RE: Occam's Razor - on request of Virender Dayal - how to calculate

[Guest guest]

Here it is, simplified for everyone's understanding.

 

It involves addition, subtraction, multiplication, division and one single

cosinus. Sorry, it can't be made simpler than this.

 

Step 1

 

"How Big Is The Earth"

 

Accepted things at the start:

 

- That the Earth is ball-shaped (not flat). If someone thinks it's flat then

the discussion will go to very wild directions...

 

- That one has mystic power of measuring the distance from point A to point

B. If anyone thinks that a conditioned living entity with his limited senses

and mind can not do that, we are in deep trouble.

 

Intentional simplifications, for the sake of simpler explanations:

 

- I assumed that North-South axis of Earth is under right angle to the plane

of ecliptic, which is not quite the truth. So real calculations would have

to be a little bit adjusted.

 

- I assumed measuring at the equator; measuring at other parts of the globe

would make this slightly more complicated, and I wanted to keep it simple.

 

- I also assumed that measuring from point A to point B is done on the

sea-level; so if someone does not have a boat and does not live at sea level

in a very flat land, he would have to be more careful about measuring.

 

- I ignored atmospheric disturbances and the fact that Sun's gravity may

curve the light.

 

First two of assumptions would not be kept in real life; they are assumed

only in order to keep this simple. As for the second two, you would need a

clear atmosphere, a flat land (or sea) and very precise measuring.

 

Needed tools:

 

- Human brain, about 1300 grams of it. Give or take 200 gr.

- A tool for checking if a line is really horizontal (in regard to Earth). A

glass pipe can be used, filled with water with one bubble of air, and a mark

must be drawn at the middle of it. When the air bubble is at the middle, it

means the entire tool is placed horizontally. Sorry, I don't know English

name for it.

- A long pole, several meters of it. Or several smaller which can connect

end-to-end, it would be easier to carry around.

An alternative would be a high building, if you can find two equal ones; or

a deep well, it was used by an ancient astronomer for the same purpose.

- A rope.

- A budget for traveling, or a set of maps. Maps would spare you the need

for measuring and also would spare you some traveling; but that would mean

that you believe to demonic, cheating, stupid scientists which of course is

not acceptable ))

 

How to do it:

 

1. Pick two places at the equator; we'll call them A and B. Farther apart

they are, more precise measuring you will get. In ideal case they would be

at opposite points of Earth, but it is not absolutely necessary to go to

that length.

 

2. Put a pole in point A so that it is at the right angle to a 2D-plane

tangential to Earth in point A. For that, attach a thin rope to the top of

the pole and a small piece of lead at the other end (the free end) of the

rope. The length of the rope should be about 90% of the length of the pole.

When the free end falls down without any distance from it and the pole, the

pole is at the proper angle.

 

3. Put an identical pole in point B. If you have an assistant, it may be

done simultaneously; otherwise, you will have to travel.

 

4. Wait for noon at point A. This is the moment in time when the pole does

not cast any shadow.

 

5. At point-A noon, measure the length of shadow at point B.

 

6. Wait at point A until the pole casts the shadow equally long as the

shadow cast by the pole at point B at the time of point-A noon.

 

7. Then calculate the amount of time passed since point-A noon. We'll call

this time "t"

 

8. Now remember that Earth rotates in 24 h. So,

24 h : total circumference of Earth = t : distance AB

 

and it is identical to

 

circumference = (24 h / t) * distance AB.

 

9. In mathematical style, we'll mark circumference as "O".

 

Since we have O = 2*r*PI, then half-diameter of Earth ("r") can be

calculated as

r = O/(2*PI)

 

 

Ok, that's it - now we know how big the Earth is, and we're ready to go to

the step 2.

 

Step 2

 

"How Far It Is To The Sun"

 

 

1. I recommend taking a paper and a pencil to follow this easier. Draw a

circle. It represents a wire-frame cross-section of Earth at equator, as

would be seen from any point of Earth's North-South axis, if you could see

through the Earth.

