|Single Speed Gears: 42:16 and 42:18 = 39:?||zmarke|
Nov 11, 2002 8:35 AM
|I am about to build up a inexpensive SS and want to keep the cost down. Since the crank I have already is a 53/39 setup (and I do not want to purchase a 42 at this point) I thought I could use the 39 small ring up front and then purchase a matching freewheel to get the correct gearing. So can anyone help me with the ratios?
42:16 = 39:?
42:18 = 39:?
|39:15 and 39:17 (nm)||Allez Rouge|
Nov 11, 2002 8:39 AM
|Here's how to figure that out for yourself||retro|
Nov 11, 2002 8:48 AM
|It's an easy formula: Divide the number of teeth in the cog into the chainring, then multiply by the diameter of the rear wheel in inches (use 27 for road, 26 for mountain.
So, on a roadie: 16 into 42=2.625, times 27=71 (the fractions are too small to matter)
That last number, by the way, is called "gear inches." It's not the most sensible way to figure gearing, but it works for comparisons like this.
|Actually, the rear wheel diameter ...||Allez Rouge|
Nov 11, 2002 10:25 AM
|... doesn't matter as long as it's not changing, too. All you have to do is cross-multiply to find the unknown value: 16 x 39 / 42 = 14.857, which rounds to 15.
You would need to deal with the rear wheel diameter if that is also changing. For example if the question had been, "I want to use my 39t chainring instead of buying a 42, AND at the same time I want to change from 700c wheels to 650c wheels. What size rear cog do I need to get the same gear inches?"
But I'm just being annoyingly and nauseatingly nit-picky here. Everything you said is correct.