--calc for converting mph to foot-pounds--

  mph to foot-pounds conversion--

  5,280 feet per-hour = 1 mph

convert 5,280 feet per-hour, to feet per-min--
( 5,280 feet divided by 60 = 88 ft per-min )

convert feet per-min, to feet per-second--
( 88 feet per-min, divided by 60 = 1.46666 feet per-second)

feet per-second x number of miles per-hour = ft per-second( at said mph)--

feet per-sec calc--

( modifier)        (mph)
 1.46666     x     60     =   27.996 ft per-sec( at 60 mph)

  calc for foot-pounds of ( wind/object)--

                             ( convert mph)
    ( pounds)          ( to ft per-sec)
      weight       -      feet per-sec          =    foot-pounds, per-sec force is applied( at 60 mph)
      4,000       -         27.996               =                     3,972.004


 --in the case of wind, the weight of 1 cubic foot of air( 0.0807 lbs ), is divided by 36 = cubic-inches, to get the weight of 1 cubic-inch of air in lbs( 0.00224 lbs)--


   wind example--

          ( lbs)
        ( weight)          -      ( 1.46 x mph)              ( foot-pounds)
  1 cubic inch of air   -       feet per-sec        =     force per sq-inch
        0.00224                    1.46 x 10                  -14.59776

we have a negative #...not good( in pounds), so, we must utilize the "decimal putting system", to regain a whole #, this seems to be a two-put( a "math-birdie"), by converting to newton meters, we can get there...

 calc to newton meters--

   ( foot pounds)          ( modifier)         ( newton meters)
      14.59776         x     1.3558      =       19.79164

note--

  by dropping the "minus" in front of 14.59776, and calcing to newton meters, we get 19.79164 newton meters, if we then bang the decimal over to the left, two places, for both values in the problem, we get .1979164 newton meters--

   ( foot-pounds)   =   ( newton meters)
     -14.59776               -19.79164
 
two-put( for birdie)--

  ( foot-pounds)   =     ( newton meters)
     .1459776                   .1979164

  in this way, we now know the amount of force applied by wind, per square-inch, to any object, at 10 mph, if we divide this # by 10, we get the value at 1 mph( .01979 newton meters, per sq-inch)...as the numbers go up( larger mph and sq-inch values), we can crunch-back-up...to foot pounds( as is the case in a wind-mill)--

note--

   by counting the square-inches of a blade, and then plugging the force expected into the calc, we can get the expected amount of rotation( in foot-pounds, watts, ect...see my "windmill prop-size/pitch/horsepower/tork computation" post, for more details--

 (cs)

  best wishes, john kruschke--