--artificial gravity( a functional system)--

  a few "heavy" words about space--

   some of those out there that read my blog may have noticed the blurb with the solution to the puzzle pertaining to "how can i guzzle coffee in outer space??"on a different post i made earlier...well, you need a good person making the stuff to start with( not everybody's coffee is the greatest, you need to validate quality standards there...), anyway, here's the gimbals on a cup-a-joe in the "outer-limits"--


  if you are familiar with star-trek( the first series)...you know about the enterprise, a ship with a "saucer" as the main "bridge", and a "tri-hull" with two engines, and a central engineering section...the saucer part is cool, if the saucer spun...on a bearing where it attached to the "tri-hull", essentially the floor of the bridge would be the edge of the huge saucer, like the bottom of a hamsters wheel, except a huge one, that was sealed from the vacuum of space...

 the thicker the saucer, the wider the edge, and the larger the living space, the crew would be essentially living and performing duties while walking sideways, on the edge of the saucer( on the inside of it)...if a landing in gravity was needed, atmospheric landing seats could be mounted on the walls( the other floor, in atmo...with gravity), getting in might be a tad weird...the saucer would need to stop spinning slowly, and in "zero-grav", the person could get in a "landing chair"...with safety belts--

  i feel we are a long way from this, but a person could have a nice bachelor-pad in space on the international space-station...in case we wanted to host an "intergalactic-kegger"( humor...kind-of, we could if we wanted, with this system)...anyway, essentially i suggest that we send up a "ferris wheel", in pieces, and enclose it in space...afterward, we could fit an electric motor to drive it...with solar panels to charge the battery to turn it...i will post the needed math to derive the necessary rotational speed to achieve gravity with a wheel of-a-certain size, later...for now, we could just send up someone's bathroom scale that's decent, and turn the wheel faster and faster in small increments, until the man/women in the space-suit weights the same as he does here on earth...

  so, why, other than comfort, should we do this?? there is a danger...if you jumped off the floor in an artificial gravity wheel you would find a wall, quickly, and at a harsh speed( once you lost contact with the floor, the wheel would turn under you, a small distance), well this would be a worth-while hazard to gain the ability to stay in space, indefinitely...as i suggest that the human body requires a central gravitational pull, towards the feet, to digest correctly...see my "digestive bio-mechanics" post, in this blog, for more about this issue...


  the math( derived from my "lateral-tork post")--

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

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

  i feel better--  :o)

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

   and so…

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

--( laying it all out, long-hand…and “mathematically throwing-down”)--

note #1--

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

note #2--

(  "lateral-tork principal"  aka “centrifugal force”   )
   weight  -   1   x    speed  =  force <<<-----a “rough estimate” as it does not account for “sq-inches”…

example( lateral-tork)--

 ( weight - 1)  x  ( mph)   =   ( foot-pounds)

   ( 2000 - 1 )     x       2       =          3,998

      1999

Note--

   Sometimes the way you “group” the terms is the key to a working formula, in the problem above, the “grouping” is “ ( 2000-1) = 1999 x 2 = 3998 “…it sounds silly, but the problem has different out-comes if you work the numbers incorrectly…in the calc below, we can see that if you input 1000 % 2000 + 2000 + 45 - 1 x 2000 – 1 into your calculator, you will not get 46.5 , you will instead get “44.5”…a value that is not the correct answer--

example( “lateral-tork interposed”)--

   (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        -    ( 1 x 2000 - 1)         =              46.5

                                                                                    1999

   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 of forward movement, after “sling-shotting” around the corner, from the initial force that was applied( 1999 foot-pounds)..

