Originally Posted by
ryth
Curious what methodology you used for this math and to come to this conclusion? If there is a generally accepted formula for this sort of thing I'd love to know it.
I think he had an error in his math. But, if you are looking for a methodology, I dug out some notes I had saved on my computer from when I built up my LHT in May 2005. I weighed the bike with racks and heavy duty wheels, it weighed 14 pounds on the front wheel and 18.5 on the rear. With me on the bike, I had a total weight (me and the bike) of 90 pounds on the front wheel and 150 pounds on the rear wheel. I did not weigh it with camping gear, but I suspect that with gear I would be adding 20 pounds to the front and 40 pounds to the rear for a total of 110 pounds on the front tire and 190 pounds on the rear.
If you use the 15 percent tire drop method described here:
http://www.bikequarterly.com/images/TireDrop.pdf
the graph only goes up to 154 pounds. But, if you put a straight edge on the 37mm line (37mm is about 1.5 inches) and extend it out to about 190 pounds, that would result in me needing roughly 84 psig in my rear tire if I follow the 15 percent tire drop theory for the 37mm tire.
I may be taking the math farther than the author would like by extending the method beyond 154 pounds, but I decided to try it and it appears to work. My 700cX37 tires have a max pressure rating of 87 psig, so my rear tire at max pressure has about a 15 percent drop when loaded with camping gear. If however my max pressure rating on the tire was significantly lower than the required 84 psig for a 15 percent drop, that would suggest that the tire won't do the job.
For off of pavement, I however prefer a 2.0 inch tire. Also, I usually run higher pressures than the 15 percent drop unladen, but I think that the 15 percent drop is a useful calculation to see if a tire would be able handle the pressure and provide the support needed on a touring bike with load and rider.