Temperature and drag relation
 

I was bored today and since I've just had an excellent tank mpg drop during the last 1/4 tank because of a cold spell I decided to play around with the temp drag relation.

All units used are metric, the only search I found was a chart with Imperial which meant very little to me.

Here's what I came up with.
drag equation
http://www.grc.nasa.gov/WWW/k-12/air...ges/drageq.gif


D=Cd*(1/2*p*v^2)*A

Then I used this link to find
past Canadian temperature pressure data for my area.

Then this calculator to calculate air density.

So using my quick calculated values for my Optima estimated Cd .28 (stock is .29 hybrid has .26) and frontal area of 2.2m^2 I compared the drag at -4C and 101Kpa and 25C 99Kpa at both 90km/hr and 110km/hr.

I found that at 90km/hr:
-4C drag is 25.18N
25C drag is 22.27N

At 110km/hr:
-4C drag is 37.60N
25C drag is 33.20N

So in the winter time going the difference from 90-110km/hr creates about 2N more drag than in the summer. Of coarse there are all the other reasons why winter fuel economy sucks too but I found these numbers interesting. I also found that last night after power washing the ice cunks off my car my avg to work was about 12% better.

Also just for fun I plugged in the EV1 and VW XL1 Cds of .18 and that would mean about 14.32 N at 90km/hr and 25C or 35% better.

This really motivates me to improve my cars aero!

Thanks for digging up the drag equation, but lets look at it as a percentage.

Since we are comparing only temperature differences to our drag, we can simplify EVERYTHING down to only a ratio of the air densities (rho). [That's because we want drag differences using the same car (Cd and A) at the same velocity.]

Rho1/Rho2 = percent drag

Unfortunately, calculating Rho isn't as easy but is necessary for both the drag equation and the equation I provided above. Using the Ideal Gas Equation of

Rho = P / (R*T) where P is air pressure (Pascals), R is a constant 287.05 (Joules/(Kilogram*Kelvin)), and T is temperature (in Kelvin).

You can get temperature and pressure from the weather channel or Weather.com. **However, this pressure is not the observed pressure but a converted pressure to compare to standard day sea level pressure. To convert it back to an observed pressure (that's what you want), use the following simplified equation which assumes constant temperature lapse rate and gravity:

Po = P*(1-0.0000225577*h)^5.257576

Where h is the elevation of the observation, Po is the observed pressure, and P is the pressure given from the weather service. You can easily get this figure from the Google search page by typing "[local airport] elevation". All units are metric. Convert in-hg to Pascals (Google search an online calculator).

Here is a live example for you.
In Huntsville, AL (where I live) Weather.Com says the temperature is 41 F and the pressure is 30.09 in-Hg. Let's teleport to Honolulu, HI. The temp there is 67F and the pressure is 30.06 in-Hg.

Calculate Observed Pressure!


Convert units!
Huntsville --- 41F -> 278.15K, 30.09 in-Hg -> 99597 Pa.
Honolulu --- 67F -> 292.59K, 30.06 in-Hg -> 101746 Pa

Plug-in Ideal Gas Equation for each City:
Huntsville air density --- 99597/(287.05*278.15) = 1.2474 kg/m^3
Honolulu air density --- 101746/(287.05*292.59) = 1.2114 kg/m^3

Ratio Time!
1.2474/1.2114 = 1.02969 or 102.97% or 2.97% increase in aerodynamic drag in Huntsville...

Which is just what I need... another excuse to move to Hawaii :D

-Ryan

Can this be generalized, all other factors besides temp being equal? 5*F = approx 1% change in drag?

A ballpark figure is just that... but...

My initial response is no because drag is a factor of density... and density is a factor of two variables: pressure and temperature.

Since there are two variables, it would be irrational to assume it is linear.

**KennyBobby's point drastically changes the outcome. KUDOS to KennyBobby**

Current Las Vegas weather compared to last night's Huntsville's weather constitutes 6.8% decrease in drag in Las Vegas compared to Huntsville at only a 6F difference in temperature.

-Ryan

Ryan et al,

You guys do know that the pressure reported from the weather channel, etc, is not the real pressure right? They report the altimeter reading pressure at the nearest airport, which is the barometric pressure that has been reduced to sea-level elevation. The pressure in HI would definitely be at sea-level, but Huntsville is about 700' above and the real barometric pressure is less. In Denver it is even lower...Temperature drops about 3 degrees per 1000' elevation, pressure drops about 1 inch Hg per 1000'.

Maybe another topic, but something else to consider is how the reduced pressure affects your mixture ratio, and also the impact of the relative humidity in the air on mixture.

:::SMACKING MY HEAD:::

You are exactly right. I can't believe I overlooked this... Thank you for bringing it up. Kudos to Kenny!!

To convert it BACK to observed pressure to calculate observed rho, use the following simplified equation (assuming constant temperature lapse rate and gravity):

Po = P*(1-0.0000225577*h)^5.257576

h is the altitude of the observation - and like Kenny said, assume the airport. You can gather the airport observation directly from Google's search page.

I'm going to edit the previous post to include this step.

-Ryan

I goofed and didn't fix the middle equation numbers. I updated the first pressure value, did the calculations in excel and posted the results. The 1.276 and 1.212 values are old and should be deleted... which I'm doing now.

Once again.. thanks!