NASA Mathematics

According to NASA’s top experts, the Arctic may be ice-free by the summer of 2012. An ice-free Arctic would require that the ice edge melts back about 20 miles a day, for the next ten weeks.

That could happen if aliens aim a giant blow torch at the North Pole. Only an evil denier would question the word of a NASA climate expert.

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6 Responses to NASA Mathematics

  1. omnologos says:

    Check it out, perhaps Lewis Pugh is really chipping the ice away as we speak!

  2. Tomwys says:

    Won’t happen, as you correctly surmise. Seems that the descriptor of “Top” needs changing, and I hope that your use of the plural is in error.

    • Summary of
      “The correct values for a circle“
      1. The correct value pages 1-7
      2. Shaded area, pages 8-9
      3. Radian, pages 10-11
      4. Relationships between Cube, Cylinder and Sphere, pages 12-14
      5. Grammar of Astronomical Numbers In English, page 15

      1. The correct values for a circle
      We have three standard figures which relate to each other. They are
      square, circle and cylinder.
      Squares, circles and cylinders each have their own grading system. Also are also sorted after their natural qualifications.
      1.1 Grading system one of squares I
      1.2 Grading system two of circles II
      1.3 Grading system three of cylinders III
      A grading system consists of areas, perimeters or circumferences attention to the number of their results. When the area increases the perimeter or circumference decreases attention to the numbers of result of solving. However, when the number of circumference or perimeter increases the area decreases. Only a square’s area is equal to its perimeter, (when from this square crossing down the course changes perimeters are bigger than areas and when crossing up areas are larger than perimeters) this rule is valid whit circles too. Cylinders page.5!
      Each grading system is divided in three groups; they are listed according to their area, perimeter or circumference:

      1.1
      Squares, Grading system I
      1.1.1 Group one: areas smaller than perimeters.
      1.1.2 Group two: area equal perimeter.
      1.1.3 Group three: areas bigger than perimeters.

      The square with the side 4 is the centre of the squares grading system I, and gives us the square squaring form and therefore quite unique.

      When we put a side of a square into the above formulae we obtain five different values. All five values are useable for calculation of a circle. Where I tell about which via that obtains 3.141592953… too. 4Q can be any value.
      With the value we are able to calculate a circle circumference and its area; via application we accomplish its diameter.
      We put number of sides …6, 5, 4, 3.929, 3.92810767, 3.9, 3.5, 3.0 … into formula which that produces for each a value. Whit values we solve circles. Diameter uses as square’s side has put into formula then circumference and area calculates. We apply the circles, because know which size of circle produces a value if we use that for general circles. Then we compare circle’s circumference whit its area after that we construct grading system II for circles.

      The side of squares puts into formula which that obtains values: see below!

      We can get definite values and their process, each of values produces a kind of diameter; that I mean values in group one produce diameters longer than standard circle’s diameter and values in group three produce diameters shorter than standard circle’s diameter.
      In group two is only the square of 4, when we put side 4 into formula obtains a value which that produces diameter equal the side and its circumference equal to its area, so it is always been an inner circle and standard circle.

      The correct value and its subordinated values ”3.125”, and 0.78125, 1.28, 0.64, 0.21875

      1.2
      Circles
      Circles grading system II constructs accordingly to the squares grading system I.

      Circles have divided into three groups:
      1.2.1 Group one: areas smaller than circumferences.
      1.2.2 Group two: area equal circumference.
      1.2.3 Group three: areas bigger than circumferences.
      The values in circles grading system II describe different lengths of diameter.
      Values in group one produce diameters greater than the standard diameter.
      The value in group two produces a correct diameter for a standard circle. This means that the diameter equals the side, consequently an inner circle.
      Values in group three produce diameters smaller than the standard diameter.
      When the values in the grading system II are applied (the side as diameter in the formula) the area and circumference can be calculated. In group two we achieve the value 3.125, putting this value in the formula will give us a circle that its diameter is equal to the side of the square or/and the correct inner circle. One can choose any square and calculate its inner circle with 3.125.
      This value is also applicable for calculation of other processes like areas, circumferences, diameters and also one can calculate the wanted diameter in per cent. Therefore, I choose 3.125 as a standard value.
      We can see the differences in theory and practice illustrated in the book, where the diameter is shorter or longer.
      One cannot change the system of the nature by choosing other values because this would be inaccurate.
      For further clarification see http://www.correctpi.com

