On your 1 gear bike, would you tell me how many teeth you got on front and rear. Do you have freewheel or fixie? I'm trying to understand if my gear ratio is proper? I tried to do ratio calculation....I cant even read **** there. So thought id ask people here. My terrain is mainly flat. I have 52t front 20t rear, freewheel. but feel like it's not fast enough to fly in the wind. I just want it to be fast. I pedal fast as I can...its just not enough. Idk if I make any senses here.

Teeth on the front chainring divided by teeth on the rear cog. 52/20 = 2.6.

It sounds like your issue is that the gear is too low, where you're turning the cranks quickly and having a hard time keeping resistance on the pedals. For this, you want a higher ratio, either a bigger chainring or a smaller rear cog. A 52T chainring is already pretty big; you could make a large difference by using a cog a few teeth smaller in back.

For instance, if you switch to a 17-tooth cog, you'd have 52/17 = ~3.06, which is about 18% higher than your current gear. So you'd be going 18% faster for a given pedaling cadence.

wow. Ok. Now I'm learning. Thank you.

One more thing! About the size of cog, let put teeth number sizing aside and let look at the cog where we insert it in the wheel, are all size come in same or they comes in different size? Even though I'm looking for 17 t. How do I know my size of rear cog so that I can get smaller teeth but same size of the cog.

For more info:

What's your gear ratio?

Your gearing is very close to mine. I have 46 in front and 19 in back. From following this forum, I have learned that my ratio is too low for a "fast" cyclist. That's OK with me. I'm not very strong, and I have an upright handlebar, so I pay an aerodynamic penalty compared to a drop-bar bike. I use my bike for riding around town, commuting to work, going to the opera, or for easy recreational rides of less than 25 miles on relatively flat terrain, and I'm happy with it.

To calculate your gear inches:

1) Teeth on chainring, divided by teeth on sprocket.

2) Multiply the result by the diameter of the wheel, including tyre.

Example 1: 49 tooth chainring, 18 tooth sprocket, 700c wheel (roughly 28 inch diameter).

1) 49 divided by 18 is 2.72

2) 2.72 x 28 = 76.2 gear inches.

This means that the combined effect of the gearing and wheel size is the same as if you were directly pedalling a 76 inch wheel (a very big penny farthing or unicycle!)

Example 2: 43 tooth chainring, 17 tooth sprocket, 700c wheel (28 inch diameter)

1) 43/17= 2.53

2) 2.53 x 28 = 70.8 gear inches.

This means that the combined effect of the gearing and wheel size is the same as if you were directly pedalling a 76 inch wheel.

You need to factor in the wheel size because a 43/17 set up on a small wheel Moulton, for example, would be a very low gear. More sensibly, in these days of 700c and 650b wheels, and 26" and 29" MTB wheels, you can't just look at the teeth.

Or you could look at a chart.

I prefer to understand things from first principles where possible, rather than just to look them up.