UniKEF analysis

[MacM]

GundamWing,

(by the way, I don't know how someone can do a integral "long-hand"... either you integrate, or you don't... there is no "long-hand" method

ANS: I was responding to this but I accept your meaning.

[MacM]

Pending a final calculus evaluation as an intrim step I am writing a program that will compute areas in 1 degree steps to be able to show the inverse square relationship.

"Attachment Workaround"

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[Persol]

If you assume that all angles follow the inverse square law, then just follow the easy way and do all the line horizontally. Not that it woul either calculation actually proves anything.

[MacM]

Persol,

If you assume that all angles follow the inverse square law, then just follow the easy way and do all the line horizontally. Not that it woul either calculation actually proves anything.

ANS: Not sure what you are saying here. It is not that all angles are inverse square. It is the integration of all angles from zero to the maximum C.O.S. that is inverse square.

[Helloween]

The Geodetic Institute in Frankfurt, Germany measured a 4.28E-9 deviation in gravity during a Lunar eclipse in Norway in 1954. A mechanical view via UniKEF yields results of 4.2E-9 deviation predicted.

Well, this is interesting. The Geodetic Institute measures a 4.28E-9 deviation in gravity. hmmm...what's wrong with this picture....ohh yeah, last time I checked, the gravitational constant was only accurately measured to 6 count it SIX significant digits. Suddenly a 4.2E-9 deviation in gravity doesnt seem so significant or even measurable for that matter.

[Vortexx]

Yo guys now that 1954 was an historic low in solar storm / sunspot activity ---> coronal size ?

http://www.dxlc.com/solar/history/hist1954.html

compare with recent historical high:

http://www.dxlc.com/solar/history/hist2003.html

I wonder just how representative this makes the gravity measurement based upon lunar eclipse

Just Joking, but who knows some jokes contain might be true....

[MacM]

Helloween,

Well, this is interesting. The Geodetic Institute measures a 4.28E-9 deviation in gravity. hmmm...what's wrong with this picture....ohh yeah, last time I checked, the gravitational constant was only accurately measured to 6 count it SIX significant digits. Suddenly a 4.2E-9 deviation in gravity doesnt seem so significant or even measurable for that matter.

ANS: Perhaps you don't realize that 4.2E-6 Gal Deviation/980 Gal Field = 4.28E-9 delta. If that isn't your problem, I have included their address such that you might inform them that they didn't achieve what they claimed.

NOTE: For some reason to make the attachment appear place your mouse in the upper left area of the screen, when the colored square appears at the lower right click on it.

"Attachment Workaround"

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[MacM]

Vortexx,

Just Joking, but who knows some jokes contain might be true....

ANS: Actually very interesting graphs. Major change in activity of the sun. Any idea what the units across the bottom are?. They call it data for a given year but there are 29 divisions across the bottom.

[Helloween]

ANS: Perhaps you don't realize that 4.2E-6 Gal Deviation/980 Gal Field = 4.28E-9 delta. If that isn't your problem, I have included their address such that you might inform them that they didn't achieve what they claimed.

Perhaps you just made that up. One Gal is approximately 0.0010197g, so a milligal (or mGal) is a very small acceleration, of about 10E-6 g (or 10E–5 m s-2). Any measurement that claims a 4.2E-6 Gal deviation would still require an accuracy on g measured of the order 10E-9. I still maintain that this is not possible given the 6 significant digit limit on the gravitational constant.

Maybe you should be more skeptical of your friends at the institute. The mean Earth gravity is about 981 000 mGal varies from 978,100 mGal to 983,200 mGal from Equator to pole due to the Earth's flattening and rotation. Variations, due to density inhomogeneities, mountain ridges, etc., range from tens to hundreds of milligals. Hundreds of milligals is a gravity fluctuation more than five orders of magntude greater than the deviation your friends claim to have detected !

[MacM]

Helloween,

Perhaps you just made that up.

ANS: THEN AGAIN PERHAPS I DIDN'T. PERHAPS YOU ARE 50 YEARS BEHIND THE TIMES. THE INSTITUTE IS WELL RECOGNIZED.