--the calc is complete...happy-time...dancing is permitted( depending on venue)--        :o)

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

Example--

5,280 %  60 =  88 feet per-sec, per sec

   since we have derived the distance value above, we can now utilize “distance to geo-sync math” to get the “lateral-force/tork” turning the corner generates on the object--

       ( pounds)                        ( feet)                          ( foot-pounds)

        ( weight)    -   1    -     ( distance)      =        ( of force per-second)       

          2000                               46.5                                   1952.5

   The calc above displays the pressure, per-second, that the object would be exerting against a rope that was tied to the object, from the central point of the vector…or, inversely, the pressure that the object would exert against the floor of a “squirrel-cage”, aka “centrifuge”( like the ones hamsters enjoy spending time in)…meaning, that if the corner was continuous, that the pressure( gravity/g-forces), would also be continuous, and provide gravity in space…this indicates to me , that rotating a huge ferris wheel at a rate of 1 mph, gives a person partial gravity in space( although I feel that they would be a tad lighter, at exactly 1 mph)…I will repeat the calc below, without the drag, to get good numbers for the “gravity module” problem, but with a 36 degree angle( the angle of a circle)--

example( “lateral tork interposed” without drag)--

--( weight + angle - mph  x  weight -1  =  foot-pounds of force)--

                                                  ( mph x weight -1)            (  foot-pounds  )
    (    lbs  )      +                       ( foot-pounds)          =        ( of force per-sec)
   ( weight  )   +   ( angle)   -        ( force)                 =             ( per-sec)     
       2000        +       36        -    ( 2 x 2000 - 1)          =               -1963  

                                            3999                                       

The same calc, expressed another way-->> 2000 + 36 - ( 3999 ) =  -1963 foot-pounds

 Note-- 

  I believe that the “minus” in front of 1963 in the calc above, is nothing to worry about, due to this representing the “trailing side” of the force…meaning, that positive force can be seen as force going in the direction of the center of the “ferris wheel”, and negative can be seen as "centrifugal forces", going outward…essentially, the “minus sign” is not relative, and 1963 is representative of both the positive force on the “leading side”, as well as the negative on the “trailing side”…I feel it to be worth mentioning that “lateral tork interposed”, without drag included, alters the meaning of the results drastically…I suggest that these results are correct, despite the calc having changed, due to there being no drag in space, and this new calc is a compilation, between “distance to geo-sync”, and “lateral tork interposed”, to deal with the special variables outer space provides…generating a value that represents the foot-pounds of force being thrown outward, against the floor of the “gravity module”( aka ferris wheel), instead of the ft per-sec, per-sec, that the hypothetical “ferris wheel” model is turning--   

The variables( utilizing an “excel” spreadsheet for the “lateral-tork interposed” calc, to derive correct “gravity-module” rotation speed )--

   and so, we need to plug different values into the formula( varying the mph), to obtain an amount of downward pressure/pull on the object( astronaut), that is as close as possible to the 1999 foot pounds of “resting weight/pressure” that the object would have on earth, to obtain good “artificial gravity” in space…to achieve this goal, I suggest, that the ideal value could be gained from utilizing a simple “excel program”, to give g-force numbers on a scale, based on changes in the “speed value”( mph)…“off the top-of-my head”, I approximate 2 mph to be very nearly correct( this value should generate about 1952.5 foot-pounds of “resting pressure” against the side of a huge centrifuge in space…in comparison, the same object, weighing 2000 lbs, would have 1999 foot-pounds of force against the floor of your living room, if it were in your home here on earth)...nearly identical values, from vastly different venues( humor, be it dry)--  :o)

g-forces( in a corner)--

   Also of note, my  “lateral-tork interposed” calc, found  above, does not require the “diameter value” to crunch the numbers, unlike the task of calcing the gimbals for the “hypothetical” ferris wheel-in space, that I suggest we build, to create artificial gravity there…this difference in requirements, to obtain the correct answer, is due to the “lateral-tork calc” deriving numbers from the weight of the object, the angle of the curve being traversed, and the mph that the object is traveling, mid-corner…and the “ferris wheel in space” problem, initially demands similar variables…the weight of the astronaut, and the speed that the wheel is turning, to derive the ultimate answer, of how much gravity is generated, from what rpm that the wheel is turning…essentially, the “ferris wheel” problem requires the "diameter", or “radius x 2” of the wheel to be known, and the “lateral-tork calc” must have the angle of the corner being traversed, to calc the problem correctly--

g-forces( artificial gravity module)--

   due to a circle being comprised of an “endless corner” that is being traversed, per-second, per-second, I feel that by simply dividing a 360 degree circle into ten pieces( like a pizza), we can say that each slice has 36 degrees on its edge…for this reason, I suggest that this angle is best for calcing artificial gravity…I believe that the other needed factor is, “at what diameter of wheel can we get a 2 mph rotation at the edge, with a shaft speed slow enough to be mechanically-viable, long-term??”( the larger the wheel, the slower the center can be turning, to obtain the desired value at the edge)--

