The Past Master Club


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ART MENU 7
BASICS & FORMULAS
  1. BB GUN
  2. MY 1ST RADIO
  3. TRISECTION OF AN ALLEY!
  4. ELECTRIC MOTORS
  5. DOUBLE YOUR VOLTAGE
  6. 2 SPEED FORWARD & REVERSE ENGINE
  7. RATIO
  8. AREA MEASURING!
  9. NORTH
  10. MUSIC
  11. ELECTRIC CIRCUIT SOLVING
  12. OVER CHARGED CAPACITOR!
  13. NON STEADY STATE
  14. VISIBLE SPECTRUM!
  15. FISSION
  16. A-BOMB
  17. ATOMIC SUBMARINE: USS NAUTILUS 1954
  18. NUCLEAR REACTOR
  19. FOLDING, EDGING & COLD BENDING
  20. MY 1ST DRAWING
  21. BEARING LOADS
  22. APPARENT HEADING 35o!
  23. FLIGHT SPEEDS!
  24. MECCANO'S UNIVERSAL JOINT
  25. LaGrange EXAMPLE
  26. PSYCHIATRIST'S ILLUSION
    OR PEACE SIGN?
  27. 3D VECTOR GRAPHICS?
  28. PERSONALITY
  29. 3D PMAT TEST
  30. 2 WAYS TO GET TRUE LENGHT
  31. ALL TRUE LENGHTS
    ARE INDICATED IN
    OR BY RED LINES
  32. TOP & SIDE VIEW WITH 2 PLANES!
  33. AN ANALYTICAL TEST
  34. ANTONYM QUIZ
  35. ARE YOUR LINES
    OR LETTERS THIN
    OR WIDE AND JUMPING
  36. CALCULATE FORCES
  37. CALCULATE FORCES ON A HINGE 'A'
  38. BUILDING A BRIDGE
  39. ENERGY CONVERSION
  40. SQUARE ROOT TO FIFTH ROOT!
  41. MY 'AUTOCAD AUTOLISP' HOOK
  42. 3D CURVE
  43. SPOT WELDING
  44. NPN TRANSISTORS
  45. GYRO PLANES
  46. HARNESS THE SUN
  47. FAST AREA?
  48. ELECTRONIC CHIPS
  49. FOCUS
  50. LOADER PLAN
  51. TRISECTION MECHANISM
  52. AMPLIFIER
  53. ORCHESTRA INSTRUMENTS
  54. CALCULUS'  VOLUME OF A CONE:
  55. CALCULUS'  VOLUME OF A SPHERE:

  56. OPTICAL ILLUSION VII

MAIN MENU
ART MENU 0 1 2 3 4 5 6 [7] 8 9 10

BB GUN
gif/bbc.gif

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MY 1ST RADIO
gif/radio.gif

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TRISECTION OF AN ALLEY!
gif/7.gif
Since: C = 6 = H/3
Therefore: sin3Æ = 18/20 = 0.9
3Æ = 64.15806o
 
Æ = 21.38602o @ 21o 23' 9.82''
 
Width W = 20cosÆ
= 20*0.931145
= 18.6229 feet
Thus the alleyway is about 18 feet 7 15/32 inches wide!
 

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ELECTRIC MOTORS
gif/electro.gif

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DOUBLE YOUR VOLTAGE
gif/voltage.gif

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2 SPEED FORWARD & REVERSE ENGINE
gif/2xspeed.gif

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RATIO
gif/ratio.gif

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AREA MEASURING!
Carefully trace the perimiter
(any) with the small blade
& the sliding long blade
will displace the area's scale!
- a cheap planimeter!
gif/knife.gif

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NORTH
Point your dial: midway between
the hour hand & 12 at the sun;
noon now points NORTH / SOUTH!
gif/time.gif

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MUSIC
(Click to enlarge)
gif/music.gif

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ELECTRIC CIRCUIT SOLVING
anim/electric.gif
gif/circuit.gif

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OVER CHARGED CAPACITOR!
gif/farad.gif

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CURRENT: NON STEADY STATE
gif/vstate.gif

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VISIBLE SPECTRUM!
anim/roygbiv.gif
gif/roygbiv.gif
NEXT TIME, LOOK AT A PICTURE PUZZLE
UNDER A LARGE, THICK GLASS, MAGNIFIER!

