Phew Based Math of "Curvature"

This time ¶ (phew) is absent, as the replacement, & (dan) - the core of phew - shows up in the equation.

^ =1000(0.4142.r(d/&r)²)

^ = sinking length
r = earth radius
d = distance
& = 0.7929
Dimension = kms

Good luck n you'll find out that "curvature" is nonsense :')

Usually when I proposed phew, this phew calculation failed first, but later on it succeed.

Input by kms, the result will be in meters. For there is 1000 multipication in the equation.

Height of observer's eyes/telescope is excluded. It is nearly nothing to give significant differences to the final results.

This time ¶ (phew) is absent, as the replacement, & (dan) - the core of phew - shows up in the equation.

And it is still just a bunch of baseless crap.

Also, you don't need any of this to find the distance to the horizon, or how much an object appears to drop below the horizon. That can be done purely with pythagoras.

To find the distance to the horizon (d), given a spherical Earth with a radius (r) and an observer height (h, I'm not going to use caret as that has a fundamentally different meaning in math meaning raise to the power of), then we construct a right angle triangle.
One line (the hypotenuse) connects the observer eye height to the centre of Earth, and is of length r+h.
Another line will be tangent to Earth (hence where the right angle comes in), touching Earth at the horizon, and going to the observer's eye height. This will have length d.
The final line connects the horizon to the centre of Earth and has length r.

Thus (r+h)^2=r^2+d^2
r^2+2*r*h+h^2=r^2+d^2
2*r*h+h^2=d^2
h*(2*r+h)=d^2.

Now r>>h, thus 2*r+h~=2*r.
Thus d^2~=2*h*r.
So h=d^2/2r

Now for some strange reason you have decided that this should instead be:
1000(0.4142.r(d/&r)²)

WHY?

Quote from: Danang on December 15, 2017, 05:16:19 AM

This time ¶ (phew) is absent, as the replacement, & (dan) - the core of phew - shows up in the equation.

And it is still just a bunch of baseless crap.

Also, you don't need any of this to find the distance to the horizon, or how much an object appears to drop below the horizon. That can be done purely with pythagoras.

To find the distance to the horizon (d), given a spherical Earth with a radius (r) and an observer height (h, I'm not going to use caret as that has a fundamentally different meaning in math meaning raise to the power of), then we construct a right angle triangle.
One line (the hypotenuse) connects the observer eye height to the centre of Earth, and is of length r+h.
Another line will be tangent to Earth (hence where the right angle comes in), touching Earth at the horizon, and going to the observer's eye height. This will have length d.
The final line connects the horizon to the centre of Earth and has length r.

Thus (r+h)^2=r^2+d^2
r^2+2*r*h+h^2=r^2+d^2
2*r*h+h^2=d^2
h*(2*r+h)=d^2.

Now r>>h, thus 2*r+h~=2*r.
Thus d^2~=2*h*r.
So h=d^2/2r

Now for some strange reason you have decided that this should instead be:
1000(0.4142.r(d/&r)²)

WHY?

what if the distance = ⅛ C which will result loss of sight of 0.4142 of earth radius?

My equation is based on ⅛ of earth's circumference = 45° twist.
No need for more than 45°.

That means the lost length = 0.4142 of earth radius.

I a few times editted my posts above. Sorry for my English n my...... kinda ... dyslexia :d

 ^ =1000(0.4142.r(d/&r)²)
d=10 kms

If r=8000 kms:


^=1000(0.4142×8000(10/0.7929×8000)²= 8.23 meters.

If r=6371 kms:


^=1000(0.4142×6371(10/0.7929×6371)²= 10.34 meters

what if the distance = ⅛ C which will result loss of sight of 0.4142 of earth radius?

Then the math is no longer that simple.
You can still use a right angle triangle, but the arc length will no longer be the same as the distance from your eye to the horizon.

