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What do you think? Give us your opinion. Anonymous comments allowed.
#1

imbartichello (06/27/2013) [+] (27 replies)
i wonder how fast you'd have to go to not fall off...but i'm to lazy to calculate it
#94

waitiknowthisone (06/28/2013) [+] (2 replies)
Speed required to make it:
let R= radius of loop
m= mass of the car
g= 9.8 m/s^2
centripetal force would need to equal gravity such that m(V^2)/R=mg
m's divide away...
V^2/R=g
multiply both sides by R...
V^2=gR
take square root of both sides and:
V= root(gR)
I'm guessing from the picture that the radius of the loop is 16.93 meters so:
V=(16.93*9.8)^(1/2)= 12.88m/s which is equivalent to 28 miles per hour...
yeah something has to be wrong there
let R= radius of loop
m= mass of the car
g= 9.8 m/s^2
centripetal force would need to equal gravity such that m(V^2)/R=mg
m's divide away...
V^2/R=g
multiply both sides by R...
V^2=gR
take square root of both sides and:
V= root(gR)
I'm guessing from the picture that the radius of the loop is 16.93 meters so:
V=(16.93*9.8)^(1/2)= 12.88m/s which is equivalent to 28 miles per hour...
yeah something has to be wrong there