Solar Dragster Calculations
The following are the theoretical governing equations of motion for a solar dragster.
You might think that the dragster would accelerate at a constant rate since the energy received from the sun is constant.
You could think this, but you would be wrong.
This is where relativity meets reality. The reality is that is takes ten times more energy to accelerate from
50 to 60 miles per hour than it does to accelerate from zero to 10 miles per hour.
The solution requires a bit of calculus, but the resulting equations are "relatively" simple. The equation is based on the
principal that the kinetic energy of the dragster at the end of the finish line will equal the amount of
solar energy that the dragster was able to collect over the 1/4 kilometer distance. While this formula ignores
friction and air drag, it at least establishes the theoretical limits.
The affect of air drag and rolling resistance can be investigated with the
Dragster_Physics.xls spreadsheet.
You will find that the faster you try to go, the less energy you will have to work with.
Variables:
S = Solar Energy, Watts. This is the power delivered to the vehicle.
S = STC Rating of Solar Panels in Watts x Actual Weather Factor x Motor Efficiency x Transmission Efficiency
t = Time in seconds
M = Combined Mass of Solar Dragster and Driver, kilograms.
x = Distance traveled, Meters
V = Velocity, Meters per Second
Governing Equations:
1. Solar Energy In = Kinetic Energy of Dragster
S t = (1/2) M V2
Therefore, velocity as a function of time, available solar energy, and mass equals:
V = (2 S t / M)1/2
Using Calculus we know that V = dx/dt, therefore, dx = (2 S t / M)1/2dt
x = ∫(2 S t / M)1/2dt = (2 S / M)1/2 ∫ t1/2dt = (2 S / M)1/2 (2/3) t3/2
Therefore, the distance traveled as a function of time, available solar power, and mass equals:
x = (2/3) [2 S / M]1/2 t3/2
Conversely, the theoretical time it takes to travel a fixed distance is equal to:
t = [(3/2) x]2/3 [M / (2S)]1/3
Sample Calculations:
Determine the time and final speed for a 100 kg (220 pounds) solar dragster with 2 meter wide by 6 meter long solar array with 20% efficient solar cells over a ¼ km distance. Neglect air and frictional drag.
Given:
Solar, S = 1000 W/m2 x 2 m x 6 m x 20% Solar Eff. X 80% Motor/Tran. Eff. = 1680 Watts
Mass, M = 100 kg
Distance, x = 250 meters (1/4 kilometer)
Theoretical time to go ¼ km:
t = [(3/2) x]2/3 [M / (2S)]1/3 = [(3/2) 250]2/3 [100 / (2 x 1680)]1/3 = 16 seconds
Theoretical speed at the end of ¼ km:
V = (2 S t / M)1/2 = (2 x 1680 x 18 / 100)1/2 = 23.3 m/s x 2.24 mph per m/s = 52 mph
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