Drag/CD scenario.......a question, a challenge..

[Craig 930 RS]

Two cars. Identical HP/TQ/gearing - everything equal incl weight.
Coefficient of drag changes.

Car 1 has a Cd of .39
Car 2 has a Cd of .29

Car 2 - with the lower CD of .29 - can top out at 190 mph.
What will car 1 top speed be - assuming top speed rpm are the same?

Trying to compile a practical, real world idea of Cd vs. top speed...

THX!

[masraum]

I think it's important to also have the frontal area, if you want to use the numbers for a 911 the '69 model had a frontal area of 1.71 m^2.

I don't think you can do the calcs without that info.

Are you saying using the same gearing on both cars or changing the gearing?

[Craig 930 RS]

Same gearing both cars.

Make the frontal area equal then? What ever works most accurately for this situation..

THEN what top speed?

[Eric Coffey]

Obviously a lot more factors to consider, but my SWAG is: 178.354mph



All I (think I) know is that drag increases @ the square of speed, and power-required increases @ (roughly) the cube of speed.

[masraum]

The drag force for a given object can be found using the formula below:

FD = CD 0.5 A v2 edit crap, forgot Rho

FD = The drag force measured in N (Newton).
CD = The drag coefficient, measured in N/ m2, i.e. the drag force per square metre frontal area of the object shape. You will have to find this figure in a table. This value is usually found through (very expensive) wind tunnel measurements.
Warning: Airfoil tables frequently give the drag coefficient per square metre wing or rotor blade area, rather than per square metre frontal area. You can recalculate that figure, since you know the relative thickness of the airfoil.
= The density of the fluid measured in kg/m3. In the case of dry air at sea surface level at 15° C the figure is 1.225 kg/m3. You can find air densities for other temperatures in the Reference Manual.
A = Frontal area of the object in m2.
v = Relative wind speed in m/s, i.e. the speed of the fluid moving past the object. If the object itself is moving while facing a headwind, you have to add the headwind speed to the speed of the object to obtain v.

[masraum]

I could be wrong, but I think the way to check this would be to first determine the force that the car with the .29 Cd is exerting at 190 mph (I think you need to convert to meters per sec first which is about 84.44 mps) then once you have that force plug the force into the equation and solve for the velocity. (since I think we are assuming that at top speed both cars can exert the same force, but I think that would mean possibly either different gearing or a really flat and/or fat power curve)

OK, so

The force that the one is exerting when going 190 is

.29 * .5 * 1.225 * 1.71 * 84.44^2 which gives 2166 I think in Newtons

then solving the next equation with that number

2166 = .39 * .5 * 1.225 * 1.71 * v^2 gives the velocity as 72.81 mps, converting that to mph you get 164 mph

I wouldn't testify in a court of law, but I think that's it assuming the equation that I got off of the internet is right and the units work out correctly. I did ok in physics in school.

edit adding rho in didn't change the results.

I suppose an easier way of doing this would have been to put the two equations back to back, that way you can skip a bunch of the numbers

.29 * .5 * 1.225 * 1.71 * 84.44^2 = .39 * .5 * 1.225 * 1.71 * v^2

That would simplify to

.29 * 84.44^2 = .39 * v^2

[masraum]

One of you engineering types tell me if I get the cookie.

Actually doing it the same way with mph I get the same results which makes sense, so I'm confident that my results are correct.

[1fastredsc]

Shouldn't the 1768 representing the drag force for car B in newtons be 2166 newtons since we are assuming maximum engine force pushing the car through the air in either scenerio or am i confused?

[masraum]

Sorry, screwed up the calcs the first time, and then when I went back and editted my first post to fix my mistake I missed changing the second instance of 1768 to 2166. Yes, you are correct, it was a typo.

[Randy Webb]

I'm a scientist type, not an engineering type, but I award you the cookie (tho I didn't check the math itself). There's also the assumption involved that the engine can generate that force...

Now -- for extra credit -- what will the maximum speeds of each car be in the rain?

[masraum]

I would think that the rain drops themselves may not actually make any difference assuming we don't get hydroplaning, and the drops are not large enough and the air isn't dense enough with them to radically alter the rho, but the temp and relative humidity would probably be different than in my assumed ideal world, which would change rho which might change the calculation some. I'll say that assuming the temp is still the same and that assuming the air isn't more humid that the top speeds will be the same, but that's just a guess.

Do I now have 102 on this test?

[kach22i]

Originally posted by masraum
I think it's important to also have the frontal area, if you want to use the numbers for a 911 the '69 model had a frontal area of 1.71 m^2.

15-20 years ago Porsche put in it's advertisements that they included their low frontal area in their CD calculations and that other manfacturers did not. Was that marketing hype?

[Randy Webb]

no, we reduce your score from 99 (before) to 97. Think about it - the car has to push the rain drops out of the way.....

kach - that doesn't sound right. Could they have been talking about the fact that many manfs. remove outside mirrors and trim when measuring Cd? Also, often it is measure with the wheels stationary instead of rotating, which is harder to do.

[masraum]

Well, you know, they always say you should leave your first answer. I had originally thought the water would slow things down by changing the rho, but then I thought maybe it was a trick question and changed my answer, but I still don't know by how much. I would assume that it would depend upon the size of the drops as well as the number of drops or drop density. I looked around the internet for more info, but couldn't find any. The closest I came was finding the rho of steam which is much different I assume due to the temp difference.

[Randy Webb]

I guess it is a trick question. The vehicle is now in a mixed media regime, not in a simple fluid (which is what all the fluid dynamics eqns. are based on).

[Wil Ferch]

Drag increases ( as noted) by the square of speed.

Horsepower required to acquire such speed, however, is a cubic relationship with speed.

I haven't plugged in the math, but if the frontal areas are the same, and you know the top speed of the CD.=0.29 car, you can also calculate the hp required to ge to 190 mph.

now...starting with that same hp, work backwards including a Cd of 0.39 and double check drag being proportional to v2.....hp proportional to v3...and see if you come up with the same answer....

Wil

[Craig 930 RS]

Can you show us the math needed (& or HP) for the .39 car to go 160, 170, 180 mph? My car has 330hp now.

THX all,

[kach22i]

Click on the 4th from the bottom, it says Audi TT, but it's about the 911. Go aerocrazy.

http://www.google.com/search?q=early+rear+spoiler+porsche+development&hl=en&lr=&start=10&sa=N

[Craig 930 RS]

GREAT STUFF!
The Rosetta stone ;-)

Thanks KACH & all -

[limble]

Originally posted by masraum

The force that the one is exerting when going 190 is

.29 * .5 * 1.225 * 1.71 * 84.44^2 which gives 2166 I think in Newtons

then solving the next equation with that number

2166 = .39 * .5 * 1.225 * 1.71 * v^2 gives the velocity as 72.81 mps, converting that to mph you get 164 mph

How does this equate to horsepower?