Attached FYI is Cover letters from the US Army Research Command and the full translated copy of the document.

There is a (5) file limit so this will be continued in subsequent posts.

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[MacM]

Continuation of attachments:

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[MacM]

More attachments

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[MacM]

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[MacM]

Last Page.


Perhaps it is you that don't seem to realise that if I can measure a 1 pound weight on a scale that can weigh a ton, you have a resolution of 5E-4, even though the scale can't measure 0.5 pounds.

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[Helloween]

READ WHAT I WROTE. THIS IS NOT ABOUT RESOLUTION.

You claim 4.2E-6 Gal deviation was detected. I claim 4.2E-6 Gal deviation is NOT POSSIBLE to detect, and I am working on YOUR scale. I will take you through the math that I explained in my last post.

10E-3*Gal = 10E-6*g

therefore:
10E-6*Gal = 10E-9*g

that was my quick 'in my head' calculation

the results of the actual calculation require the equipment your institute has to be able to detect fluctuations in g of 4.197E-8 m/s^2. This I claim is not possible because the gravitational constant has only been confirmed to 6 significant digits! In case you dont know about significant digits and the gravitational constant, that's 6.67XXXE-11. This means that any calculation of the earth's gravity can only be accepted to 6 digits! Fluctuations in g of 4.197E-8 m/s^2 are too small to detect now and were certainly too small to detect in 1954.

Your predicament is worsened by my next statement in my last post.

Maybe you should be more skeptical of your friends at the institute. The mean Earth gravity is about 981 000 mGal varies from 978,100 mGal to 983,200 mGal from Equator to pole due to the Earth's flattening and rotation. Variations, due to density inhomogeneities, mountain ridges, etc., range from tens to hundreds of milligals.

This means the earth's gravity fluctuates by as much as .1 Gal due to stuff that happens on earth. A 10E-6 Gal shift as you claim would be vastly overshadowed by fluctuations of that size.

I DONT WANT TO READ YOUR MAIL, I WANT YOU TO RESPOND TO WHAT I HAVE SAID.

[MacM]

Helloween,

I DONT WANT TO READ YOUR MAIL, I WANT YOU TO RESPOND TO WHAT I HAVE SAID.

ANS: I understand YOUR predicment.

You would rather argue with me and make unsupported claims than to argue with the results claimed by a recognized institute.

If you actually read their work you would see they have been very careful to take into account all pertabations before and after the eclispse by several days. The pertabation was infact recorded, as indicated, during the eclipse.

There is a lot of scientific information available on this just search the web for the 1954 eclipse.

This is not my mail, it is scientific data which you seem to want to disavow. Let me suggest it takes more than your word to undermine the work of this Institute. This finding is well recognized and accepted world wide. What is your problem.?

FYI I am adding Chapter 8, which is my vision of what caused the measured deviation and as approved by the letter from the Institute for publication referencing their work.

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[MacM]

THe last page of Chapter 8.

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[Persol]

Once again, nobody cares that you've sent mail and recieved a response from the Geo. Inst..

The issue here was supposed to be your theory. Specifically, JamesR directed this thread towards any math that may exist. Currently, you have stated that:

1) Your theory reproduces an inverse square equation similar to that observed for gravity.
2) You have numercially solved this equation
3) The fact that the COS is not multi-source, multi-directional is an acceptable simplification.

The problem starts with number 3, and moves to others...
1) As we've discussed, you have not demonstrated that the COS is an acceptable simplification.
2) considering that the field is supposed to be omni-directional, there will be field lines that travel through both spheres horizontally.... where F is not related to an inverse square... but has a constant component.
3) The field lines in your COS example do not take into account that a substaintial percentage of the 'force' produced is not in the horizontal direction. You can not simply see how much mass it has gone through. You would need to account for the direction the field 'impacts' the mass to see which direction the force is in. This is a BIG deal, as any COS with large angles is very very wrong.

but, most importantly...

