--calculating lateral tork/speed increases-per apex encounterd--

  this is the concept i had last night, and lost...due to not finding a piece of paper, with some math i was working on a few days earlier, that  i needed to calc this prob right then--

a rant--

  i got this one...just need to find all my stuff, and post it all at once, this way someone reading over my shoulder has to try and figure it all out for themselves( it's a terrible world sometimes, and someone might want to up-load some of this silly stuff to a copyright website...you never know), so they can retire in the bahamas( the numbers do need to crunch-out correctly for that)...joke--

  i feel better--  :o)

  anyway, this concept is also known a "centrifugal force", and i have loved this one since i was young...calcing the total amount of force, for a given weight, at a given speed, and knowing, without doing any work in the lab...( crunching it on paper, and having it be just right, in real-world practice)...is the fun--


  lateral tork/foot-pounds--(centrifugal force)--

( laying it all out, long-hand)

--sq inches  %  weight  +  weight + angle - mph x 1.46  =  distance( shot-put/sling-shot)--

   note--

(  "lateral-tork principal"     )
 speed x weight - 1 =  force 



example--

    (sq inches)                                           ( mph x weight -1)          (    ft per-sec  )
   (  % weight)  +  (    lbs  )   +                    ( foot-pounds)      =     ( force is applied)
   (    drag     )  +  ( weight  )+  ( angle)  -        ( force)             =         distance
  1000%2000  +     2000     +     45  -             1999               =            46.5

   the above, shows a 1000 square-inch object, weighing 2000 lbs, turning a 45 degree corner, at 1 mph...generating  46.5 feet per-second, that the force is applied( 1999 foot-pounds)...the calc is complete...happy-time...dancing is permitted( depending on your venue)--   :o)

 ( "acceleration/quantification principal" )--
 
      ( feet)                         ( miles per hour modifier)
  --distance % seconds =  feet per-sec force is applied--

 "thrust/yield calc"--

example--

   ( ft per sec)                                          ( feet)
 (force is applied)         ( seconds)        ( distance)
       46.5                x         60         =       2790  

 this shows the total distance gained from 1999 foot-pounds of force being applied to the 2000 lb object mentioned above, with 1000 sq-inches of it's surfaces, at the forward face of travel( drag), going a total of 2790 feet after 60 seconds of thrust applied--


  if we use "distance to geo-sync math" within my "distance-to-geo-sync" post...i feel we can obtain mor data by inverting the "lateral-tork calc", to find all the values present within the situation being studied( see distance-to-geo-sync post)--

more later--   :o)