Modulus of elasticty, brittleness, strength, stiffness, elastic limit, are standard terms of the theory of elasticity. They are, however, more often than not misinterpreted when used in "technical" debates about bicycles. Many times they are abused as a "scientific evidence" of some natural feel like: "steel is softer then aluminium". Softness, by the way, is not a standard quantity in the theory of elasticity. It can be interpreted as a reciprocal quantity to stiffness. The measure of stiffnes of a material is its modulus of elasticity, or Young modulus, E, which is defined as a ratio of stress vs. strain. In that sense the common bicycle myth that steel is softer than aluminium is a total nonsense. It is not true that steel has relatively low modulus of elasticity, i.e. is a "soft material". In fact its modulus of elasticity (Young modulus) is the highest of all commonly used engineering materials: it is 210 GPa, which is 3 times greater than aluminium (with 69 GPa), 1.75 times greater than titanium (120 Gpa) and at least 1.4 greater than carbon fiber (150 GPa). You can find these figures in a table on Wikipedia page: http://en.wikipedia.org/wiki/Young%27s_modulus. Steel is the stiffest material to make a frame of, unless you are considering a tungsten or a diamond one.
It is perfectly OK with me if someone feels a steel frame to be softer than aluminium. But please don't try to prove this by misinterpretation of some half-understood theory.
P.S. I am not a "steel-hater". Steel is the best engineering material for most applications because of its strength, stiffness, ductility, fatigue strength, weldability and price - but certainly not because of it's softness.
Thursday 10 December 2009
Thursday 5 November 2009
Why is riding in mountains slower than on flat ground?
It may be obvious to most of you, but it wasn't to me. After all, there are always two sides of a mountain. While you are slower going up, you are much faster going down. The average should be equal to riding on flat. Hmmm, this calls to some math.
Let's say we are riding up and down a symetrical hill of road length 2S. We cycle up with average velocity v1 and go down with average velocity v2. The up and down parts have equal length S. What is our overall average speed v?
The enticing answer is v=(v1+v2)/2, but, as you may presume, the world is not that simple. Combining the formulae: v1=S/t1, v2=S/t2, t=t1+t2, v=2S/t, we have: v=2S/t=2S/(t1+t2)=2S/(S/v1+S/v2)=2/(v2/v1v2+v1/v1v2)=2v1v2/(v1+v2).
So the answer is: v=2v1v2/(v1+v2) and it's independant on length S. For example, if you climb with 10 km/h and descend with neck breaking 70 km/h, your overall speed is not 40 km/h as you might have wished, but only 17,5 km/h, as you will find out looking incredibly at your computer at the end of the downhill. The only reasonable way to increase the overall speed is to climb faster. If you climbed just 50% faster, with 15 km/h instead of 10 km/h, then the overall speed increases to respectable 24,7 km/h. If, instead of climbing half faster you descended twice faster, i.e. 100% faster, at the world record breaking 140 km/h, your overall speed would still be miserable 18,7 km/h.
The point of this story: instead of risking your neck at speeds where there is no margin of error, beter have an ultralight pack which will enable you to climb a little faster.
We'll make some more use of mathemathics. We first express climbing speed as a percentage f of descending speed, v1=f v2. The average overall speed is then v=v2 2f/(1+f), as compared to intuitive (and wrong) average speed (v1+v2)/2=v2(1+f)/2. The difference, together with the ratios of climbing/descending speeds from the above example (10/70, 15/70 and 10/140), is indicated in the graph.
Let's say we are riding up and down a symetrical hill of road length 2S. We cycle up with average velocity v1 and go down with average velocity v2. The up and down parts have equal length S. What is our overall average speed v?
The enticing answer is v=(v1+v2)/2, but, as you may presume, the world is not that simple. Combining the formulae: v1=S/t1, v2=S/t2, t=t1+t2, v=2S/t, we have: v=2S/t=2S/(t1+t2)=2S/(S/v1+S/v2)=2/(v2/v1v2+v1/v1v2)=2v1v2/(v1+v2).
So the answer is: v=2v1v2/(v1+v2) and it's independant on length S. For example, if you climb with 10 km/h and descend with neck breaking 70 km/h, your overall speed is not 40 km/h as you might have wished, but only 17,5 km/h, as you will find out looking incredibly at your computer at the end of the downhill. The only reasonable way to increase the overall speed is to climb faster. If you climbed just 50% faster, with 15 km/h instead of 10 km/h, then the overall speed increases to respectable 24,7 km/h. If, instead of climbing half faster you descended twice faster, i.e. 100% faster, at the world record breaking 140 km/h, your overall speed would still be miserable 18,7 km/h.
The point of this story: instead of risking your neck at speeds where there is no margin of error, beter have an ultralight pack which will enable you to climb a little faster.
We'll make some more use of mathemathics. We first express climbing speed as a percentage f of descending speed, v1=f v2. The average overall speed is then v=v2 2f/(1+f), as compared to intuitive (and wrong) average speed (v1+v2)/2=v2(1+f)/2. The difference, together with the ratios of climbing/descending speeds from the above example (10/70, 15/70 and 10/140), is indicated in the graph.
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