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4

Generally, the tire pressure range marked on the tire is somewhat loose - the marketing department wants a big range since it will sell better, the legal department wants a small range so liability is lower and the engineers have something probably bigger than whats marked on the tire as safe. Your tire pressure should be high enough that you don't get pinch ...


2

There are three good reasons why professional racing teams care about weight Hills Margins of victory are often VERY small Racers run in packs for aerodynamic reasons. If you can out-climb an opponent enough to drop him from your pack on a climb, then he probably won't catch up on the flat - because the he is working against air resistance and the power ...


1

The following is junk science: Dynamic friction on moving surfaces comes into play - I'd put $ on it that the 8kg has better quality components and therefore bearing than the 11kg bike..... Nothing to do with weight, but in cycling weight and quality is inextricably linked Utter junk. Mechanical losses in a bike powertrain are tiny - and when you're ...


1

I am trying to understand the physics of bike weight. I have just made a switch from an 11Kg road bike to an 8Kg one (obviously there are many more factors and differences between the 2 bikes other than weight), and have noticed the benefit of reduced weight. However I am not sure I understand it. ..So you haven't noticed a difference due to weight at all. ...


4

First, I'll give you an estimate on the weight of a cardboard box. Then, you can read the side notes at the end of this answer to see why the weight is a relatively irrelevant quantity. EDIT: This link sells a bike box and lists the weight as 7.40 lbs. The rest of this answer gives you a way to estimate this, as well as tells you why this whole problem is ...


0

As usual, I think the best way to consider this is through energy; in moving from rest at the top of a hill (height h), the conservation of energy applies between potential energy at the top and kinetic at the bottom: Mgh = MV^2 + losses(due to aero and rolling resistance) therefore V = sqrt( gh - losses/M ) as losses are not proportional to mass, then ...


3

Thought I might add some extra comments to those very good and comprehensive examples of the aero v weight scenarios that Robert provided last year. In particular the dynamic scenario of accelerations on flat terrain, which is a little more complex than steady state cycling. Some might think light wheels would accelerate better than heavier aero wheels, ...



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