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-   -   Broken Presta valve (https://www.bikeforums.net/singlespeed-fixed-gear/613580-broken-presta-valve.html)

elTwitcho 01-08-10 10:12 PM


Originally Posted by happypills (Post 10240074)
Here's something to keep in mind..... the maximum psi includes you on the bike too in addition to the air pressure in the tire already.....

I believe you're mistaken

carleton 01-09-10 01:48 AM

http://www.treefortbikes.com/imgDsp....staadapter.jpg

$2 at any bike shop in the US

http://www.treefortbikes.com/542_333...a-Adapter.html

Problem solved. NEXT.

bigvegan 01-09-10 04:58 AM


Originally Posted by elTwitcho (Post 10245464)
I believe you're mistaken

Seriously. Otherwise I'd have to invite a friend over to pump up my tires while I sat on the bike every time, and that would be tedious for all concerned.

TejanoTrackie 01-09-10 09:05 AM


Originally Posted by bigvegan (Post 10246061)
Seriously. Otherwise I'd have to invite a friend over to pump up my tires while I sat on the bike every time, and that would be tedious for all concerned.

Dang! I guess I need to start making friends. :(

rogwilco 01-09-10 11:35 AM

How much does the pressure in the tires change with the rider sitting on the bike/not sitting on it anyway?

Scrodzilla 01-09-10 11:38 AM

Everyone's weight varies but it averages around 45,000 psi.

PlatyPius 01-09-10 11:42 AM


Originally Posted by carleton (Post 10245938)

Not at any bike shop, no. They are 94 cents + tax ($1.00) at my shop, and several others in the area.

elTwitcho 01-09-10 12:41 PM


Originally Posted by rogwilco (Post 10246905)
How much does the pressure in the tires change with the rider sitting on the bike/not sitting on it anyway?

I imagine that depends on things like tire volume, how stiff the sidewalls of the tire are and the like, but I'd be pretty confident that the answer is not a significant amount at all

TejanoTrackie 01-09-10 12:51 PM


Originally Posted by rogwilco (Post 10246905)
How much does the pressure in the tires change with the rider sitting on the bike/not sitting on it anyway?

It's inversely proportional to the cube root of the Q-factor.


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