3.60 “the perfect slice”( "PI"…but not exactly 3.1415)--

       ( feet)                              ( modifier)                 ( feet)

     ( Radius)      x   2     x            3.60          =       ( diameter ) 

        1000         x   2     x            3.60          =          7200 feet

The values--

5280 feet  =  1 mile

Mph x 1.46 =  feet per-sec

The math--

 ( mph)     x     ( modifier)               ( feet per-sec)

      2        x          1.46          =                 2.92

( feet per-sec)    x   ( sec)   =     ( feet per-min)

         2.92            x      60     =            175.2

175.2 feet per-min  =   ( “modifier”)

Describing the problem( with good gimbals)--

   7200 foot circumference wheel  %  175.2 feet per-min( modifier) =  41.095890411 minutes to complete 1 rotation--

  So, I suggest that a ferris wheel with a radius of 1,000 feet, would only need to complete 1 rotation per 41.09 minutes, to provide gravity equal to the earth’s, at the edge of the hypothetical “ferris wheel” that we are imagining to be turning in space--

A word of caution--

   The main down-side to this system, seems to be, the issue of a person jumping off of the floor( the edge of the spinning disk), due to the floor moving under the astronaut before they regain contact with it, from centrifugal force throwing them outward, and back to the “deck”…as for the exact forces this error in judgment would provide an individual…the lateral movement is about 2 mph, meaning, that the astronaut would appear to move laterally…at 2.92 feet per-second, I estimate the average jump to take about 1 second( up and down), so, we can expect a yield of an approximately 3 foot lateral movement, in a direction that is counter( the inverse), to the rotation of the disk/capsule/”gravity module”…I suggest that the force of impact is also an expected value, if a wall or other object is incurred, before regaining contact with the “deck”, allowing the astronaut to once again, rotate with the edge of the “module”, and experience gravity...here are the gimbals on jumping, for your inspection--

             ( lbs)
          ( weight)                              ( 1.46 x mph)                     ( foot-pounds)
    ( of astronaut)       -   1     -     (  feet per-sec )       =      ( force per sq-inch)
             200               -    1     -     (    1.46 x 2  )            =                 196.08
                                                            2.92    

   and there we have it( almost)…an encounter with a wall, or other object, would appear to have the force of 196.08 foot-pounds, if they were “air-borne” for 1 second…some may question the reasoning here…but, I am utilizing a “weight - distance  = force” principal…meaning, that if we know the distance the object traveled, and the weight that the object was, before moving, we can derive the “expected force” that was initially applied to the object, to get it moving, and, if the distance is short, we know…with minimal degradation, the “peak-force” value, at impact…


note--

  a clever person might increase the value of the speed, to make the calc appear to be nonsense...yet, we must keep in mind the fact that we are calcing variables related to a rotating centrifuge...and this means that a faster rotational value must increase the weight, in a congruent manner...hence, the outcome remains the same as a 2 mph rotation...if we increase the mph to 20, we get 29.2 ft per-sec...and we then have a weight value of 2000...generating 1960.08 foot-pounds of force--

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  Here we see( below), the same result, from deriving the values a different way, I feel that this indicates correct results(  in the case of the astronaut whom is 200 lbs, x 2.92 feet traveled if he jumped off the deck, % 2…we get 292 foot-pounds of force at “bird-feeder”, or “impact”, if he struck a stationary object…for me, this shows that the “quick-way” below, is within 100 foot-pounds of being exact, at lower weights, leading me to conclude that this calc would be very useful when dealing with larger objects, traveling at higher speeds)--    200 x 2.92 % 2 = 292 foot-pounds

Peak force at “bird-feeder” calc( a quick method to get close, for a crash into a wall…a rough estimate)--

  ( foot-pounds)                ( feet)                                   ( foot-pounds)

      ( force)               x     ( distance)    %   2  =    ( peak-force at “bird-feeder”)

          99                   x           100           %   2  =                     4,950

