First post here, need a little help please.
I last rode this race in 1984, now 30 years later, at age 54, I am riding it again.
I am a rider, not a racer. The winners are sub one hour, I'll be VERY happy with sub two hours.
So here's the question: What is the gearing I need? Everybody and everything I read says I need 1:1 at a minimum. Equal teeth front and rear with 700C x 25 tires nets a 26.5" gear, so that means that every crank rotation moves the bicycle forward 26 and a half inches. (At least, that is what I understand, maybe my error is right there. Please read on.)
So, I am going to work gearing backwards.
I want to ride the 7.6 miles in under two hours. What is the AVERAGE gear I need to move forward 7.6 miles at an 80 cadence in two hours?
Here's my math, tell me where I am wrong, because the answer doesn't make sense.
63,360 inches in a mile (12 * 5280)
481,536 inches in Mt Washington hillclimb (63,360 * 7.6 miles)
80 cadence per minute
9,600 pedal strokes in two hours (80 * 120 minutes)
So divide 481,536 inches in the ride by 9600 pedal strokes, and you get 50 inches per pedal stroke.
The lowest gear on my bicycle now is 34 front and 30 rear which is 29.9", and I have am pushing hard to ride a 10% grade with that, being an old man and all. A 50" gear can not be correct, there is NO WAY I can climb that mountain, or any mountain for that matter, on my 34 tooth front and 18 tooth rear. On a 10% grade, I need my 28 or 30 tooth rear regularly.
Please tell me what I am doing wrong with the math. I already know I'm going to put a single 24 or 26 tooth chainring up front and leave the 11/30 in the rear, but want to understand the math.
And thanks in advance! (PS I can do this in four languages, but math is not my strong suit!