Gear ratio question. ugh i'm confused.

i dont think there would be a huge loss. you can freely spin a turbocharger with ur hand. i figured running the hot side of a charger through the auxilery belt to drag out the exhaust would be pretty nifty. maybe if there were some changes in design of the motor this could be useful at high rpms.

When you "freely spin a turbocharger with ur hand" you are hardly doing anything. A turbocharger does work when it's plumbed into an engine that provides a resistance. That, and when it's spinning at something like 100,000rpm...which is much more than you can hope to accomplish with your hand.

Again, you are underestimating how much brute force it can take to move air around.

so in the end, god i swear i've re-written this 4 times. so i'll keep it simple

more low end torque will increase mpg with smarter driving because...the engine is working more efficiently and using less fuel?

When you "freely spin a turbocharger with ur hand" you are hardly doing anything. A turbocharger does work when it's plumbed into an engine that provides a resistance. That, and when it's spinning at something like 100,000rpm...which is much more than you can hope to accomplish with your hand.

Again, you are underestimating how much brute force it can take to move air around.

i wasnt underestimating, i was making an observation, there are "fake" superchargers called prochargers that use the same design i stated only shove air into the motor,

i've been told the suction idea wont work so the idea is irrelevent.

so in the end, god i swear i've re-written this 4 times. so i'll keep it simple

more low end torque will increase mpg with smarter driving because...the engine is working more efficiently and using less fuel?

Yes, thats correct. But you are suggesting methods that require lots of power to achieve very little (like the 'reverse supercharger'). Spending a pound to save a penny so to speak.

Frankly smarter driving makes a bigger difference to MPG than technology. If you drive with a lead right foot, then you are going to have low mpg no matter what technology you use. If you drive with MPG in mind, you'll be slower (not necessarily by that much though) but have fantastic mpg.

i wasnt underestimating, i was making an observation, there are "fake" superchargers called prochargers that use the same design i stated only shove air into the motor,

i've been told the suction idea wont work so the idea is irrelevent.

What on earth is a procharger?

i'm lost on something though.

for a 20" wheel to travel one mile per minute it needs to rotate 2933 times right?

with all the gear reduction of the rpm it doesnt make sense.

like an engine at 2000rpms in a transmission overdrive gear of .83:1 and a differential of 3.75:1 the wheel is only rotating 642rpm

what am i not getting..

Redbelly98
Staff Emeritus
Homework Helper
A 20" diameter wheel has a circumference of 63" or 5.2 feet. So it's about 1000 revolutions per mile:
1000 x 5.2' = 5200' = 1 mile​
At 642 rpm, it takes about 1.5 minutes to get to 1000 revs or 1 mile.

A 20" diameter wheel has a circumference of 63" or 5.2 feet. So it's about 1000 revolutions per mile:
1000 x 5.2' = 5200' = 1 mile​
At 642 rpm, it takes about 1.5 minutes to get to 1000 revs or 1 mile.

did u get the circumfrence by multiplying 20" by pi?

how much torque is required to keep a car at 60mph?

like 3000lb weight, 0.30 coefficient of drag, 63" wheel circumfrence

All it is is a centrifugal supercharger, fundamentally no different to any other type of supercharger. Why did you describe it as a 'fake' supercharger?

Redbelly98
Staff Emeritus
Homework Helper
did u get the circumfrence by multiplying 20" by pi?
Yes.

jack action
Gold Member
how much torque is required to keep a car at 60mph?

like 3000lb weight, 0.30 coefficient of drag, 63" wheel circumfrence

$$T_{eng}=\left(\frac{1}{2}\rho C_d A_f V^{2}+ fW \right)\frac{r_{tire}}{\eta GR_{trans} GR_{axle}}$$

Where:

$$T_{eng}$$ = torque of engine (N.m)
$$\rho$$ = air density = 1.23 kg/m³
$$C_d$$ = drag coefficient
$$A_f$$ = frontal area of vehicle (m²)
$$V$$ = vehicle speed (m/s)
$$f$$ = rolling resistance coefficient = 0.015
$$W$$ = weight of the vehicle (N)
$$r_{tire}$$ = tire radius (m)
$$\eta$$ = efficiency of drivetrain = 0.85-0.90
$$GR_{trans}$$ = gear ratio of transmission
$$GR_{axle}$$ = gear ratio of axle