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THE ELECTROMAGNETIC SPECTRUM!
gif/e/tele1.gif

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FISSION
Electrical repulsion tween the protons rips the nucleus apart! gif/fission.gif
The 'MOLE' is Chemisty's "Rossetta Stone"!

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A-BOMB
gif/abomb.gif
  1.  Altimeter trigger.
  2. Explosive charge.
  3. Thick casing.
  4. United forms a complete 
       sphere of uranium of
       greater-than-critical mass!
anim/atom1.gif

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ATOMIC SUBMARINE: USS NAUTILUS 1954 gif/asub.gif
  1. Reactor.
  2. Engineering room.
  3. Missille compartment: 16.
  4. Missille control center.
  5. Control center of sub.
  6. Gyro room: 50-ton gyro stabillizers.
  7. Tanks nagative.
  8. Tanks 2.
  1. Batteries.
  2. Main ballast tank.
  3. Torpedo room.
  4. Control center of sub.
  5. Missille control center.
  6. Quarters of crew & officers.
  7. Mess room.
  8. Periscope.

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NUCLEAR REACTOR
gif/reactor.gif

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FOLDING, EDGING & COLD BENDING
  Up to thickness of s:
Tensile strength of: 11.5 2.53456 7 8101214161820
-> 40 kg/mm311.6 2.53568 1012162025283640
-> 50 kg/mm31.22 3 458101216202528364045
-> 65 kg/mm31.62.5 4568101216202532364560
  Correction factor q:
Ratio R:s5.03.02.01.20.80.5
Correction factor:1.00.90.80.70.680.5
An example for computing the length before bending:
gif/bendng.gif
Length of legs:
a=50, b=130, c=240, d=50
Bend radius:
R1=20, R2=20, R3=32
Thickness: s=10
ie. 6 mm @ 63 kg/mm3
Bend angles:
α1=90o, α2=45o, α3=135o
Correction factors:
q1=0.8, q2=0.8, q3=0.96
COMPUTATIONS:
L = a + (R1 + q1 * s/2)π α1/180
    + b + (R2 + q2 * s/2)π α2/180
    + c + (R3 + q3 * s/2)π α3/180 + d

    = 50 + (20 + 0.8 * 10/2) 22/7 * 90/180
        + 130 + (20 + 0.8 * 10/2)22/7 * 45/180
        + 240 + (32 +0.96 * 10/2)22/7 * 135/180 + 50

    = 50 + (24) 11/7 + 130 + (24)11/14 + 240 + (36.8)33/14 + 50

    = 50 + 24 * 1.57 + 130 + 12 * 1.57 + 240 + 18.4 * 4.71 + 50
    = 50 + 37.68 + 130 + 18.84 + 240 + 86.664 + 50
    = 613 mm. in length.

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MY 1ST DRAWING
gif/dwgs.gif

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BEARING LOADS
gif/bearings.gif
A load of 100 lbs. (down) is exerted on the left pulley
and a load of 50 lbs. (up) is exerted on the right pulley.
SM from left = 0 = -100 x 10 + RB x 20 + 50 x 28
The right bearing RB supports: (-100x10 + 50x28)/-20 = -20 lbs. (down).
SM from right = 0 = LB x 28 - 100 x 18 - 20 x 8
The left bearing supports: (-100x18 - 20x8)/-28 = +70 lbs. (up).
CURVES & SLOPES
y = f(x): any point on a curve defined by f(x)
slope = (y2 - y1)/(x2 - x1)= delta y /delta x
delta y /delta x = d(y) / dx = f '(x)
slope = d(y) / dx = f '(x): at any point x1 on line f '(x1) = y'1 = V
acceleration = d2(y) / dx = f ''(x): at any point x1 on line f ''(x1) = y''1 = M
MOMENT & SHEAR
Moment = Force x Distance = Shear x Length
or M = F x D = V x L
V = M/L = d(M)/dx
ie. V or shear = slope of Moment curve. 
A constant Shear of +70 gives a constant slope for M of @ +45o
A constant Shear of -30 gives a constant slope for M of @ -25o
A constant Shear of -50 gives a constant slope for M of @ -40o
or vice versa!
Moment at point (1) is +70 x 10 = 700 in-lbs. from the left.
Moment at point (2) is +70 x 20 -100 x 10 = 400 in-lbs. from the left.
or
Moment at point (1) is +50 x 18 - 20 x 10 = 700 in-lbs. from the right.
Moment at point (2) is +50 x 8 = 400 in-lbs. from the right.

