[JRMichler]

I have not removed my radio antenna, which was commented on in at least one thread. Now I know why because I finally did the calculation for drag from my radio antenna. Here's the calculation:

INPUTS
Antenna diameter 5/32"
Antenna length 29"
Air density 0.075 lb/ft3
Viscosity of air 3.8E-7 lb-sec/ft2
Speed 80 ft/sec (just under 55 MPH)

CALCULATIONS
Reynolds number = 0.156" / 12 in/ft X 0.075 lbs/ft3 / 32.2 ft/sec2 / X 80 ft/sec / 3.8E-7 lb-sec/ft2 = 6400

Cd for a cylinder at Re 6400 = 1.0

Drag = 0.5 X Cd X A X rho X V2 =
= 0.5 X 1.0 X 0.156" / 12 in/ft X 29" / 12 in/ft X 0.075 lb/ft3 / 32.2 ft/sec2 X 80^2 ft2/sec2 = 0.234 lbs drag from the radio antenna

At 80 ft/sec, this is 0.234 lbs X 80 ft/sec = 19 ft-lbs/sec = 0.034 hp (1 hp = 550 ft-lbs/second). The radio antenna air drag requires 0.034 hp (0.025 kw)

I averaged 35 MPG last year. If we simplify by assuming all driving was at that speed, then fuel flow was 55 MPH / 35 MPG = 1.57 gallons/hour = 9.4 lbs/hour. Power is estimated by assuming 0.5 lbs/hp-hr, so 9.4 lbs/hour / 0.5 lbs/hp-hr = 18.857 hp. Removing the radio antenna would reduce this to 18.857 - 0.034 = 18.823 hp. The resulting estimated gas mileage would be 35 MPG X 18.857 / 18.823 = 35.06 MPG.

I would gain an estimated 0.06 MPG. I drove 10,588.6 miles last year. That would be 302.53 gallons at 35 MPG, and 302.01 gallons at 35.06 MPG. The radio antenna cost me 0.52 gallons of fuel last year. The actual savings would be less because part of my driving is at slower speeds. I will pay that cost in order to listen to the radio once in a while.