      1.3
      Cylinders
      The mantle area and perimeter of cylinders are three different forms when rules one, because of that cylinder also divide into three groups of their patterns. Cylinders diameter are 1 u.l. and their height are different. Where only a cylinder’s mantle is square which that has relation whit circle and square in groups two.
      Also all figures in group II have relations to each others.

      1.3.1 The mantle relations are; mantle’s perimeter is equal circumference and mantle area 0.78125 of circle’s area. (Circle squaring form)
      1.3.2 The mantle’s perimeter is 0.78125 of square and mantle’s area is 0.781252 of square’s area in group II. (Square squaring form)
      1.3.3 The cylinder in group II mentions Cylinder squaring form.

      Via their relations we can find the cylinder in group two. For further see the book on the http://www.correctpi.com

      1.3.4 Cylinders in group one produce horizontal rectangles.
      1.3.5 Cylinder in group two produces only square.
      1.3.6 Cylinders in group three produce vertical rectangles.

      1.4
      The values for π or 3.141…
      The value of 3.141592653… is incorrect value for a circle.
      Comment:
      See pages: 113,114,115 in the book!
      Thinking Reality, if we use a value instead for pi, number as 3.75, what happen to circle, well circumference and area get to be bigger and diameter get be bigger too how we can know this fact.
      When we have the precious formulae which that can tell you even the length of diameters. When we use each value, the formulae tell us the absolute exactly correct, of the decimal to decimal when a diameter changes whit different values.

      See the places of 3.14… and the square which that gives 3.14 on the grading system I, II, III.

      You can calculate a circle with these values but the circle is not an inner circle or standard circle! See below, values of π:
      Other approximated values and their subordinated values for example: 3.14… in group three
      3.141592653… and its subordinated 0.785398163…, 1.273239545…, 0.6366197724…, 0.2146018367

      2
      SHADED AREAS

      The area and perimeter of the shaded areas around a square with an inner circle are to be calculated.

      S = side of square squaring form s = sides of all squares

      The area of the shaded parts and their perimeter are to be calculated. Assume that the area of a square is x units and the perimeter is y units. The area of the circle will be 0.78125 x. If the perimeter of the square is y units, the circumference of the circle will be 0.78125 y.

      The shaded parts:

      B indicates the percentage of the square covered by the shaded parts.
      1 – M = B = 0.21875 = 21.875 %
      B * y = perimeter of shaded parts
      B * x = the area of the shaded parts

      Thus the shaded parts b =

      The calculation of the area and the perimeter of the shaded parts are described below:

      3
      Radians

      A radian: 180º / 4M = 57.6º

      A radian is part of a circle that forms an isosceles triangle where the length of the chord is equal to the radius of the circle.

      Radian: 18º / 4M = 57.6º 4M / 180º = 0.01736111…

      The chord of the radian is part of the circumference of the circle.
      The length of the chord is equal to that of the radius of the circle. This means that an equilateral triangle will be formed if the chord is made into a straight line.

