Recently I have been trying to figure out a model of how aerodynamics affect energy consumption at higher speeds. I was looking up info on this when I found the page on the coefficient of drag.

So I decided to do a quick and somewhat dirty test. I used a huge parking lot near me that’s roughly a quarter mile long. It’s dead flat and it was wide open. I recorded two runs in the same direction. I capped the RPM to limit the board to 27mph for a little bit more consistency.

On the first run, I was in a very tight tuck. The second run I did in a normal riding position.

Run 1, tuck

Run 2, normal riding position

After I reached cruise speed, I drew a minimum of 16.1 amps while tucked, and 27.1 in a normal riding stance.

Sometimes when I’ve been riding for a long time I get pins and needles in my back leg. I turn it slightly inward for relief until I get to the next coffee shop.

To maintain steady velocity your thrust has to equal your drag. The drag equation is:

Drag = Cd * 0.5 * rho * V^2 * A

Cd is your drag coefficient. The coefficient of drag for a skydiver is 1.0 when horizontal and 0.7 feet first. Your tuck can’t improve your coefficient of drag by a lot. Assume it is a constant 0.8. V is your velocity. Your velocity was constant at 27 mph which is ~40 ft/s. Rho is the density of air. The correct English units for density of air are slug/ft^3 and the value is .002378.

With the torque constant of your motor(s) and the amps you were drawing you can calculate the torque you were producing. Multiply it by the wheel radius to get thrust. From that you can back calculate the area when you were standing and the area in your tuck.

To(rque) = Kt * amps
Th(rust) = R * To

R is the wheel radius. Kt is your torque constant.

Thrust = Drag

R * Kt * amps = 0.8 * 0.5 * .002378 slugs/ft^3 * (40 ft/s)^2 * A

A = 0.66 ft-s-s/slugs * R * Kt * amps

keep your units consistent feet and seconds

I went through that quick. I can’t vouch for perfect math but that is how you do it.

Edit: You probably want an efficiency factor on the torque depending on your power train. Divide the right side of the equation by your efficiency, E.

for that equation I believe to calculate the drag in newtons you have to use 1.225kg/m^3 for the air density, and “A” above- the frontal area has to be in square meters, and the V^2 velocity squared needs to be in meters per second. for the air drag in watts use V^3 instead of V^2.

for the thrust you have to multiply the motor torque in newton meters by the gear ratio to get the wheel torque, then to get the thrust in newtons, you have to treat the wheel as a lever with a 1 meter distance to fulcrum through which you are supplying a certain amount of newtons, and the other end of the lever is the radius of your wheel.

most of the equations are in this post:

^at the bottom of this post is a chart and spreadsheet which shows the estimated wind drag force in newtons, watts & pounds based on 0.75 drag coefficient and 0.6m^2 frontal area, and 1.225kg/m^3 air density. these numbers are roughly based on @MoeStooge ‘s reported power consumption during one of his 75mph attempts.

I do skate slower than most people here using a shorter deck for mobility on the train has the downside of being less stable at speed, but being light dude and not cranking the throttle too much I do get pretty good efficiency:

Was actually about 17 min ride but forgot to shut metr off when I got home, distance checks out on Google maps though and pretty typical numbers with my single drive setup.

This was the Max watt used getting to 63mph on a slight uphill grade with no headwind. Note the volt sag from the 10c lipos. 1v = ±3mph. This was our 5th and final pass getting Max Capps acclimated to a powered board.

Its uncertain to determine drag. The motor rpm was still accelerating when Max lifted the throttle. He crested but never leveled air speed to get a decent mph flatline watt measurement. Still a good real world of how many watts it takes to get there. Accel, Drag, Ele. loss, drive train loss and grade.