I did a time trial with my college team this past weekend. The location is
11 miles away, and during the ride there, I got 2 flats! I managed to get
enough air in the two wheels, and managed to get there in time to do the TT
before everyone left. I ran a 20:19 for the 7-mile course, which was a lot
better than the 22:00 I had gotten the last time I tried it. It wasn't
until I got home and discovered that my tires were only inflated to 30F/35R
that I realized I could have broken 20 minutes. Oh well.
So I pumped up the tires to 120 later that night, and within a few minutes,
psssssssss. The 3-mm slit the glass made in the tire tread was too big for
the tube. So I cut out some water bottle and used duct tape for a boot.
Works well.
Today in class I was mulling over whether I should repair it, not because
I'm worried about the tire failing in the future, but whether I could
actually sew the casing back together. Then I thought about the kind of
tension the thread would be under if I sewed it. I remembered that I have
some Kevlar thread from a digital camera fix a year ago, and it was rated at
25lbs. I did some calculations here:
http://plaza.ufl.edu/phillee/tireslit.jpg
and determined that I would need one thread to hold at least 14.5 lbs to
keep the tire closed, given a 3mm slit parallel to the direction of travel
in the center of the tread of a 23mm road tire at 120psi.
Ignoring gross uncertainties such as actual tire size, the fact that the
slit isn't so much parallel as diagonal to the tread, am I anywhere near a
correct answer?
Ignore the secant and bisection methods in the picture... that was class
notes. The FBD in the bottom right of the pic is the pressure distribution
perpendicular to the tread. I wasn't sure how to calculate it though.
--
Phil, Squid-in-Training