      See pages describing chords for the method of producing a straight line from a chord.
      http://correctpi.com

      A radian can be represented in different ways:

      Or

      But
      Correct values:

      M = 0.78125 4M = 3.125
      R = 0.64 2R = 1.28

      M = 0.78125 2M = 1.5625
      4M = 3.125 8M = 6.25
      R = 0.64 2R = 1.28
      3R = 1.92 4R = 2.56

      Calculating a radian:

      360° / 8M = 57.6º 180° / 4M = 57.6º 90° * R = 57.6°
      45° / M = 57.6° ® 45° / 0.78125 = 57.6°
      90º * R = 57.6º ® 90º * 0.64 = 57.6º 45° * 2R = 57.6º ® 45º * 1.28 = 57.6º

      A circle = 360º
      360º = 8M = Radian
      Radian: 57.6º = (360º / 8M)
      Radian: 57.6º = (180º / 4M)
      Radian: 57.6º = (90º / 2M)
      Radian: 57.6º = (90º * R)
      Radian: 57.6º = (180º * R/2)
      Radian: 57.6º = (270º * R/3)
      Radian: 57.6º = (360º * (R/4)

      4
      RELATIONSHIPS
      between
      CUBE, CYLINDER and SPHERE
      By the use of M & R

      Notice
      M = 0.78125
      4M = 3.125
      R = 0.64, 2R = 1.28 & 3R = 1.92
      M*2R*1.5= 3/2
      M*2R *(2/3) = 0.666666… ˂ 7

      Provided that the diameter, height and side of a sphere, cylinder and cube respectively are equal, the following relationships are obtained. The area of the cylinder is equal to 2/3 of the mantle area of the cube “mantle area = four sides of a cube”. The surface area of the sphere is equal to 2/3 of the surface area of the cylinder.
      The volume of the cube is (2R)= 1.28 times larger than the volume of the cylinder. The volume of the cube is (3R) 1.92 times the volume of the sphere.

      Note! The above principle is not valid if a value that is larger or smaller than M is used to calculate these characteristics.

      Relationship between a cube, cylinder and sphere:
      A cylinder is placed in a cube with a side of 7u.l.
      The diameter of the cylinder is 7u.l. and the height is 7u.l.
      In the cylinder is a sphere with a diameter of 7u.l.
      The relationships between the different bodies can be seen below.
      The values of the cylinder and sphere can be calculated by using the cube as follows:

      Total area of cube: s2 * 6 → 72 * 6 = 294 cm2
      Mantle area of cube: s2 * 4 → 72 * 4 = 196 cm2
      Volume of cube: s3 → 73 = 343 cm3

      Mantle area of cylinder:

      Mantle area of cylinder: 4 * 7 * 7 * M = 153.125 cm2
      4d * h * M

      Total area of cylinder:

      ((2 * (d2 * M ) + (4d * h * M )
      ((2 * (72 * M ) + (4 * 7 * 7*M ) = 229.6875 cm2

      Volume of cylinder:
      Volume of cylinder: 72 * 7 * M = 267.96875 cm3
      d2 * h * M

      Area of sphere ® d2 * 4 *M
      Area of sphere ® 72 * 4 * M = 153.0125 cm2

      Volume of sphere ® d3 * 2M / 3
      Volume of sphere ® 73 * 2M / 3 = 178.6458333…cm3

      The relationship between the total area of the cube and the total area of the cylinder is
      2R =1.28.
      Total area of cube and mantle area of cylinder is 3R = 1.92
      Mantel area cube and mantle area cylinder is 2R = 1.28
      Volume of cube and volume of cylinder is 2R = 1.28
      Volume of cylinder and volume of cube is M = 0.78125

      The relationship between cube total area and sphere area is 3R =1.92.
      Mantle area of cube and sphere area is 2R= 1.28
      Volume of cube and volume of sphere is 1.92 = 3R

      The relationship between the mantle area of the cylinder and the area of the sphere is 1.
      Total area of cylinder and area of sphere is 1.5 = 3/2.

