The usual measure of aerodynamic performance is the coefficient of drag. As a measure of the slipperiness of the shape, that's fine, but it isn't exactly relevant to consumers.

"Nice car, how much aero drag does it cause per frontal area square foot?"

- No one

More relevant for comparisons would be area * C_d. Or something like "brake horsepower needed to mix air at 60 MPH". That number would be immediately comparable to rolling resistance, cabin electronics, and numbers for other vehicles with similar interior space.

Consider, for example, that motorcycles have awful coefficients of drag, because it's such a complicated and messy surface that air flows over. They have great mileage, though, because of the size. A larger motorcycle might have a smaller drag coefficient but more drag.

So what is the frontal area going to be? I read somewhere that it's about 20 square foot, which is what you would expect of a small car. Fair enough, in that case the comparisons to Model 3 etc. are reasonable. It would also make the math work out in terms of battery pack size and estimated range (0.1 kWh/mile).

(In a recent interview though, someone mentioned that the vehicle has the drag of a 12 inch cube flying on the road. In that case, and assuming 20 square feet, the C_d would be 0.05. Which it obviously isn't. Maybe he meant to say 20 inches, or he only accounted for the body and excluded the wheels, or something.)

Anyway, drag at a given speed is more interesting to me. At highway speeds, specifically, because that's where the big energies are normally spent, and where efficiency and range matter.

Why? Seems to me the drag is only part of the equation. Your question means less to me than the drag coefficient. Never heard that asked before for any car.

What people really want to know is its energy efficiency under various conditions. We already have been told that simulations show the Aptera can go 10 miles on the EPA test cycle using only a single KWH of energy. Wow! I remember seeing an estimate of how much energy it will take to drive faster than the EPA test cycle, but I don't remember where I saw it.

I'd really like to know how much energy it would take for long distance traveling at a steady 65, 70, 75, 80 or 85 mph. Anyone got those numbers?

There is an rough efficiency calculation here: http://aptera.nu/?p=67

Unfortunately, Aptera has only published the drag coefficient, which is useful for optimizing a vehicle of a given size, but does not tell you the actual drag without the effective area, which they have not published yet. If we knew the area, then we could easily calculate the energy needed at higher speeds where the drag dominates .

Here is a method for Poor Man's integration that only requires a $25 scale:

If anyone wants to make their own estimates of total drag, and you don't know the frontal area, just print out a front view image of the item on a piece of paper. Weigh the paper on a good scale. Cut out the shape with scissors and weigh just that piece. Divide the shape weight by the paper weight. This gives you a fraction.

Example: say I'd printed a 1:6 scale front view image of an Aptera on an 8.5" x 11" piece of paper. Pretend like the paper weighs exactly 10 grams (for simplicity's sake). Say the 1:6 scale piece weighed 5 grams. That gives me a fraction of 5/10 or 1/2. Multiply that fraction by the area of the paper, (which is 8.5*11 = 93.5 square inches or 0.6493 square feet)--that number is .46.75 square inches of paper. Next multiply that value by the reciprocal of the scale factor, or multiply by 6, which gives the fully scaled area in square inches. Multiply that by 144 (12*12) to get square feet. Now times that frontal area by Cd to get the effective pure drag area.

I didn't have time to work out the actual math to reflect the case of the Aptera which as Cd 0.11, frontal area ~19.1 ft^2 and total drag about 2.10 ft^2 or about 17.4 inches * 17.4 inches effective cross sectional area. but the method works with surprisingly good accuracy. I've gotten this method to work to better than 99.5% accurate when I blow the image to take up most of the paper, and cut the shape precisely. Poor man's integration.

Last I heard Aptera was quoting a Cd of 0.13. How the EPA calculated Cd and effective area from frontal photographs some years ago is at https://nepis.epa.gov/Exe/ZyPURL.cgi?Dockey=94003COV.txt

I thought I saw a 0.11 Cd quoted somewhere. You may be right. In any case, it's the total Cd *times* Area that matters most, but most people falsely only pay attention to Cd. I say this because vehicle functionality and usefulness is somewhat divorced from its size, *especially* in the case of electric motor-in-wheel designed vehicles with clever seating and packaging.