You have not shown a connection between your theory and the COS calculation (other then both of them being supported by you). Any credible scientist would show the steps needed to get from point A to B. You have not done this. "It's my vision" is not a valid step, and I think that you understand this.

[MacM]

Persol,

Once again, nobody cares that you've sent mail and recieved a response from the Geo. Inst..

ANS:Of course not we wouldn't want any sort of confirmation that this isn't made up, especially if it comes from reputable sources.

The issue here was supposed to be your theory. Specifically, JamesR directed this thread towards any math that may exist.

ANS: Indeed, then lambast Helloween for introducing this aspect of the theory since it is not falsifiable. I addressed his assertion that "I made it up".

You have not shown a connection between your theory and the COS calculation (other then both of them being supported by you). Any credible scientist would show the steps needed to get from point A to B. You have not done this. "It's my vision" is not a valid step, and I think that you understand this.

ANS: Not a true statement. I have explained the calculation done. I have provided calculus done by a highly qualified physicist. The fact that it is difficult to follow since it wasn't prepared as a formal paper doens't alter the fact that it was done and has been made available.

Current efforts here are to duplicate the calculation in a manner that can be followed by all.

[Persol]

ANS: Not a true statement. I have explained the calculation done. I have provided calculus done by a highly qualified physicist. The fact that it is difficult to follow since it wasn't prepared as a formal paper doens't alter the fact that it was done and has been made available.

You completely didn't address my point. Yes, you showed the COS work. No, you didn't explain how this represents the field. Well, since I'm stuck in a country by myself where I don't speak the language... I had some time on my hands... and since it seems that you aren't going to simulate a multi-directional / multi-source field, I did it myself. I don't have any programming tools on this laptop, so I had to do it in Excel.

The data shows that the values do not approach zero, and level off way above.

The results follow, calculations were taken for distances of 20 to 90 at steps of 10. A 0 means no simulation was run (to save time).

For a size of 10:
2.817457995
0
1.441430195
1.246450474
1.067436176
1.016585752
1.070528463
1.143130426

For a size of 30:
0
16.8341404
12.12778174
9.66103224
8.241228907
7.379189245
6.965528769
0

For a size of 50:
0
9.954927498
9.445978488
7.882929662
6.479220824
5.612773714
0
0


Here's the actual code... since you'll ask for it anyhow. The size is changed by changing the value of 'dia1' and 'dia2'. The distance is prompted for, as is the row to save the results on. If you find a problem, explain it and fix it yourself. But don't completely ignore that your theory is supposedly multi-directional/sourced.

'----START OF MACRO----'

Sub Macro1()

'Simulation Variables
simbxmin = -100
simbxmax = 100
simbymin = -100
simbymax = 100

dia1 = 50
dia2 = 50
dist = 40
dist = InputBox("Input distance", "Distance?", "40")


Linear = 10 '10
Angular = 0.5 '2
Length = 0.5 '10

masseffect = 1
bstrength = (dia1 + dia2) * 2 * masseffect

'Scale
x0 = 300 / 2 + 200
y0 = 300 / 2 + 10
xscale = 300 / (simbxmax - simbxmin)
yscale = 300 / (simbymax - simbymin)

'Place Borders and Masses
'ActiveSheet.Shapes("Oval 1").Select
'Selection.ShapeRange.Height = dia1 * yscale
'Selection.ShapeRange.Width = dia1 * xscale
'Selection.ShapeRange.Left = x0 - dist * xscale / 2 - dia1 * xscale
'Selection.ShapeRange.Top = y0 - dia1 * yscale / 2
'ActiveSheet.Shapes("Oval 2").Select
'Selection.ShapeRange.Height = dia2 * yscale
'Selection.ShapeRange.Width = dia2 * xscale
'Selection.ShapeRange.Left = x0 + dist * xscale / 2 ' dia2 * xscale
'Selection.ShapeRange.Top = y0 - dia2 * yscale / 2
'ActiveSheet.Shapes("Rectangle 4").Select
'Selection.ShapeRange.Height = (simbymax - simbymin) * yscale
'Selection.ShapeRange.Width = (simbxmax - simbxmin) * xscale
'Selection.ShapeRange.Left = x0 + simbxmin * xscale
'Selection.ShapeRange.Top = y0 + simbymin * yscale