So with h= (1/8) C or 2*pi*r/8=pi*r/4, you have:
2*h*r+h^2=d^2
2*(pi*r/4)*r+pi^2*r^2/16=d^2
d^2=pi*r^2/2+pi^2*r^2/16
=r^2*(pi/2+pi^2/16)
Thus the horizon will be at a distance of 2.19*r

But now the arc length will be much smaller.
Continuing with our right angle triangle, and letting alpha be the angle at the centre of Earth, we have:
cos(alpha)=r/(r+h)=r/(r+pi*r/4)=1/(1+pi/4)=4/(4+pi)=~0.976290843 (in radians).
Now the arc length will be alpha*r, i.e. 0.976290843*r.

So the horizon is now twice as far as the arc length along the surface.

But this still isn't your formula.

My equation is based on ⅛ of earth's circumference = 45° twist.

Which would make it a specific case, not a formula.
A formula should be general which can apply to a wide variety of cases.

Or did you mean it the other way around, where you have a distance along Earth of 1/8 C?
In that case, alpha =pi/4.
Cos(alpha)=r/(r+h)
cos(pi/4)=1/sqrt(2)=r/(r+h)
r+h=sqrt(2)*r
h=sqrt(2)*r-r
h~=0.412*r.

This still isn't your "formula" and again, it applies for a specific case.

^ =1000(0.4142.r(d/&r)²)
d=10 kms

If r=8000 kms:


^=1000(0.4142×8000(10/0.7929×8000)²= 8.23 meters.

No, in this case, you would be quite close to Earth so the small h approximation applies.
Thus h=d^2/(2*r)
So if r=8000 km, and d=10 km, you get:
h=d^2/(2*r)=100 km^2/2*8000 km)=0.00625 km = 6.25 m.

Again, making up a formula and sticking numbers in doesn't make you correct.
You need to be able to justify it.

Jack changed the formula as he wished. #ThumbsUp

Okay, what about d=1 km? Should its formula be changed?

Jack changed the formula as he wished. #ThumbsUp

Okay, what about d=1 km? Should its formula be changed?

Why do you persist with your idiotic "Phew Based Math" garbage? It's no wonder everybody thinks that you are an imbecile!

Start using ordinary math and arithmetic if you want to be treated seriously.

"Currently Phew is the fastest growing formula to destroy pi." - John Lennon

Jack changed the formula as he wished. #ThumbsUp

Okay, what about d=1 km? Should its formula be changed?

No.

The more general formula uses cosine.
So if you have an arc length (l), which is the distance to the horizon along the surface of Earth with radius r, from an observer with eye height h, you have a right angle triangle where the angle at the centre is a.
This gives:
d=r*a
Thus a=d/r.

It also means
cos(a)=r/(r+h).
Thus r+h= r/cos(a)=r/(cos(d/r))
Thus h=r/cos(d/r) - r=r*(1/cos(d/r)-1)

This is the physical formula which holds as long as refraction isn't an issue.

But again, this can be simplified if the angle is small enough.
cos(x)~=1-x^2/2, for small x.
So taking it in the form:
(r+h)*cos(d/r)=r
And applying the simplification we get:
(r+h)*(1-0.5*d^2/r^2)=r
r*(1-0.5*d^2/r^2)+h*(1-0.5*d^2/r^2)=r
r-0.5*d^2/r+h*(1-0.5*d^2/r^2)=r
h*(1-0.5*d^2/r^2)=d^2/2r

Now, as d<<r, then d/r<<1, and thus (d/r)^2<<1, and thus 1-0.5*(d/r)^2~=1
This again gives us:
h=d^2/2r.

One is an approximation, one is not.

To determine which one you need, you need to see if d/r is small, and what your cutoff is.

You can also put both of these into google (normalised to r) to see the difference, e.g. searching for x^2/2 and 1/cos(x)-1, or the difference between them, 1+x^2/2-1/cos(x).