[The gravity pt of a triangle is 1/3 from larger end]
Z1EI = moment area of A 1, A2 & A3 about point (2)
  = (10 x 700/2) x (10 + 10/3) + 10 x 400 x 5 + (10 x 300/2) x (20/3) = 78,667 lb-in3
Z2EI = Z1EI/(10/20) = 38,333 lb-in3
Z3EI = Z1EI/(28/20) = 107,334 lb-in3
Z4EI = moment of area A1 about point (1) = (10 x 700/2) x (10/3) = 11,667 lb-in3
Z5EI = moment area of A1, A2, A3 & A4 about point (3)
  = (10 x 700/2) x (18 + 10/3) + 10 x 400 x 13 + (10 x 300/2) x (8 + 20/3) + (8 x 400/2) x (16/3) = 157,216 lb-in3
ε2EI = Z2EI - Z4EI = 38,333 - 11.667 = 26,666 lb-in3
ε1EI = Z5EI - Z3EI = 157,216 - 107,334 = 49,882 lb-in3
[ E = 3 x 107 psi (steel) ] and [ I = (pi x d4)/64 = (22/7 x 24)/64 = 0.785 in4 ]
ε1 = 49,882/(3 x 107 x 0.785) = 0.002118 ins.
ε2 = 26,666/(3 x 107 x 0.785) = 0.001132 ins.
CRITICAL SPEED 

W1ε1 = 50 x 2.118 x 10-3  = 10.59 x 10-2

W2ε2 = 100 x 1.132 x 10-3 = 11.32 x 10-2
                      S1 = 21.91 x 10-2

W1ε1² = 2.243 x 10-4

W2ε2² = 1.282 x 10-4
  S2 = 3.525 x 10-4


      ________       _______________________

wc = ÖgxS1 / S2  =  Ö 386x0.2191 / 0.0003525 

= 490 rad. / sec. = 490 x 60 x 7 / 22  = 9,300 RPM. max.

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APPARENT HEADING 35o!
gif/heading.gif

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FLIGHT SPEEDS!
gif/e/flightspd.gif

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MECCANO'S UNIVERSAL JOINT
1 wheel power!
gif/union.gifgif/gear.gif

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LaGrange example:
Problem:
F = grad ø = d/dx i + d/dy j + d/dz k = Force Potential [1]
F = (5zSINx + y)i + (4yz + x)j + (2y² - 5COSx)k [2]

Solution:
From [1] & [2]:
Since: d/dxø = 5zSINx + y
Therefore: ø = - 5zCOSx + yx + f1(y,z) [3]
f1 is an arbitrary constant or function without 'x'.
Since: d/dyø = 4yz + x
Therefore: ø = 4y²z/2 + xy + f2(x,z) [4]
f2 is an arbitrary constant or function without 'y'.
Since: d/dzø = 2y² - 5COSx
Therefore: ø = 2y²z - 5zCOSx + f3(x,y) [5]
f3 is an arbitrary constant or function without 'z'.
From [3] & [1]:
Since: d/dyø = x + d/dyf1(y,z)
and since: d/dyø = 4yz + x
Therefore: d/dyf1(y,z) = 4yz
Thus: f1 = 4y²z/2 + c1 [6]
c1 is an arbitrary constant.
Substituting [6] into [3] gives:
ø = - 5zCOSx + yx + 4y²z + c1 [7]
From [4] & [1]:
Since: d/dzø = 2y² + d/dzf2(x,y)
and since: d/dzø = 2y² - 5COSx
Therefore: d/dzf2(y,z) = - 5COSx
Thus: f2 = - 5zCOSx + c2 [8]
c2 is an arbitrary constant.
Substituting [8] into [4] gives:
ø = - 5zCOSx + yx + 4y²z + c2 [9] = [7]

The answer in work units W = F x D = F x 2ør = f(ø):
From [7] & [9]:
ø = - 5zCOSx + yx + 4y²z + c
c is an arbitrary constant.