   I have divided the force to go the entire distance( 100 feet), by 2 to give the “peak-force” value, representing the amount of force needed per-sec, per-sec, to maintain the velocity, or the “peak-force” of an impact at “bird-feeder”, if the thrust was continual( a sustained value) after the “halfway-point”, until the total distance was traveled( thrust per-sec, per-sec…)-- 

Another method( that is in relation to a “straight-drop”)--

  Example--

  ( foot-pounds)                ( feet)                                   ( foot-pounds)

      ( force)               x     ( distance)    %   2  =    ( peak force at “bird-feeder”)

          99                   x              1            %   2  =                     49.50

                                                                                                                            ( total foot-pounds)

( peak force at bird-feeder)    +    ( resting foot-pounds of force)  =    (  of force at “bird-feeder” )

               49.50                            +                     99.00                             =                   148.5

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 --( The method above is in relation to an object moving, or “dropping”, 1 foot, and hitting a stationary object, and i feel this is the correct “angle-of-attack”)--

On second thought( the same thing, in a different way)--

  if the distance was long, we might want to divide the foot-pounds of force by 2, to get the “peak” of the curve( regarding foot-pounds of force), before it “tapered off”, and eventually, the object stopped…so, I will show this math for your inspection( this may be the best numbers at any distance, and also factors mph into the problem)--

            ( lbs)                                                                                             ( additional)
         ( weight)                                 ( 1.46 x mph)                               ( foot-pounds)
     ( of astronaut)      -   1     -     (  feet per-sec )    %    2      =     ( force per sq-inch)
              200                -    1     -     (   1.46 x 2    )      %    2       =             197.54

         Grouped as “ 200 -  1   -  2.92  % 2  =  197.54 “

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  Above, we see the “peak value” of the amount of force "in-play" before contacting the wall…I feel that the initial foot-pounds of “resting force” must be re-factored in, after we know the “additional foot-pounds of force” generated by the movement, before contacting  a stationary object( see below)--

    ( lbs)

 ( weight)         -1              =      ( resting foot-pounds)

     200             - 1              =                     199

                                                         ( additional)                    ( total foot-pounds)

                                                       ( foot-pounds)                     ( of force at )

( resting foot-pounds)            (  force per-sq inch)      =       ( bird-feeder)

             199                     +               197.54                  =             396.54

Note--

  As we can see, from observing all three ways I have shown, to calc the foot-pounds of impact into a stationary object, if an astronaut jumped off the deck in the “gravity module”…that the results from all three of the calcs are within 100 foot-pounds of each other…although there is a difference between the results, I feel that the margin is small, considering that the values for weight, and speed, are low…

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Problem summary--

   I have shown the problem many ways, it would seem that gaining good numbers is very much dependent on the angle that you are attacking the puzzle, utilizing mph seems best in the case of the spinning “gravity module”…due to the astronaut not being in “free-fall”, and also not having reached “terminal velocity”, if he/she was in free-fall…simply put, “throwing your weight into something”, means the foot-pounds in pressure of your “resting weight”, added to the foot-pounds of force generated from your weight moving, at a given mph…equals the “total yield” in foot-pounds, from the effort--

Note--

  And there we have it…396.54 foot-pounds of force( I am going with the highest value derived from the three ways I have shown), against any solid object, if you jumped off the floor for 1 second, in the “gravity module”, and contacted a stationary object…for me, putting up signs that depicted a “stick-figure’s” head hitting the wall, might be better than working the problem “long-hand”, to display the foot-pounds incurred from this error in judgment, although seeing the exact figures puts a really fine point on things( it should be remembered, that this surprisingly-high amount of force, “396.54 foot-pounds”, is also representative of a 200-lb person walking, full-force into their patio-door, while it was closed, at 2 mph…which is a “steady walking speed”, but not “power-walking”…dangerous, but not horribly)--

 "thrust/yield calc"--

example--

     ( ft per-sec)                                                  ( feet)
 ( force is applied)    x     ( seconds)    =     ( distance)
           46.5                 x            60           =          2,790  

   the calc above shows the total distance gained from 1999 foot-pounds of force being applied to the 2000 lb object mentioned earlier, with 1000 sq-inches of its surfaces, at the forward face of travel( drag), going a total of 2,790 feet, after 60 seconds of thrust was applied--

