I may have got this wrong but the
force on the rider of air drag is proportional to the
square of velocity (through the air).
The
power needed to overcome air drag is proportional to the
square of velocity through the air and then multiplied by the actual velocity of the rider. So in still conditions: power needed to overcome air drag is proportional to the
cube of velocity.
https://www.gribble.org/cycling/power_v_speed.html
"One of the scary implications of this equation is that at high speed, the power you have to produce is proportional to the
cube of your velocity. So, to increase your speed by 25%, you need to nearly double your wattage!" [1.25^3]