More import is what is the constant watt draw at 65mph?

6500 watts or some other number?

It's in Biff's link above. Quoted here:

"The power necessary to overcome the drag is defined as follows:

With:

Pd = power in Watts

ρ = density of the air in kg/m3

v = speed in m/s

Cd = coefficient of drag (unitless)

A = the frontal surface area in m2

According to

this source, at IUPAC standard temperature and pressure (0 °C and 100 kPa), dry air has a density of 1.2754 kg/m3. With the earlier mentioned assumptions of no wind and driving speeds of 100 km/h, this leads to the following power for overcoming drag:Let’s use this power to calculate the energy per unit of distance (Wh/km or Wh/mile), that’s commonly used. If the Aptera uses

3731 Watt, for an hour, driving 100 km/h, it will have used 3.69 kWh over a distance of 100 km or 62 miles. This breaks down to only 37 Wh per km, or 60 Wh per mile.Note that the efficiency decreases at larger speeds, since the drag force increases quadratic with speed. So when traveling say 150 km/h (91 mph), the energy to overcome drag more than doubles to 84 Wh per km, or 135 Wh per mile."

That doesn’t match the stated highway efficiency of the vehicle which is rated at 10miles per kwhr

which means the aero portion is purely academic and of no use considerin all the other power wasters like RR and even vampire draws like screens, bms and pack management

a real world measure of instantaneous usage at 65mph is needed to know the real worl range

I'm not sure why you say the aero portion is academic. Aptera has said that they have <100Wh/mile for the EPA highway cycle which has an average speed of 48mph, I believe. So the aero drag is about half of the total power draw at 50mph and about 3/4 of the total power needed at 75mph.

Someone please double-check my simple math

@Biff :based ONLY onUsing 100Wh/mi at 48mph avg, with half of that aero, that's 50Wh aero.

Let's say we're having a good day, using 100Wh/mile at

50mph (for easier math with his last sentence)...The other 50Wh stays relatively constant regardless of speed (lights, screens, BMS, RR, etc).

If the power draw

is 3/4 aero, that means the "constant" 50Wh is 1/4 of the total, so the total is 50 x 4 =at 75mph.200Wh/mile, 5mi/kWh, 1kW "instantaneous usage"That would mean

, so instead ofat 75mph you get HALF the range250/400/600/1000 miles, it's

125/200/300/500 miles.

It doesn't quite scale that way since we are looking at energy used/mile. The non-speed dependent stuff (HVAC, electronics, lights, etc) is constant with time, not miles, so you use 2/3 the kWh per mile at 75mph vs 50mph for that stuff. The rolling resistance does have speed dependence though, so it gets a bit complicated. I had to make an estimate for the ratio of RR vs constant in the remainder kWh after subtracting the drag, so a lot of uncertainty here. What we need is better numbers for the various losses, then we can calculate kWh/mile for whatever speed. We do know that Aptera will have a much smaller falloff in range with speed than most EV's.

Also, the estimate for the projected frontal area is higher than for the Aptera 2e, which I don't understand, so we might be overestimating the drag contribution to the kWh/mile used.

All good points, but it's probably in the ballpark.

Someone could run the steady-state formula with the numbers above (0.13 Cd x 20 sq ft, etc), then tweak the non-aero if needed to match the 100wH/mile at 48mph, & then do the calcs for 65, 70, 75, 80 & 85 mph.

The fact our data is for 48mph @75F no wind with accel/decel cycles makes the mathematical side academic

Ford used a mathematical model on their CMAX and was sued, most decent companies validate off the real world And add a fudge factor for temperature and weather.

If Aptera wants to claim a 1000 mile range on a vehicle class that is exempt from EPA MPG certification they really better be using the pants down realistic highway range like Toyota and Honda do Or people will be disappointed.

From previous automotive experience I can say a 30% loss of range is very realistic off the EPA highway metric in 60F 65mph operations which are very normal conditions .

amd the EPA allows vehicle makers to legally down rate their expected MPGS/range on the Monroney Toyota and even Hyundai have been know to do this