The cars are about 10 pixels long, the loop about 200 pixels in radius. Assuming the cars have a typical length of 4.7 m, this would give a loop radius of 94 m. To create a centripetal force large enough to counter-act gravity at the top of the loop, you need a speed vtop such that v2/r=g. With g=9.8 m/s2 and r=94 m, this gives a required speed at the top of vtop =sqrt(gr)=sqrt(9.8⋅94)=30,4 m/s=109 km/h=68 mph at the top.
However, the car would lose speed while while ascending the loop. Assuming that the kinetic energy of the car is converted to gravitational potential energy losslessly, and that no energy is added while ascending (giving some safety margin), it would need the kinetic energy at the bottom:
mvbottom2/2=mgh+mvtop2/2 =>
vbottom=sqrt(2gh+vtop2)=sqrt(4gr+gr)=sqrt(5gr)=sqrt(5⋅9.8⋅94)=67,9 m/s = 244,3 km/h = 151,8 mph.
This would be a recommended minimum speed on this road.
The cars are about 10 pixels long, the loop about 200 pixels in radius. Assuming the cars have a typical length of 4.7 m, this would give a loop radius of 94 m. To create a centripetal force large enough to counter-act gravity at the top of the loop, you need a speed vtop such that v2/r=g. With g=9.8 m/s2 and r=94 m, this gives a required speed at the top of vtop =sqrt(gr)=sqrt(9.8⋅94)=30,4 m/s=109 km/h=68 mph at the top.
However, the car would lose speed while while ascending the loop. Assuming that the kinetic energy of the car is converted to gravitational potential energy losslessly, and that no energy is added while ascending (giving some safety margin), it would need the kinetic energy at the bottom:
mvbottom2/2=mgh+mvtop2/2 =>
vbottom=sqrt(2gh+vtop2)=sqrt(4gr+gr)=sqrt(5gr)=sqrt(5⋅9.8⋅94)=67,9 m/s = 244,3 km/h = 151,8 mph.
This would be a recommended minimum speed on this road.