 

2. Mark point A1. Wait till sunrise and put the "horizontalness-measurement"

tool (the longer the better) horizontal to the Earth and towards the center

of the Sun. The aim is to catch the very first time (up to a second, if you

can) when the center point of the sun is visible.

 

3. Repeat the Sun-center-catching at sunset. The aim is to get the very last

second when the Sun's center is visible.

 

4. Now the Earth has rotated, and you should mark your new point on the

circle - let's call it A2.

 

5. Draw lines from A1 and A2 to the center (which you may mark as "C").

 

6. The angle between A1C and A2C will be marked as "alpha" (sorry, no true

Greek letters on COM). You may calculate that angle using the time

difference from sunrise to sunset: 360 (full circle) : alpha = 24 h :

measured daytime.

 

7. The center of the Sun we'll properly mark as "S". Draw straight line A1S.

 

8. CA1 is a half-diameter from the center of the circle to point A1. A1S is

determined by the horizontalness-measurement tool and thus it is a

tangential line to the circle in point A1. This makes the angle CA1S a right

angle (90 degrees) and entire triangle CA1S is a right-angled triangle.

Angle A1CS is half of A1CA2, so it is equal to alpha/2.

 

9. Thus we have CA1/CS = cos(alpha/2) which resolves to CS =

CA1/cos(alpha/2).

 

Satisfied?

 

The only problem in practice is to make a horizontalness-measurement tool

precise enough, and to catch the precise center of the Sun.

 

Entire system devised by me and a colleague of mine this Friday afternoon.

 

Of course, this also ignores Earth's revolution (from sunrise to sunset).

With a little more bit of math we may have a system which is not dependant

on that. How?

 

Step 2 version 2 : "How Far Is To The Sun Again"

 

1. We continue after the Step 1 with it's data

 

2. We need the Sun exactly between A and B. Thus, we need to be at point A

at time "noon of A"-t where t is the time period measured in Step 1.

 

3. Place a horizontalness-measurement tool horizontal but toward the Sun.

Place your head so that your eye is near the tool. Place a stick so that

it's one end touches the tool where your eye is and target the other end

toward the center of the Sun. Measure the angle between the tool and the

stick (ever heard about the sextant?)

 

4. Draw straight lines AB, AC, BC (C is the Center of the Earth), AS, BS (S

is the center of the SUn as you perceive it).

 

Note that AS line represents the stick.

 

5. Draw a straight line, tangential to Earth circle at point A. This

represents the horizontalness tool.

 

(Lacking such a tool, you may use 3 pieces of rope, its proportions should

be 3:4:5, this gives you a right-angled triangle. Make medium side hang in

the air ant it will be vertical to Earth (when there is no wind). Fix it in

that position and pull & fix other two sides until they are straight. Lower

side of the triangle will now be perfectly horizontal. Aim the corner you

just got toward the Sun, but only in horizontal plane - it's vertical

position shouldn't change. Now add the stick and you are ready to do

sub-step 3 without fancy equipment...)

 

6. Where this straight line crosses CS is the point X. Where AB crosses CS

is the point T.

 

7. Since the Sun is in the middle of AB, angle TCA is half of the angle BCA.

(BCA calculated in Step 1)

For the same reason, CS crosses AB at the 90 degrees angle.

 

This gives that angle CAT is 90 degrees - angle TCA

 

Now since the tangential line to a circle is at the right angle to

circle's diameter at tangential point,

we have that the angle TAX is 90 degrees - angle CAT.

 

Since you measured angle XAS, you may calculate angle TAS (or BAS, it's

the same)

as angle TAX + angle XAS.

 

ST/AT = tg(angle TAS), thus ST = AT*tg(angle TAS)

 

CT/AT = tg(angle TAC), thus CT = AT*tg(angle TAC), or

CT/r = sin(angle TAC), thus CT = r*sin(angle TAC)

 

Finally CS = CT + ST.

 

 

 

I hope you enjoyed this. If anyone actually thinks about doing this ))) ,

note that achieving a precise measurement is the greatest problem of all.

 

yours,

 

DVD