If I have a motor that spins at 3000rpm at 1200 watts according to the label, how do I calculate the gears I need to slow down to 15mph. I'm just guessing that will still give me enough HP. I don't really want to calculate that now.

1200 watts is 1.6 hp (there are 748 watts per horsepower). That should be more than enough to move a light vehicle 15 mph on level ground. It might even be enough to climb a shallow grade. Need to know the mass of the vehicle and contents to say more. You have not given enough information to answer your question. Need to know the diameter of the wheel that is rolling on the ground. I am assuming that you want to know the gear ratio between the motor shaft and the wheel of a vehicle you are constructing.

formula needed Well I don't know the mass of the vehicle yet.... but it will be a 20" tire most likely. This why I need a formula. I know it cant be complex as you are trying to make it. Try 450 lbs on a 16 percent grade, very short, maybe only 20ft? with not much momentum. In that case how big of a motor would I need.

Ok. A couple of formulas for you. Speed of a wheel Velocity of car= Circumference of wheel * Rev/min of wheel. RPM of wheel= RPM of motor/Gear ratio HP to climb a grade: HP= Vehicle Wt (pounds)* Grade (for 16% use 0.16)* Car Velocity (Feet/Sec)/550 550 is a conversion factor from Ft-lbs/sec to horsepower. This Formula calculates the horsepower required to raise the weight of the car up the grade at any chosen velocity. It ignores the horsepower required to overcome wind resistance, rolling resistance of tires, friction losses in your gearbox. 16% is a fairly steep grade. Freeways max grade is 6%. 8% is a comfortable grade for a hiking trail. 16% is a steep hiking trail.

It's not as simple as you are trying to make it. Tire diameter * pi * rpm = speed, in inches per minute. Gear ratio = motor rpm / tire rpm. Power required to climb equals vertical speed times weight. This is in addition to the power required to overcome friction in the drivetrain, rolling resistance in the tires, and acceleration. 1HP = 550 lb-ft/s. 1.6HP is not enough for a 450# vehicle to climb a 16% grade at 15 mph. It doesn't matter how long or short the slope is unless you're going to get some speed up and slow down as you coast up the hill. The math is left as an exercise for the student.

what gear ratio do I need??? So I need to know what gear ratio I do need for a 1.6 hp motor to climb a 16% grade with 450 lbs. And that would be with a two stage gear reduction. Don't bother with the rolling resistance, not much there. The speed does not matter.

1.6HP = 880 lb-ft/sec. This means you can raise 450# at a rate of 1.96 vertical ft/s. On a 16% grade, that's a vehicle speed of 12.22 ft/s or 8.3 mph. Your 20" diameter wheel has a circumference of 62.8" or 5.24 ft, so your wheel is turning 2.33 revolutions/sec or 140 rpm. 3000 motor rpm divided by 140 is a gear ratio of 21.4:1. However, that's assuming no losses, and there are always losses and friction so you need to go slower... say 35:1 which would give you 5 mph. That would probably be a multiple stage reduction. One other thing to consider is that if your motor nameplate says 1200W, that's the electric power it draws. Its actual output will be less, possibly considerably less depending on the motor design.