Note--

  if we use "distance to geo-sync math" within my "distance to geo-sync" post...i feel we can obtain more data by inverting the "lateral-tork calc", to find all the values present within the situation being studied above( see “distance to geo-sync post”), also, we can modify the ”thrust to yield” calc with drag included, to get nice numbers in relation to a “racing scull”( see below), like the ones they have at Harvard, Yale, Princeton( hmmm??), ect… with a modification to my “distance to geo-sync” calc-- 

The gimbals( racing sculls)--

   Initially, we will need to treat the issue being calc’d as a situation within “outer-space”, where there is no drag what-so-ever…to do this, we will be utilizing my “interposed distance to geo-sync calc”, to get the amount of force required to go the distance that we are studying, without drag( you may note, that within this, and other calcs of mine, if we know two of the values in the problem, we can derive the third…similar to the concept "in-play" when utilizing “the pythagorean theorem”, for deriving numbers relating to the isosceles triangle)--

  interposed distance to geo-sync( for “sculling”…in space…)--

   (pounds)           (feet)              (foot-pounds)
   (weight)    -   (distance)   =           (force)
     2000        -        100         =            1900

  calc complete  :o)  ….giving us our “base-line” amount of force with no drag( 1900 foot-pounds), and we can now plug this value into the calc below, that includes drag as a variable…

 ( calc’d w/“distance to geo-sync w/drag”)

      (pounds)      ( foot -pounds)             ( pounds % sq-inches)                 (feet)       
      (weight)   -        ( force)            +     ( weight % sq-inches )     =        (distance)  
        2,000       -        ( 1,900            +     ( 2,000 % 2,000  = 1 )     =              99
                                  ( 1,900             +                   1                   )

                                                              1,901

grouped as  2,000 -  1,901 =  99

note--

  I have decided to group the calc above as “ 2001 - 1900 =  99”   since adding the “drag exponent”( 2,000 % 2,000), to the force value, seems best in this case, to get a good distance number...also, if a person looks at this calc being utilized earlier in this study, you will find that the calc is different, the sq-inches are divided by the weight in the earlier work...in the calc directly above, we see that the situation is "inverted"( the pounds are divided by the sq-inches)...the question...why??, simply put, i am not adding the drag to the weight in this calc( requiring sq-inches to be divided by pounds...a method that usually generates a "negative number", from a smaller value being divided by a larger one)...instead, i am increasing the value of the "force" category, with the exponent of the "drag computations"( 2000 % 2000, requiring that the weight be divided by the sq-inches), to make the calc viable...essentially, if adding the "drag-exponent" to the "force value", that will then be subtracted from the weight, to get the "expected distance" form the force applied( 1900 in the case of the calc directly above), to give a distance of 99 feet of travel after drag is considered, the value needs to be a positive number, or a number above "zero"--

Summary--

   as we see above, the “sq-inches value” reduces the “distance value” a slight amount, and the racing scull in question has a reduction of the distance yield, of 1 foot, per stroke of the oars( it would seem that sculling in space is best, “on-paper”)…but, in the case of a “racing scull” in the water, a reduced draft appears to be an excellent thing( “by the numbers”), due to there being less surface area contacting the water, to slow the boat from friction…meaning, the lighter the boat…the shallower the draft, the less surface area will be contacting the water…the faster the boat shall be….and we all like fast boats in schools of higher learning, don’t we??( humor, be it dry…)
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P.S.  :

--( these calcs can continue “ad nauseam”, feel free to vary the numbers you plug-into the calc, and see for yourself how well they apply in real-world situations…I believe I am on to something here, let me know if you feel I am not)--

 Best wishes, john kruschke--

 note--

  i believe that the math above should settle the issue( "artificial gravity"), in its entirety, i am looking forward to hearing if the numbers "crunch" well, or go in the waste-basket best, after being "crumpled" into something fairly aerodynamic( a "wad")...anyway, i have fore-gone spell-check in the name of scientific progress here...i only get an hour at the library each day--


  P.S. : if NASA throws a "kegger" in space with this one...i would like a chair near the tap...

 best wishes, and god bless, john kruschke--