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Old 12-03-2008, 11:53 AM   #1 (permalink)
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Help my understanding of Cd...

Is the co-efficient of drag related to how much of the front profile actually catches air dead on instead of deflecting it in one direction or another? To be more specific, can it safely be assumed that a flat-faced RV has a cd of 1.0?

Also, are there any calculators available that could help me plot the energy consumption of a vehicle at various speeds, based on it's weight/cd/inclines/etc? Keeping in mind that the math portion of my brain is retarded and my eyes glaze over at mathematical formula?

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Old 12-03-2008, 12:21 PM   #2 (permalink)
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Quote:
Is the co-efficient of drag related to how much of the front profile actually catches air dead on instead of deflecting it in one direction or another? To be more specific, can it safely be assumed that a flat-faced RV has a cd of 1.0?
No.... Allow me to explain each component of the drag equation - you don't need to use it, but knowing what each bit does isn't math and explains much.



Fd is the drag force - the thing we're calculating
rho (that p looking thing) represent the density of the flowing fluid, in our case - air
v is the velocity
A is the frontal area, I'll explain more in a moment
Cd is the drag coefficient - it is a unit less number that must be experimentally determined.
that v with the hat is the unit direction


Frontal Area
Basically put, it's the area that's in the way. For your car, if you were to take a picture of the front and measure the area - that would be the frontal area if we assume no cross winds. If your car is parked, and there's 15mph pure cross wind - the "frontal" area would be the area of the side of your car.

Cd
Cd is a phenomena - which means it's observable but beyond intuition. One really can't guess the Cd of an object. Think of Cd as the form constant - the shape (not size) will change the value of Cd. And Cd can be greater than 1.

A few things to note
1. A change in velocity increases force exponentially
2. A reduction in A is just as effective as an equivalent reduction in Cd
3. Rho will be dependent on the environment - altitude, temperature, humidity (even rain).
4. The negative sign comes from that v with a hat, because drag force is opposing the direction of motion.

Here's some selected Cd values for you
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Old 12-04-2008, 09:35 AM   #3 (permalink)
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Wow

Thanks Treb. I had a pretty good grasp on Cd before, but your reply shed even more light on it. I guess they're teaching you some pretty good stuff down in FL.
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Old 12-05-2008, 07:56 AM   #4 (permalink)
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Another pair of "intro to aerodynamics" posts well worth a read:

http://ecomodder.com/forum/showthrea...-knox-840.html

http://ecomodder.com/forum/showthrea...-knox-841.html

The second one introduces the concept of Cd, minus the formula (covered in a later seminar). It makes a good companion to Treb's succinct post, above.
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Old 12-05-2008, 04:30 PM   #5 (permalink)
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are there any?

Quote:
Originally Posted by The Atomic Ass View Post
Is the co-efficient of drag related to how much of the front profile actually catches air dead on instead of deflecting it in one direction or another? To be more specific, can it safely be assumed that a flat-faced RV has a cd of 1.0?

Also, are there any calculators available that could help me plot the energy consumption of a vehicle at various speeds, based on it's weight/cd/inclines/etc? Keeping in mind that the math portion of my brain is retarded and my eyes glaze over at mathematical formula?
Atomic,I saw your post the other day and didn't want to chime in until I'd pulled my wagons in a circle.The members have tossed out a lot of good for you to chew on,and I thought I'd add to that.

With respect to your RV question,I know of no vehicles of any type with Cds as high as 1.0.The Volkswagen van of 1951 had Cd 0.75 and with rounding of the sides and roof leading edges,the drag dropped to Cd0.42.

Modern busses have Cds on the order of 0.20.With sculpting of the rear roof-line,and sides,that number can be reduced to Cd0.16.

18-wheelers of the 1970s,with Cd0.85 can and have been lowered to 0.24.

As far as formulas which would solve for the various forces as you would encounter on a real road ( curves,hills,wind spectra,temperature and weather variations,etc.),would require partial differential equations simultaneously solving for all those variables,something the big-boys won't even attempt.

If you want to get a "feel" for what's possible,use the force equation Trebuchet provided for the aero portion.Think straight and level roads,with no wind or very light wind and no cross-wind.That will get you out of a lot of calculus and trig.

The following will give you some extra things to consider:

Any testing is conducted only after 30-minutes of continuous driving
at 50-mph ( to get everything up to equilibrium temps).

70-degrees F is considered "COLD',and testing below this temp can be problematic.

Air drag varies directly as a function of frontal area ( reduce area 50% drag is reduced 50%).

Air drag varies directly with Cd ( cut your Cd in half and you've cut your drag in half).

Aero drag force varies as the square of the velocity.

The power to overcome aero drag varies as the cube of the velocity.

MPG can vary 14% depending on your rate of acceleration.

It takes 6X more fuel to accelerate a car from a dead stop than it does from 2-mph.

A 3% uphill grade can cost you 32% mpg.
A 7% grade can cost you 55% mpg.

Wet roads can cost a full mpg.-

Optimum mpg will occur at about 30-mph.-
At 40-mph you'll lose 11%.
At 50-mph you lose 20%.-
At 60-mph/ 31%.
70-mph/41%.-
80-mph/52%.

From 70-degreesF,driving 50-mph at 50-degrees F will cost you 5% mpg.
Driving at 20-degrees F will cost you 11% mpg.

Also,a stuck-open thermostat can cost you 7% mpg.-

Stop-and go driving can cost you 50% mpg.

