that engine power is measured at the clutch. However, the drivetrain losses are NOT included in the rolling resistance......

Good evening, Lsos.

I know what you mean, but I used an empirical methos of getting the rolling resistance. It was 2 or 3 years ago, now I do not remember the calculations nor numbers, but, anyway, what I did:

On a no wind conditions and a leveled road, driving at 45/50 kms/hr in the 4th gear I step down the clutch pedal keeping the 4th gear engaged. The car then slows down and I start my timer just when the speed is 40km/h. Car keeps going decreasing speed and then I stop the timer just when my velocity is 15 km/hr. I did this several times back and forward to have a good average in case of slight wind or road slope I would not had noticed. I also logged the lap-times of 35,30,25 and 20 km/hr passing-by in order to get a better model.

So according to the time of decelerating from 40 down to 15 km/h and suppresing the small drag interference, I figured out my mean rolling resistance between 15 and 40 km/hr.

What I guess is that if I had the 4th gear engaged all the time I accounted the 4th gearbox rolling resistance as well, did I not?.

I probably made some mistakes, mainly because I was concious that a slight deviation would not cause large errors in the final calculations. But everything is subject to be improved.

I want to thank you very much your help in this issue.

....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.

I think you are right. Those would be the final relationships between the engine crankshaft and the traction wheels.

... 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.

Upsss !!, what I have just done is going down to my garage and have the actual distance between the center of the wheel and the floor. It is 28 cms for a 1,8 bar tyre pressure.

(Now I start of thinking this tiny and basic car does not deserve so many calculations !!!, but I like it)

Thanks again, ST.

rcgldr
Homework Helper
It is not clear to me how to find the tire radius for 135 tyres.
What you want is the effective radius. Determining the effective circumference can be done directly by putting a thin strip of water on the pavement between the tires (so that only the front or rear tires roll over the strip), then going forward or backwards and measuring the distance between the wet stripes the tire leaves behind.

What you want is the effective radius. Determining the effective circumference can be done directly by putting a thin strip of water on the pavement between the tires (so that only the front or rear tires roll over the strip), then going forward or backwards and measuring the distance between the wet stripes the tire leaves behind.

You mean the tire circumference? On the R/C boards they calculate the rollout but I don't use it in my models. In this case we're only looking at top speed equilibrium, so driveline intertia is neglected. The best case force at the drive axle as a function of velocity is then approximately:

F(v) = G*(T/r)

where G is the overal gear ratio, T is the torque-speed curve, and r is the tire radius. I plug this in to plot F(v) in kilometer per hour for each gear and we'll compare to the Newton's of drag force on the nose of the car to estimate the equilibrium speeds.

(Now I start of thinking this tiny and basic car does not deserve so many calculations !!!, but I like it)

I'm working on educational modules to solve interesting differential equations using case study and reasoning by analogy. My web site is still in the idea phase but I might want to use your better than average data to illustrate gearbox design in an Internal Combustion Engine car. I've already done some models for 1/10 scale electric and the Tesla Roadster, single speed gearbox cars.

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rcgldr
Homework Helper
Determining the effective circumference can be done directly by putting a thin strip of water on the pavement between the tires (so that only the front or rear tires roll over the strip), then going forward or backwards and measuring the distance between the wet stripes the tire leaves behind.

You mean the tire circumference?
The effective tire circumference. Due to deformation at the contact patch, the effective circumference is a bit less than the unloaded tire circumference, and the water strip measurement is an easy way to determine the effective circumference.

The effective tire circumference. Due to deformation at the contact patch, the effective circumference is a bit less than the unloaded tire circumference, and the water strip measurement is an easy way to determine the effective circumference.

Hi Jeff, you are right. The effective tyre circumference is a bit less, that is why I provided System Theory with the actual radius I measured from the center of the wheel to the floor. It is 28 cms whilst from this wheel center to any other point but the contact patch is about 30 cms.

Regards.

rcgldr
Homework Helper
Hi Jeff, you are right. The effective tyre circumference is a bit less, that is why I provided System Theory with the actual radius I measured from the center of the wheel to the floor. It is 28 cms whilst from this wheel center to any other point but the contact patch is about 30 cms.
Although you measured 28 cms, the effective radius will be larger than that because the tire acts a bit like a tank tread. The deformation I mentioned is due to linear compression of the tread, not vertical.