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PSYCHIATRIST'S ILLUSION OR PEACE SIGN?
gif/2.gif
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3D VECTOR GRAPHICS?
gif/gallows.gif
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PERSONALITY
gif/psyche.gif

Psychiatric oath?!:
I am so biased that I can't see staight!
If I weren't forced to believe what I'd read
I wouldn't be able to get my job done!
Ideal Psychiatric oath?!:
I swear an oath to CRETE or turn into concrete;
I dont need concrete evidence that they're NUTS!
- uncles excluded!
- call me MANDRAKE, a weed!
Psychiatry!:
Good thing my kids are NUTS;
parents are s-s-o-o-o lousy at raising their kids;
but it proves I can do my "JOB"!
Psychiatric Ads!:
Just because the ad in the Psychiatric Journal said so;
and I don't think so; doesn't mean I have to wrap the
NUT up!
- why kill the 'cashcow'!
- who deserves 'payola' more!
Ideal IQ question or questions?!:
  1. Lord?
  2. 1st. born?
  3. Santa Claus?
  4. Angel?
  5. Parent?
  5. In control of one's steering wheel?
  6. Baby sitter?
  7. Tug-of-war?
  8. Building-block?
  9. Cross-section-of-Society?
10. At the front line defeating satan & his angels?
11. Missionary?
12. Trained in Heaven to entertain?
13. Trained in Heaven to Crusade?
14. Mean Esper?
15. Waving a sheet of his works like a 'white flag'?
16. All the above?

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  3D PMAT TEST  
  HOVER YOUR CURSOR OVER CIRCLES  
  TO SEE ANSWER ON STATUS LINE BELOW!  
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  ANTONYM QUIZ  
  HOVER YOUR CURSOR OVER CIRCLES  
  TO SEE ANSWER ON STATUS LINE BELOW!  
gif/test3.gif
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2 WAYS TO GET TRUE LENGHT
gif/trulnght.gif
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ALL TRUE LENGHTS ARE INDICATED IN OR BY RED LINES
gif/trutran.gif
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TOP & SIDE VIEW WITH 2 PLANES!
gif/objctpln.gif
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AN ANALYTICAL TEST:
Which group does not belong?
CLICK for a HINT
gif/analytcl.gif
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ARE YOUR LINES OR LETTERS THIN OR WIDE AND JUMPING;
ESPECIALY WHEN DRAGGED BY A MOUSE!
IT'S BECAUSE VECTOR GRAPHICS CAN ONLY EMULATE HIGH RESOLUTION!
ROY G BIV: RGB
gif/bytes.gif
HOW IS A PICTURE STORED (my opinion):
EVERY OTHER LINE & BY COLOR, WITH BLEEDING INBETWEEN FOR A LARGER PICTURE!
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CALCULATE FORCES
gif/forces.gif 		SOLUTIONS: [S F=0]
1. GRAPHICS:
Drawing all known forces to scale and
drawing intersection point 'P' with given angles for T & C.
Tension = 8.5582(+) & Compression = 2.8880(-) [measured]

2. x'-y'AXIS PARALLEL TO MEMBER 'T':
[S Fx'=0] -Ccos10 -3cos40 -8sin40 +16sin40 = 0
C = 2.8880
[S Fy'=0] T +8cos40 -16cos40 -3sin40 -Csin10 = 0
T = 8.5582

3. SIMULTANEOUS EQUATIONS:
[S Fx=0] +8 + Tcos40 +Csin30 -16 = 0
0.7660T +0.5C = 8          [1]
[S Fy=0] Tsin40 -Ccos30 -3 = 0
0.6428T -0.8660C = 3     [2]

