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To calculate air flow capacity of a fan in cubic feet per minute (cfm): multiply the average air speed you measured in feet/minute (fpm) by the area of the fan face in square feet. (Area of circle =þ d2/4; where d = diameter in feet). Example: you calculated an 524.93 ft/min average air speed across the face of a 12 inch (1 foot) diameter fan. Air flow (cfm) = speed (fpm) * area (sq ft)= 524.93fpm*þ(1)2/1 sq ft =1048.78 cfm?

The fan in question is visible from this link:http://greglemond.com/#!/revolution

The diameter of the Fan is 30cm or 12inchs in diameter

The website claims the maximum airflow in meters/second^2 is 160 (in imperial 524.93ft) and the maximum newtons of resistance is 40N (I am aware that is probable not the correct phrasing?). All airflow increases in a linear fashion to this maximum point.

I fear I have got a basic fundamental wrong whilst trying to figure out the airflow capacity of this resistance trainer correctly as meters/second^2 and cubic feet per minute are not directly comparable. Can I compare the too figure or would I be better off converting the 160 meters/second^2 and cubic feet per minute to lbs/minute so I can compare this fan to others available on the market?

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    $\begingroup$ The unit of airflow is $\frac{m}{s^2}$? $\endgroup$ – Wojciech Mar 24 '14 at 20:02
  • $\begingroup$ yes I asked some one why this was in bicycle.stack and they said 'm/s^2 is the base unit for acceleration, so perhaps it is because the resistance (measured in newtons), varies as a function of the acceleration which you put into the pedals/trainer.' Others added 'the x-axis label should be read as (m/s)^2, not m/s^2.' $\endgroup$ – user95786 Mar 24 '14 at 20:09
  • $\begingroup$ It seems that you need some mass/volume units to measure airflow: en.wikipedia.org/wiki/Airflow (unless you use another definition). $\endgroup$ – jinawee Mar 24 '14 at 20:22
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If I understand the advert correctly the aim of the fan is to simulate the force the bicycle rider feels from aerodynamic drag. Aerodynamic drag increases with the square of the speed, so if you plot drag against speed squared you get a straight line. In the graph they show:

Drag

and the $x$ axis is speed squared i.e. $(m/sec)^2$ not $m/(sec^2)$. Also this isn't the wind speed through the fan, it's the speed of the cyclist (squared) so $160$ means the cyclists speed is $\sqrt{160} \approx 12.6$ m/sec.

Actually $40N$ seems a very low drag - is the drag at $12.6$ m/sec ($\approx 28$ mph) really only $4$ kg?

Anyhow, I don't think their article is at all concerned with the wind speed through the fan. The fan is just a convenient way to generate a load that varies as speed squared and therefore gives a realistic impression of drag.

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  • $\begingroup$ 4kgs does seem low after using this online calculator (though I have little experience with drag numers ):analyticcycling.com/ForcesPower_Page.html which states that the wind resistance is 9.8kgs which is featured in bicycles.stack answer: bicycles.stackexchange.com/questions/9938/… So if I wanted to build my own lemond revolution I would be best starting with the speed of a cyclist and his/her drag and find the equivalent resistance in a fan? $\endgroup$ – user95786 Mar 24 '14 at 21:42
  • $\begingroup$ @user95786: Yes, that's the way I would do it. You have two variables: (1) the size/shape of the fan and (2) the revolution speed i.e. the gearing you use. I suspect Lemond established the best values for these by experiment. $\endgroup$ – John Rennie Mar 25 '14 at 6:43

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