I sampled the curve with a software tool to make this table. Read the table into an array for speed w {rad/s} and array for torque T {N-m} (note 9.807 Conv factor applies):
229.475 3.93939
252.324 4.02597
272.63 4.11255
290.392 4.19913
308.154 4.28571
325.883 4.43723
343.612 4.58874
366.473 4.65368
391.91 4.65368
417.347 4.65368
442.784 4.65368
468.222 4.65368
493.659 4.65368
516.585 4.58874
534.468 4.43723
552.352 4.28571
570.235 4.1342
590.673 3.96104
This code builds arrays for Force and Power versus velocity:
CV2 Chassis
rho = 1.15 {kg/m^3}
Af = 1.69 {m^2}
Cd = 0.51 {#}
r = 0.28 {m}

3rd Gear Transformer
G3 = 7.367 #
F3 = G3*(T/r) {N}
v3 = (w/G3)*r {m/s}
vW = 60/3.6 {m/s} headwind
R3 = 250 + 0.5*rho*Cd*Af*(v3 + vW)^2 {N}
P3i = F3*v3 {W}
P3o = R3*v3 {W}

4th Gear Transformer
G4 = 5.429 #
F4 = G4*(T/r) {N}
v4 = (w/G4)*r {m/s}
R4 = 125 + 0.5*rho*Cd*Af*v4^2 {N}
P4i = F4*v4 {W}
P4o = R4*v4 {W}

Note I added 250 {N} rolling resistance estimate in 3rd gear and 125{N} rolling resistance in 4th gear just to make the curves "look about right." Not sure if this factor of 1/2 is reasonable. In lower gears (higher reduction ratios) the friction in the engine tends to dominate the equilibrium. In higher gears (lower reduction ratios) the friction in the load starts to dominate the steady state equilibrium. Actual driveline friction is the hardest part of the model.

Also note the torque-speed curve got truncated a bit so plot does not go out to 120 kph top speed. Graphs attached with conversion to HP and kilometer per hour horizontal axis.

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On a no wind conditions and a leveled road, driving at 45/50 kms/hr in the 4th gear I step down the clutch pedal keeping the 4th gear engaged. The car then slows down and I start my timer just when the speed is 40km/h. Car keeps going decreasing speed and then I stop the timer just when my velocity is 15 km/hr. I did this several times back and forward to have a good average in case of slight wind or road slope I would not had noticed. I also logged the lap-times of 35,30,25 and 20 km/hr passing-by in order to get a better model.

So according to the time of decelerating from 40 down to 15 km/h and suppresing the small drag interference, I figured out my mean rolling resistance between 15 and 40 km/hr.

What I guess is that if I had the 4th gear engaged all the time I accounted the 4th gearbox rolling resistance as well, did I not?.

"Keeping the car in 4th gear" is actually a damn good idea (I don't think I would think of that), but unfortunately I don't think it would accurately show how much power is lost in the drivetrain. This is because drivetrain losses depend, approximately, on the power transmitten through them (although some people would argue this). Your test essentially spins the drivetrain under no load, which would significantly decrease it's internal friction.

What I'm saying is if you're going 40kph down hill, your drivetrain losses will be MUCH less than if you're going 40kph up hill (under load)...whereas rolling resistance will be almost the same throughout.

Also, I didn't quite understand if you accounted for the air resistance when figuring the rolling resistance. I figure the air resistance would start at around 65N and end at around 10N when doing the coast down test. This sounds pretty significant...

After that, you'd probably be going to deep into the exercise...

Yes, drivetrain frictional losses depend upon load. So with the engine at full-throttle, you will lose more than if you were coasting at the same speed. This is independent and separate from rolling-resistance (tyres & bearings).

Ok, I've added to your gearing table and calculated the RPMs your engine is actually running at 60km/h:

Plot that on your dyno-chart's power-curve and we get:

As you can see, at 60km/h in 4th-gear, your engine is really ony generating 17-18hp, not 30hp. This is less power-output than using 3rd-gear @ 60km/hr where the engine is putting out 25hp. Regardless of the actual power-required to travel at 60km/hr into a headwind, you have more reserve power to accelerate to a higher-speed from 60km/h in 3rd-gear than 4th.

Also 120km/hr top-speed in 4th gear with no wind isn't fully optimized. The gear is too low and you are revving the engine past its maximum power-peak at 5100rpm. At 6000rpm, the engine's only generating 26-27hp; less than the 32hp @ 5100rpm. If you had 20% taller 4th-gear, you can reach a higher top-speed by matching the engine's maximum output to its calculated top-speed. In which case, you may see 125km/h.

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Sorry gentlemen for my absence during the last two months.

I would like to thank you all for your high level responses which I have checked in the "real world" and all of them are quite accurate.

Thanks again.