Top Speed Of Motorized Mountain Bike?

for example, a normal bicycle has wheels that are 26 inches in diameter. The "lowest" gear ratio on the bike might be a front chain wheel with 22 teeth and a rear gear having 30 teeth. That means that the gear ratio is 0.73-to-1. For each pedal stroke, the rear wheel turns 0.73 times. In other words, for each pedal stroke, the bike moves forward about 60 inches (about 3.4 mph / 5.4 kph at a 60-rpm pedaling rate). The "highest" gear ratio on the bike might be a front chain wheel with 44 teeth and a rear gear having 11 teeth. That creates a 4-to-1 gear ratio. With 26-inch wheels, the bike moves forward 326 inches with each pedal stroke. At a 120-rpm pedaling rate, the speed of the bike is 18.5 mph. now that is human power. Add in a 5 HP lawn mower engine that spins at 2000 rpms and 6pounds of tourque PSI. there thinking that wind resistance and drag the bike will not Top 50. By my calc this nike could reach 616 mph. i think it could top 150 MPH

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wingstwo said:

1 HP on a motorized bike will generally give 20 to 30 mph. 5 HP should get 30 to 50 mph, depending on rolling friction, drive chain efficiency and aerodynamic drag.
Your calcs are wrong because the engine doesn’t have enough torque turn the crank directly. Engines are geared down to provide more torque, and as a result loose top speed.
Be careful if you mess with this stuff; be safe and don’t get hurt.
Edit: 62 mph is most impressive, I would question the accuracy of your speed measurement. It is possible though. To get highest speeds, use racing bike with high pressure tires rather than mountain bike, use efficient drive chain, and most key, severe aerodynamics like bike racers use: low racing position, special aero helmets, etc.
Edit again: to run vehicles, engines run through a drive chain that gears the speed down and the torque up.
Power required to propel your bike = force * velocity.
Power generated by an engine is
power = torque * angular velocity.
Torque is critical because your engine won’t have enough torque to push the bike, just as you can’t push your pedals hard enough to accend a steep slope in high gear.
Final edit:
You want calcs? Cool!
Your engine puts out 6ft-lb @2000rpm.
Power = torque * omega
= 6ft*lb * 2000 * 2*pi
= 75,400 ft*lb
= 75400/33000 = 2.3HP
For comparison, a typical B&S 5 HP puts out max power at 3600 rpm:
P = 5*33000 ft*lb/min
Omega = 3600RPM * 2*pi,
T = P/omega = 7.3 ft*lb
Now gears are not the magic you think. Using gears, power is conserved (conservation of energy or 1st Law of Thermodynamics). Since power is conserved, if you use gears to speed up the output from an engine up 10x, the drive chain now has only 1/10 the torque (neglecting friction, which makes things worse).
To apply conservation of energy for a typical mountain bike, drag is about 12 lb @30mph (or 44 ft/sec):
Power = F*V = 12lb * 44 ft/sec = 528 ft*lb
divide by 550 ft*lb/sec to convert to HP:
528 ft*lb = 0.96HP required from engine, neglecting friction.
Assuming all drag is aerodynamic for simplicity, power required is proportional to V^3, so going from 30mph to 60mph requires 8x the power or 7.7HP. Drag increases by V^2, so 4x12lb = 48 lb. Doing 120 mph, 4x the 30 mph, requires 4^3 or 64x the power, or 62.4 HP, and a force of 192 lb. Using F = P/V, 5 hp or 2750ft*lb/sec / 160 mph or 235 ft/sec, = 11.7 lb. Since some 350 lb is needed to push your bike to 160, obviosly it’s not going to go that fast.
To easily do the calcs yourself, google an online bicycle calculator. An OK one is listed below, but there are many others.

# 17 February 2010 at 6:03 am

Angela D said:

Utter rubbish!
If what you claim was true we’d all be blatting around on these things instead of motorbikes.
In reality bikes getting to that kind of speed need around 80bhp or more.
You haven’t factored any real world forces into your equation at all….in short, ain’t going to happen.
Classic engineer.

# 17 February 2010 at 10:58 am