T = 8.5582 & C = 2.8880
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CALCULATE FORCES ON A HINGE 'A'
gif/beam.gif 		SOLUTIONS: [S F=0 & S M=0]
1. MOMENT ABOUT 'A':
[S Ma=0]
-.25Tcos25 +(5 -.12)Tsin25 -10(5 -1.5 -.12) -4.66(2.5 -.12) = 0
Tension = 19.61(+)

2. EQUATING x & y FORCES ABOUT 'A':
[S Fx=0] Ax -19.61cos25 = 0
Ax = 17.77
[S Fy=0] Ay +19.61sin25 -4.66 -10 = 0
Ax = 6.37
From which 'A' = 18.88 = Ax^2 + Ay^2 = 17.77^2 + 6.37^2

3. MOMENT ABOUT 'A' FOR COMBINED WEIGHT 14.66=10+4.66:
[S Ma=0] -19.61(5)sin25 +14.66(DISTANCE) = 0
DISTANCE = 2.2756

4. GRAPHICS:
Drawing all known forces to scale and
and closing the yellow triangle gives
'A' = 18.88 compression(-) & b = 20o                   [measured]
Locating point 'P' also gets DISTANCE = 2.2756  [measured]
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BUILDING A BRIDGE
gif/bridge2.gif
TABLE
#LASugε g=LSug/Aufε f
1122-P0000
2182-.75P0000
31531.25P0000
492-2P001-9P
5155-1.25P0000
6183-.5P001-3P
7155-2.5P-5P/2/13½18.75P/13½00
81820-3P/2/.5000
915305P/2/13½000
1093-2P-3P/13½18P/13½00
111220-2P/13½000
S=    36.75P/13½ -12P
Δg=ε g/E=36.75*12*2000/13½/29/106=0.008435''
Δg=0.54/64''
Δgx=0.45/64'' - horizontal displacement
Δgy=0.30/64'' - vertical displacement
Well within limits!!
Bridge will but against roadway!
Δf=ε f/E=12*12*2000/29/106=0.000000''
E=10^6 psi STEEL
P=1 TON truck
L in feet
A is normaly 20x larger for I BEAMS
PROBLEM:
Construct the bridge described in note 1. on the left and
find out the deflection at the rightmost point, or overhang,
to merge with a roadway.
Cross section may be 20 to 50 times larger than given.
- Note 2. shows released structure with references.
- Note 3. shows all elements numbered.
- (+) = TENSION & (-) = COMPRESION
SOLUTIONS: [S F=0 & S M=0]
IN NOTE 2.:
At point 'D': [S Md=0 & S Fy=0] At point 'G': [S Mg=0 & S Fy=0]
-3P(2*9') +Fy3(3*9') = 0
Fy3 = 2P 		Fy5(4*9') +2P(1*9') -3P(2*9') = 0
Fy5 = P
At point 'F': [S Fx=0] At point 'D': [S Fx=0]
Point 'F' is free to move.
Thus Fx2 = 0 		Point 'D' is pinned.
Thus Fx4 = 2P
IN NOTE 4.:
Member #12 is cut and a TENSION of 1 lb. force is exerted on it. [S F=0]  
At point 'G': [SF=0 & F²=Fx²+Fy²]
Fgy=F[11]=-F[12]*(L[11]/L[12])
F[11]=-1*(2/13½))
F[11] = -2/13½ 		Fgx=F[10]=-F[11]*(L[9]/L[11])
F[10]=-1*(3/13½))
F10] = -3/13½
At point 'C': [SF=0 & F²=Fx²+Fy²]
F[9]=-F[11]*(L[9]/L[11])
F[9]=-(-2/13½)*5/4
F[9] = 5/2/13½ 		F[8]=-F[9]*(L[8]/L[9])
F[8]=-(5/2/13½)*3/5
F[8] = -3/13½
At point 'F': [SF=0 & F²=Fx²+Fy²]
Fy[7]=-Fy[9]=-F[11]
Fy[7] = 2/13½
F[7] = (2/13½)*5/4
F[7] = -(5/2/13½) 		Fx[6]=-Fx[7]+F[10]
Fx[6]=-F[10]+F[10]=0
Fx[6] = 0
At point 'B': [SF=0 & F²=Fx²+Fy²]
Fy[12]=F[12]*(2/13½)=1*2/13½
Fy[7]=F[7]*(4/5)=
-5/2/13½*(4/5)=-2/13½
Fy[10]+Fy[7]=0 		Fx[12]=F[12]*(3/13½)=1*3/13½
Fx[7]=F[7]*(3/5)=
-5/2/13½*(3/5)=-3/2/13½
Fx[2]+Fx[7]>+Fx[12]>+Fx[8]=0
Fx[2] -3/2/13½ + 3/13½ -3/2/13½=0
F[2]=0
IN NOTE 5.:
Point 'F' has a TENSION of 1 lb. force is exerted on it.[S F=0]  
At point 'F': [SF=0 & F²=Fx²+Fy²]
By inspection F[4]=F[6]=1
IN NOTE 6.:
Release structure has member #12 removed. [S F=0]  
At point 'A': [SF=0 & F²=Fx²+Fy²] At point 'B': [SF=0 & F²=Fx²+Fy²]
F[3]=5P/4
F[2]=-F[3]x*3/5=-3P/4
F[1]=-F[3]y*4/5=-P4 		F[7]y=-F[7]*4/5=+5P/2*4/5=2P
F[5]y=-F[5]*4/5=+5P/4*4/5=P
F[7]y+F[5]y-3P=0
At point 'E': [SF=0 & F²=Fx²+Fy²] At point 'F': [SF=0 & F²=Fx²+Fy²]
F[3]y=-F[3]*4/5=-5P/4*4/5=-P
F[5]y=-F[5]*4/5=+5P/4*4/5=P
F[3]y+F[5]y=0