Fluctuating between 40 and 45-mph can cost 1-mpg.

A 1-mile trip versus a 40-mile trip will cost you 75% mpg.

A 10% change in weight can effect mpg by 4.3%(urban),2.4%(@70-mph).

For an 18-mph tailwind,you gain 19% mpg HWY.For 18-mph crosswind you lose 2%,and for a 18-mph headwind,you lose 17%.

Think about these things,and if you want to go "deeper" we'll steer you into the big calculations. Happy noodling!
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Old 12-06-2008, 04:46 AM   #6 (permalink)
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Wow... I really don't understand any of this.

But yeah, I don't want to get too complicated in my measurements... Mostly what I'm trying to achieve is a chart showcasing power requirements at increasing speeds and increasing, but otherwise simplistic inclines, to show hill-climbing requirements. I'll just add a fudge factor to over-calculate for real world conditions. Wind, temperature and other errata are off my checklist as this is basically designed to scale the battery pack (did I mention this is a pipe dream of mine, an all electric RV?), to a reasonable usefulness.

Now how does a modern bus manage a Cd of around 0.20? Being that it's for all intents and purposes a brick wall, I find that a bit difficult to swallow. Even the newer "chrome dome" style buses seem like they would have massive frontal drag.
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Old 12-06-2008, 01:35 PM   #7 (permalink)
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inclines and bricks

Quote:
Originally Posted by The Atomic Ass View Post
Wow... I really don't understand any of this.

But yeah, I don't want to get too complicated in my measurements... Mostly what I'm trying to achieve is a chart showcasing power requirements at increasing speeds and increasing, but otherwise simplistic inclines, to show hill-climbing requirements. I'll just add a fudge factor to over-calculate for real world conditions. Wind, temperature and other errata are off my checklist as this is basically designed to scale the battery pack (did I mention this is a pipe dream of mine, an all electric RV?), to a reasonable usefulness.

Now how does a modern bus manage a Cd of around 0.20? Being that it's for all intents and purposes a brick wall, I find that a bit difficult to swallow. Even the newer "chrome dome" style buses seem like they would have massive frontal drag.
Atomic,for the incline part of your quest,use the following rule: (and this is US Standard units),

1-horsepower is defined as lifting 550-pounds,one foot,in one second.

Weigh your RV at a truck scale to get within 20-pounds of its actual weight.

Pick any grade your interested in,and calculate the vertical rise over ,say a mile.

Pick a speed you'll be driving up that grade.

Converting your speed into feet-per-second,and knowing the rise over a mile,you can calculate the weight you lifted in one second.

Dividing by 550-lb-ft per second,gives you your horsepower necessary to lift the RV up that grade.

If you stay with that particular velocity,by plugging in other grades (rise over run),you can calculate the lifting horsepower for any grade to construct your graph.

Add in the aerodynamic drag horsepower requirements along with rolling resistance power requirements at the same velocity,and you'll have a pretty good picture of your electric motor requirements,given their particular efficiency ratings.

Your chosen driving range criteria will help you select a battery pack,etc. based on its allowable depth of discharge.

As to the modern bus with Cd 0.20,that's out of Hucho's book and other SAE Papers.The bus is long enough such that flow velocities and pressures playing over the body of the bus can yield these numbers.The fact that they are also rear-engine,with no openings in the front of the body doesn't hurt either.

And from the Jaray/Klemperer research of 1922,as I mentioned above,with aft-body taper (boat-tailling) the Cd can easily be dropped to 0.16.
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Old 12-06-2008, 02:19 PM   #8 (permalink)
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Cd simply put is the "smoothness" of your shape. so a tear shape is going to have the best Cd because only at a very few places does either the direct impact of resistance take place(the front) or the concurrent drag resist the flow of the object. The airfoil shape allows the fluid to be pushed out of the way. . .smoothly.

If you've taken calculus there are no points where it is not continuous.

If you haven't then there is no place where the object forces the fluid to do something at a different angle than the angle that preceeded it. so if its going to force it away faster it does it gradually and not with a flat edge, or if its going to allow it past(the tail section) it allows the drag to avoid pulling the shape and expand against the surface.
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Old 12-06-2008, 05:43 PM   #9 (permalink)
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A sheet to calculate your consumption based upon Frontal Area, Cd, Crr, weight and speed :
http://www.geocities.com/frontsidesk8er/delsolaero.xls
(not mine)
On flat and in ideal conditions, but can be great to see how mpg changes when you change some parameters.

Have fun,

Denis.
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Old 12-07-2008, 07:11 PM   #10 (permalink)
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Quote:
Originally Posted by groar View Post
A sheet to calculate your consumption based upon Frontal Area, Cd, Crr, weight and speed :
http://www.geocities.com/frontsidesk8er/delsolaero.xls
(not mine)
On flat and in ideal conditions, but can be great to see how mpg changes when you change some parameters.

Have fun,

Denis.
According to that calculator, assuming I've got all of my vehicle parameters correct, I need ~2900hp to make 200mph. And roughly 2 megawatts of power.

By my calculations I'm still under 1C at 200mph, hooray!

Also, on gasoline, I would be getting roughly 1.5 gpm.

And a drop-dead range at 55mph of over 1,500 miles! Success!

Now for hill-climbing calculations...

I'm left with more than enough brunt to muscle up a 50% grade at 100mph...

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