'GoTo endit

'Reset 'force' felt on mass
m1x = 0
m1y = 0
m2x = 0
m2y = 0

'ActiveSheet.Shapes("Line 3").Select

'Process Field comming from left
For y = simbymin To simbymax Step Linear
Call upd(y)
For Angle = 0 To 360 Step Angular
'Set position and direction
xd = Cos(Angle * 3.14 / 180) * Length
yd = Sin(Angle * 3.14 / 180) * Length
x1 = simbxmin
y1 = y
x2 = x1 + xd
y2 = y1 + yd
strength = bstrength
'Before doing this loop, I should check to see if the line intersects both spheres
'I'm lazy though and don't care if the sim takes longer to run
While strength > 0
'Make sure it is within the borders. Otherwise finish with this field line
If x1 < simbxmin Or y1 < simbymin Or x1 > simbxmax Or y1 > simbymax Then strength = -1
'Draw the field line
'Selection.ShapeRange.Height = (3) * yscale
'Selection.ShapeRange.Width = (3) * xscale
'Selection.ShapeRange.Left = x0 + x1 * xscale
'Selection.ShapeRange.Top = y0 + y1 * yscale
'doevents
'See if the field line has intercepted mass1
checkcollision = Sqr((x1 + dist / 2) ^ 2 + (y1 - 0) ^ 2)
If checkcollision <= dia1 / 2 Then
'Beep
m1x = m1x + Cos(Angle * 3.14 / 180) * strength
m1y = m1y + Sin(Angle * 3.14 / 180) * strength
strength = strength - masseffect
End If
'See if the field line has intercepted mass1
checkcollision = Sqr((x1 - dist / 2) ^ 2 + (y1 - 0) ^ 2)
If checkcollision <= dia2 / 2 Then
'Beep
m2x = m2x + Cos(Angle * 3.14 / 180) * strength
m2y = m2y + Sin(Angle * 3.14 / 180) * strength
strength = strength - masseffect
End If
'Update location
x1 = x1 + xd
x2 = x2 + xd
y1 = y1 + yd
y2 = y2 + yd
Wend
Next
Next

'Beep

'Process Field comming from right
For y = simbymin To simbymax Step Linear
Call upd(y)
For Angle = 0 To 360 Step Angular
'Set position and direction
xd = Cos(Angle * 3.14 / 180) * Length
yd = Sin(Angle * 3.14 / 180) * Length
x1 = simbxmax
y1 = y
x2 = x1 + xd
y2 = y1 + yd
strength = bstrength
'Before doing this loop, I should check to see if the line intersects both spheres
'I'm lazy though and don't care if the sim takes longer to run
While strength > 0
'Make sure it is within the borders. Otherwise finish with this field line
If x1 < simbxmin Or y1 < simbymin Or x1 > simbxmax Or y1 > simbymax Then strength = -1
'Draw the field line
'Selection.ShapeRange.Height = (3) * yscale
'Selection.ShapeRange.Width = (3) * xscale
'Selection.ShapeRange.Left = x0 + x1 * xscale
'Selection.ShapeRange.Top = y0 + y1 * yscale
'doevents
'See if the field line has intercepted mass1
checkcollision = Sqr((x1 + dist / 2) ^ 2 + (y1 - 0) ^ 2)
If checkcollision <= dia1 / 2 Then
'Beep
m1x = m1x + Cos(Angle * 3.14 / 180) * strength
m1y = m1y + Sin(Angle * 3.14 / 180) * strength
strength = strength - masseffect
End If
'See if the field line has intercepted mass1
checkcollision = Sqr((x1 - dist / 2) ^ 2 + (y1 - 0) ^ 2)
If checkcollision <= dia2 / 2 Then
'Beep
m2x = m2x + Cos(Angle * 3.14 / 180) * strength
m2y = m2y + Sin(Angle * 3.14 / 180) * strength
strength = strength - masseffect
End If
'Update location
x1 = x1 + xd
x2 = x2 + xd
y1 = y1 + yd
y2 = y2 + yd
Wend
Next
Next