F[3]x=-F[3]*3/5=-5P/4*3/5=-3P/4
F[5]x=-F[5]*3/5=+5P/4*3/5=3P/4
F[4]=-2P

F[3]x+F[5]x+F[4]+F[6]=0
F[6]=-F[3]x-F[5]x-F[4]
F[6]=3P/4-3P/4+2P
F[6]=2P
[too large: stationary point] 		F[7]y=-F[7]*4/5=+5P/2*4/5=2P
F[9]y=-F[9]*4/5=0
F[7]y-2P=0

F[7]x=-F[7]*3/5=+5P/2*3/5=+3P/2
F[9]x=-F[9]*3/5=0
F[10]=-2P

F[6]+F[7]x+F[9]+F[10]=0
F[6]=-F[7]x-F[10]=-3P/2+2P=P/2
F[6]=P/2 [TENSION]
By inspection NOTES 5 & 4 do not react with each other so DELTA centre = 0
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ENERGY CONVERSION           by Medard Gabel
  Anthracite Coal
tons 		Oil
barrels 		Natural Gas
cu ft 		U-235
grams 		Deutrium
grams
1 kwh = 129E-6 608E-6 3.51 43.5E-6 15.1E-6
1 joule = 35.9E-12 172E-12 1030.9E-12 12E-12 4.21E-12
1 metric ton (10E6 BTU)
abthracite coal = 		1 5 25,400 .33 .12
1 barrel (42 gal)
oil = 		.21 1 5,470 .071 .026
1 cu ft (dry)
natural gas = 		39E-6 180E-6 1 13E-6 4.54E-6
1 gram
U-235 = 		3 14 79,300 1 3.5
1 gram
deutrium = 		8.6 40 227,000 2.9 1
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SQUARE ROOT TO FIFTH ROOT! gif/htmath.gif
CALCULUS APPROXIMATIONS FOR:
(124)1/3 = sin(60o 1') =
y = x1/3, dy = (1/3)(x)-2/3dx
x = 125 = 53, dx = -1,
dy = (1/3)125-2/3(-1) = -1/75 = -.1033
(124)1/3 = y + dy = 5 - .0133 = 4.9867 		y = sin(x), dy = cos(x)dx, x = 60o, dx = +1' = .0003 rad.
y = (3.5)/2 = .86603
dy = cos(x)dx = .5(.0003) = .00015
sin(60o 1') = y + dy = .86603 + .00015 = .86618
CALCULUS WEIGHT FOR: CALCULUS ROOT FOR:
.5''Dia. copper tube 8' long 1/8''thick 2cosx - x2 = 0 or y = 2cosx = x2 or y = 2cosx - x2
p = 3.1415926536
Specific weight of copper is 550 lb/ft3        
V = 8pr2
dV = 16prdr
dr = (1/8)/12 ft
r = (1/4)/12 ft
dV = 16p(1/48)(1/96) = p/288 ft3
Weight is 550(p/288) = 6 lb. 		Curves y = 2cos(x) & y = x2 give intersects » ±1              
because of symetry let x1 = 1, first approximation
x2 = x1 + dx1 = x1 - y1/f'(x1) = x1 - f(x1)/f'(x1) =
        1 - (2cos(1) - 12)/(-2sin(1) - 2(1)) =
        1 + (2(.5403) - 1)/(2(.9415) + 2) = 1.02
x3 = 1 - (2cos(1.02) - 1.022)/(-2sin(1.02) - 2(1.02)) =
        1.02 + (.0064)/(3.7443) = ±1.0217
to 4 decimal places.
 
