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Post Info TOPIC: Why they could ride so fast 為什麼他們能騎得那麼快?


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Why they could ride so fast 為什麼他們能騎得那麼快?
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I have noticed that the elite cyclists could ride much faster than the ordinary people. According to physics doubling the speed means quadrupling the air resistance so by riding so much faster their energy output must be even a lot higher. How could they manage to do that? Only by stronger physically or is there any secret to their riding skill?


我注意到精英單車運動員能比普通人踩單車的速度快很多。根據物理學兩倍速度等於四倍空氣抵抗因此他們的能量消耗必定是更加厲害。他們怎麼能做得到? 只憑更強體能或有是任何秘密騎乘技巧嗎



-- Edited by cyclingnut at 20:54, 2005-05-23

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Both. That is why we need to train hard and also need to learn how to pedal efficiently. There have been a lot of studies on this subject which is often considered as the single most difficult technique to master.


Studies include:


(1) pedalling with longer or shorter crank is more efficient?


(2) sitting more forward or backward is more efficient?


(3) pedalling a lower gear faster or pedalling a higher gear slower is more efficient?


(4) is pedalling a circle the most efficient way? (it is commonly agreed to be more efficient than pedalling square)


(5) is the body position the lower the better?


(6) is pushing (downstroke) and pulling (upstroke) at the same time better than pushing with one foot and unweighting with the other?


(7) is climbing while seated better or standing better?


(8) which ankling technique is better, toes down, heels down, toes down on downstroke (and toes up on upstroke) or heels down on downstroke (and heels up on upstroke)?


(9) is there any other technique to push oneself further (alternate the foot with harder effort)?


There seems to be no firm answer to many of the above questions yet. I will try to dig out what I have read and share with you later.


  



-- Edited by kwyan at 10:36, 2005-05-24

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都話有冇中文lo~!睇唔明啊


我估佢地成日都有練ga



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對不起,只因中文輸入需時,會盡快...

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中文版本


 


兩者都有。 那是為什麼我們需要努力地訓練以及需要學習該如何有效率地踩踏板。在這一個踩踏板的題目上已經有許多研究並常常被當作為單一最難掌握的技術


 


研究的項目包括:


 


 (1) 曲柄的長短對踩踏板的效率有何影嚮?


 


 (2) 坐位的前後位置對踩踏板的效率有何影嚮?


 


 (3) 快速地踩踏一個較低的齒輪比或慢慢地踩踏一個較高的齒輪比何者更有效率?


 


 (4) 圓型的踩踏方法是否最有效率?(普遍同意比方形踩踏方法更有效率)


 


 (5) 是否身體的位置伏得愈低則愈有效率?


 


 (6) 是否採用一腳推時(向下行程)一腳拉(向上行程)的踩踏方法勝於一腳推時和一腳吊(減磅)的踩踏方法?


 


 (7) 是採用坐姿還是採用站姿的爬坡方法較好?


 


 (8) 腳踝技術比較好的方法,是足趾向下推、腳跟向下墬、向下行程時足趾向下推(和向上行程時足趾向上挑)、或是向下行程時腳跟向下墬(而向上行程時腳跟向上揚)?


 (9) 有任何其他的技術可更進一步地用盡自己嗎? (用交替重腳法?)


 


 許多上述的問題似乎仍然沒有絕對的可靠答案。 我將稍後試著搜出我曾見過的論述與大家分享。


 



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根據物理學兩倍速度 等於四倍空氣抵抗力.同時輸出功率更為前者八倍.


空氣阻力相對速度二次方,而輸出功率是相對速度三次方關係.


    功率(P) = 力(F) X 速度 (V)



-- Edited by TH CHAN at 14:11, 2005-05-29

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According to the data that I have read (see http://www.coe.uky.edu/~brian/pdfs/ejap-2003.pdf) doubling the speed of cycling will not require increasing the power output as much as 8 times.


Speed                    Power


4 m/s (14.4kph)     23.1W (the power for riding on


                                         cycling track, barely enough


                                         to lit up some headlight)


8 m/s (28.8kph)     112.2W (the power for an easy ride)


12 m/s (43.2kph)   333.2 W (now I know how much 


                                            cooling I need for an


                                            hour's ride on the trainer)


 


Increasing the speed from 4 m/s to 8 m/s doubles the speed, but the power output is only 4.86 times, not 8 times.


 


Air resistance is roughly proportional to the square of speed but the internal resistance is probably more like a linear relationship. Total resistance come from air PLUS internal resistance (hub bearing etc.) and at contact between tyres and roads, so the power output is the SUM of air resistance which is proportional to the square of speed and other resistance which is proportional to the speed, and the total is only slightly more than the square of speed.


根據資料,踩單車的兩倍速度並不需要八倍的能量輸出。


將速度由每秒四公尺提升至每秒八公尺,兩倍速度但能量輸出是4.86倍而已,衹是稍高於四倍。


空氣阻力相對速度二次方,但其他阻力如輪軸的軸承等,相對速度一次方,而總輸出功率是空氣阻力和其他阻力的總和,故衹是稍高於速度二次方而小於速度三次方。



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單車行使時主要阻力是滾動阻力+空氣阻力.在底速行車時滾動阻力佔總阻力比例較高,


但高速行車時空氣阻力佔總阻力比例會相對題高,例如將速度由每秒六公尺提升至每秒十二公尺,


兩倍速度但能量輸出已是六倍多,速度越高空氣阻力佔總阻力比例越大,就越接近理論所示.



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-- Edited by TH CHAN at 15:37, 2005-06-02

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The total power of cyclist Pt= power of roll resistance Pr + power of air resistance Pa


So   Pt = Pr + Pa


Pr = roll force x speed    V = Speed ( meter / second)


Pa = 0.63 x C x A x V^3     C  is shape index is take 0.8


A = area (meter square)    


I refer to eiap2003.pdf result If  take


roll force = 3N  C=0.8 A=0.35


Speed         Pt        Pr       Pa       By formula Pa


   4           23.1      12       11.1          11.3


   6           55.3      18       37.3          38.1    


   8          111.2     24       87.2          90.3


  10         202.1     30       172.1        176.4


  12         333.2     36       297.2        304.8


The result is similarity to calculated



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RE: Why they could ride so fast 為什麼他們能騎得那麼快?
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TH CHAN wrote:


  根據物理學兩倍速度 等於四倍空氣抵抗力.同時輸出功率更為前者八倍. 空氣阻力相對速度二次方,而輸出功率是相對速度三次方關係.     功率(P) = 力(F) X 速度 (V)-- Edited by TH CHAN at 14:11, 2005-05-29


最初的題目是空氣抵抗力與速度的關係,倒沒有注意後來說的是輸出功率,原來大家都同意空氣阻力是速度二次方關係,真是好笑。


希望遲一點有時間再探討原有的題目:踩車的技術與效率的關係。



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