Show us.
Well Howard, let's take a look-see.
For aircraft design the historical D over q for a wheel in a streamlined fairing is 0.13. D over q has units of ft^2 ...although we usually divide buy the reference area and express it as a coefficient.
So lets say a faired wheel has a frontal area of 2.5 in^2, dividing by 144 we get 1.736E-2 ft^2.
The CD for each wheel is then (1.736E-2ft^2 x 0.13)/3.72ft^2 = 6.07E-4 (where 3.72 ft^2 is the reference area - $ - for a Brodack Vector 40 we used in the turn radius analysis...turns out it won't have an effect on the outcome but I wanted to put the CD in coefficient form.)
Next, since we are astute aircraft designers we add 20 percent for interference factors between the struts and fairings...7.28E-4 is the Drag Coefficient we use.
We know the drag is calculated as 1/2 x rho x V^2 x $ x CD <- you see here the reference area isn't really required to compute the drag. We divided by it and later multiplied by it.
Using .2308E-2 slugs/ft^3 for rho, and 79 ft/sec for V we plug and chug to get the drag per wheel.
(.5)(.002308)(79^2)(3.72)(.000728) = 3.90E-2 lbs drag per wheel. If you do the dimensional analysis you see we end up with slugs x ft/ sec^2, so that's a good check. Now multiplying by two and then by 16 to get ounces we end up with 1.25 oz drag from the wheels.
Let's say the wheels are 6 inches from the vertical CG, that's roughly 7.5 inch-ounces of moment.
And I should note, an unfaired wheel will have about twice the drag of a well faired one, so make that 15 inch-oz for unfaired wheels.
And actually, to simplify things we left the drag from the struts out and they too make a contribution. I'm too lazy to integrate over their length today. Interestingly, round wire gear have substantially less drag than aluminum fuse-mounted do.
Also, the wheel drag isn't constant. It decreases on inside turns and increases on outside turns due to the wing's flow field effects. Betcha didn't see that one coming!
Chuck