THE # IS DIVISIBLE BY: THE # IS DIVISIBLE BY:
2, if it is even
3, if the sum of its digits is div. by 3
4, if the last 2 digit # is div. by 4
5, if the last digit is a 5 or 0
7, if the # is 100a+b and a+4b =>0mod7
ie. 257 => 2+4*57=238 => 2+4*38=154  		  6, if the # is even and div. by 3 =>0mod3
  8, if the last 3 digits is div. by 8
  9, if the sum of its digits is div. by 9
10, if the last digit is 0
11, if the diff. of the sums of odd-even =>0mod11              
ie. 92,213 => 3+2+9-1-2=14-3=11
THE ABACUS HAS BEATEN EVERY COMPUTER IN SPEED!
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MY "AUTOCAD" HOOK
IF YOU KNEW EVERYTHING ABOUT ELLIPSES
AND AUTOLISP YOU WOULD HAVE NO TROUBLE
DRAWING UP THIS CABLE HOOK, IN 'AUTOCAD'!
AS I'VE DONE IT, THERE IS NOTHING TO IT!
IN ProENG. I WAS FORCED TO USE VERTEX POINTS!
gif/hook1.gif

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3D CURVE
gif/3dcurve.gif
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SPOT WELDING
gif/e/sptwldng.gif
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GATE FUNCTIONS
gif/e/gates.gif
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NPN TRANSISTORS
gif/e/npn.gif
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GYRO PLANES
gif/e/gyro001.gif

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HARNESS THE SUN
gif/e/ecosystm.gif

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FAST AREA?
gif/e/areamn.gif

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ELECTRONIC CHIPS
gif/e/chips.gif

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FOCUS
gif/focus.gif

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LOADER PLAN
gif/excavtr.gif

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TRISECTION MECHANISM
gif/3.gif

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AMPLIFIER
gif/e/amplifier.gif

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ORCHESTRA INSTRUMENTS
gif/e/orchstra.gif

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CALCULUS'  VOLUME OF A CONE:

  height = h with base (x/R)2 + (y/R)2 = 1  
gif/e/cone1h.gif
r/R = (h - z)/h or r = R(h - z)/h
dr/dz = -R/h or dr= -(R/h)dz
Area = (r*dz - dr*dz/2) = R/h((h - z)*dz + dz2/2)
Volume = Area*ds = Area*r*d
Volume = Area*ds = (R/h)[(h - z)*dz]r*d
Volume = (R/h)2[(h - z)2*dz]d
Volume = (R/h)2(h - z)2*dz
Volume = (R/h)2[(h - z)3/3]
Volume = R2h/3


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CALCULUS'  VOLUME OF A SPHERE:


gif/e/cone4.gif
Volume of a sphere:
Volume = dV = ds*ds1*ds2
Volume = (p*d)(psin*d)(dp)
Volume = 4p2sin*dp*d*d
Volume = (4/3)R3sin*d*d
Volume = (4/3)R3sin*d
Volume = (4/3)R3(-cos)sinb0290.gif
Volume = 4R3/3


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OPTICAL ILLUSION VII
gif/puzz7o.gif
Will the 3 rd solid pass thru all the 1 st holes?

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