      Area of sphere and total area of cylinder is 2/3.
      Area of sphere * M* 2R* 3/2 = total area of cylinder
      Total area of cylinder * M* 2R* 2/3 = area of sphere

      Cylinder volume and sphere volume is 1.5 = 3/2.
      Volume of cylinder * M* 2R* 3/2 = volume of sphere

      Volume of sphere and volume of cylinder is 2/3
      Volume of sphere * M* 2R* 2/3 = volume of cylinder

      Compare even their volumes with the value with the 3.141592653…

      5
      Grammar of Astronomical Numbers In English
      Numbers and alphabets are connected in a very simple way
      Everybody, young or old can learn and remember the numbers quite easily
      Author: Mohammadreza Mehdinia / http://www.correctpi.com
      1 One 1 = 1 -
      10 Ten 1 * 10 = 10 1
      100 Hundred 10 * 10 = 100 2
      1000 Thousand 100 * 10 = 1000 3
      Million 1000 + 000 = 1000,000 6
      1. A Aillion (ai)= (idea) Million + 000 = 1000,000,000 9
      2. B Billion (bi) as (bill) Aillion + 000 =1000,000,000,000 12
      3. C Cillion (ci) as (cinema) Billion + 000 = Cillion 15
      4. D Dillion (di) as (differ) Cillion + 000 = Dillion 18
      5. E Eillion (ei) as(ea of eat) Dillion + 000 = Eillion 21
      6. F Fillion (fi) as (fifth) Eillion + 000 = Fillion 24
      7. G Gillion (gi) as (gi of give) Fillion + 000 = Gillion 27
      8. H Hillion (hi) as (hid) Gillion + 000 = Hillion 30
      9. I Iillion (ii) as (i=illness) Hillion + 000 = Iillion 33
      10. J Jillion (ji) as (jib, Jimmy) Iillion + 000 = Jillion 36
      11. K Killion (ki) as (kilo) Jillion + 000 = Killion 39
      12. L Lillion (li) as (liver) Killion + 000 = Lillion 42

      13. M Milliard (mi) as (mid) Lillion + 000 = Milliard 45
      14. N Nilliard (ni) as (ni of nick) Milliard + 000 = Nilliard 48
      15. O Oilliard (oi) as (oil) Nilliard + 000 = Oillhard 51
      16. P Pilliard (pi) as (pi of pitch) Oillhard + 000 = Pilliard 54
      17. Q Qilliard (qi) as (quill) Pilliard + 000 = Qilliard 57
      18. R Rilliard (ri) as(ribbon) Qilliard + 000 = Rilliard 60
      19. S Silliard (si) as (signal) Rilliard + 000 = Silliard 63
      20. T Tilliard (ti) as (till) Silliard + 000 = Tilliard 66
      21. U Uilliard (ui) as (juilliard)ju≈ you) Tilliard + 000 = Uillard 69
      22. V Villiard (vi) as (Viking) Uillard + 000 = Villard 72
      23. W Williard (wi) as (will) Villard + 000 = Willard 75
      24. X Xilliard (xi) as (ksilliard) Willard + 000 = Xillard 78
      25. Y Yilliard (yi) as (yie of yield) Xillard + 000 = Yillard 81
      26. Z Zilliard (zi) as (zi of zigzag) Zillard + 000 = Zillard 84
      U=uniform, ui= uniform. Xi=ksi

  3. omanuel says:

    NASA’s mathematics successfully brought in government funds for decades.

    Only recently a few NASA scientists have discovered that scientists at NASA and the US NAS were not half as clever as they thought they were in leading society into an Orwellian, one-world government by twisting experimental data and observations to deny the fountain of energy that Copernicus discovered at the center of the solar system in 1543 . . . 469 years ago. http://tinyurl.com/7qx7zxs

    The sad tale of deceit for short-term advantage is here:

    http://omanuel.wordpress.com/about/#comment-132

    Here’s evidence Copernicus’s fountain of energy at the center of the solar system

    a.) Made our elements
    b.) Sustains our lives
    c.) Controls our climate

    The Apeiron Journal 19, 123-150 (2012) http://tinyurl.com/7t5ojrn

  4. Lance says:

    Need to buy by row boat soon and start paddling to the pole!!

  5. Stefan Landherr says:

    Something for the alarmists to ponder:

    He who refuses to do arithmetic is doomed to talk nonsense.
    — John McCarthy, 1995

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