As an example, using r=6378, you get these values (note, d is in km always, h (and dif) is in m up until and including d=100, after than (from d=500 onwards) h and diff are in km).

d	h simp	h act	diff
0	0	0	0
1	0.0782840144042587	0.0782840157944609	1.39020221723296E-09
5	1.95710036010647	1.95710086003076	4.99924291963438E-07
10	7.82840144042587	7.82840943626795	7.99584208177606E-06
50	195.710036010647	195.715033579948	0.00499756930139483
100	782.840144042587	782.920111136535	0.0799670939479711
500	19.5710036010647	19.6211029091228	0.050099308058094
1000	78.2840144042587	79.0916566366371	0.807642232378484
2000	313.136057617035	326.460781429205	13.32472381217
5000	1957.10036010647	2622.56488352141	665.46452341494
10000	7828.40144042587	1242045.27828033	1234216.87683991

Even when d=100 km, the difference between the 2 is only 8 cm.

"Currently Phew is the fastest growing formula to destroy pi." - John Lennon

Is that what you call yourself?
You are yet to destroy pi in any way.

As an added bonus, here is a comparison between reality, and you:

d	h act	Your bs	diff
1	0.0782840157944609	67850319779864	67850319779863.9
5	1.95710086003076	2714012791194.56	2714012791192.6
10	7.82840943626795	678503197798.64	678503197790.812
50	195.715033579948	27140127911.9456	27140127716.2306
100	782.920111136535	6785031977.9864	6785031195.06629
500	19621.1029091228	271401279.119456	271381658.016547
1000	79091.6566366371	67850319.779864	67771228.1232274
2000	326460.781429205	16962579.944966	16636119.1635368
5000	2622564.88352141	2714012.79119456	91447.907673155
10000	1242045278.28033	678503.19779864	-1241366775.08253

The only time the match is when d=5036.735786

.......

I think the equation notation can be change a little bit >> ^_^ =1000(0.4142.r(d/&r)²)

Sorry for my English n my...... kinda ... dyslexia :d

Dyslexia in mathematics isn't really a problem.   Your dyscalculia is the problem.

I think the equation notation can be change a little bit >> ^_^ =1000(0.4142.r(d/&r)²)

So you are just trolling?

Better trolling than making up a formula and replacing the numbers.

"Paul, I think I've told you: I'm a lover, not a troller" ~

Better trolling than making up a formula and replacing the numbers.

But that is what you do, make up formulas and play with numbers.

I think the equation notation can be change a little bit >> ^_^ =1000(0.4142.r(d/&r)²)

'.' and '&' mean nothing in equations.  As you know.

Quote from: Danang on December 17, 2017, 04:49:23 AM

I think the equation notation can be change a little bit >> ^_^ =1000(0.4142.r(d/&r)²)

'.' and '&' mean nothing in equations.  As you know.

"." is sometimes used to indicate multiplication, but it is meant to be a middle dot.

"&" is part of his BS for pi.
He seems to like using strange symbols instead of letters.

Any symbol can be used as long as been communicated before. 

Curvature??
Look at the sun in the morning before bright enough.
Why there are TWO type of sights at the same time: Bright AND Dark.

Bright indicates its altitude is higher than the sun.

Dark (underneath cloud) indicates its altitude is lower than the sun.

The sun lies between both layers.

......

AND...

The sun's light CAN reach 12 hours distance, or half of earth circumference i.e. from east to west.

IMPOSSIBLE for globe model.

The sun's light CAN reach 12 hours distance, or half of earth circumference i.e. from east to west.

IMPOSSIBLE for globe model.

The sun's light always reaches slightly more than half the earth....        Why slightly more?   1:  The sun is much bigger than the earth,  and 2:  the earth's atmosphere refracts some light around the edges.
The illuminated portion grows slightly towards our perihelion (currently early January)  and reduces towards the aphelion (currently early July).
This all happens because the earth is a sphere.
The more light coverage at perihelion and less light coverage at aphelion is one of the proofs that the sun is much bigger than the earth. If it were smaller then the opposite would happen.