'Beep

'Process Field comming from top
For y = simbxmin To simbxmax Step Linear
Call upd(y)
For Angle = 0 To 360 Step Angular
'Set position and direction
xd = Cos(Angle * 3.14 / 180) * Length
yd = Sin(Angle * 3.14 / 180) * Length
x1 = y
y1 = simbymin
x2 = x1 + xd
y2 = y1 + yd
strength = bstrength
'Before doing this loop, I should check to see if the line intersects both spheres
'I'm lazy though and don't care if the sim takes longer to run
While strength > 0
'Make sure it is within the borders. Otherwise finish with this field line
If x1 < simbxmin Or y1 < simbymin Or x1 > simbxmax Or y1 > simbymax Then strength = -1
'Draw the field line
'Selection.ShapeRange.Height = (3) * yscale
'Selection.ShapeRange.Width = (3) * xscale
'Selection.ShapeRange.Left = x0 + x1 * xscale
'Selection.ShapeRange.Top = y0 + y1 * yscale
'doevents
'See if the field line has intercepted mass1
checkcollision = Sqr((x1 + dist / 2) ^ 2 + (y1 - 0) ^ 2)
If checkcollision <= dia1 / 2 Then
'Beep
m1x = m1x + Cos(Angle * 3.14 / 180) * strength
m1y = m1y + Sin(Angle * 3.14 / 180) * strength
strength = strength - masseffect
End If
'See if the field line has intercepted mass1
checkcollision = Sqr((x1 - dist / 2) ^ 2 + (y1 - 0) ^ 2)
If checkcollision <= dia2 / 2 Then
'Beep
m2x = m2x + Cos(Angle * 3.14 / 180) * strength
m2y = m2y + Sin(Angle * 3.14 / 180) * strength
strength = strength - masseffect
End If
'Update location
x1 = x1 + xd
x2 = x2 + xd
y1 = y1 + yd
y2 = y2 + yd
Wend
Next
Next

'Beep

'Process Field comming from bottom
For y = simbxmin To simbxmax Step Linear
Call upd(y)
For Angle = 0 To 360 Step Angular
'Set position and direction
xd = Cos(Angle * 3.14 / 180) * Length
yd = Sin(Angle * 3.14 / 180) * Length
x1 = y
y1 = simbymax
x2 = x1 + xd
y2 = y1 + yd
strength = bstrength
'Before doing this loop, I should check to see if the line intersects both spheres
'I'm lazy though and don't care if the sim takes longer to run
While strength > 0
'Make sure it is within the borders. Otherwise finish with this field line
If x1 < simbxmin Or y1 < simbymin Or x1 > simbxmax Or y1 > simbymax Then strength = -1
'Draw the field line
'Selection.ShapeRange.Height = (3) * yscale
'Selection.ShapeRange.Width = (3) * xscale
'Selection.ShapeRange.Left = x0 + x1 * xscale
'Selection.ShapeRange.Top = y0 + y1 * yscale
'doevents
'See if the field line has intercepted mass1
checkcollision = Sqr((x1 + dist / 2) ^ 2 + (y1 - 0) ^ 2)
If checkcollision <= dia1 / 2 Then
'Beep
m1x = m1x + Cos(Angle * 3.14 / 180) * strength
m1y = m1y + Sin(Angle * 3.14 / 180) * strength
strength = strength - masseffect
End If
'See if the field line has intercepted mass1
checkcollision = Sqr((x1 - dist / 2) ^ 2 + (y1 - 0) ^ 2)
If checkcollision <= dia2 / 2 Then
'Beep
m2x = m2x + Cos(Angle * 3.14 / 180) * strength
m2y = m2y + Sin(Angle * 3.14 / 180) * strength
strength = strength - masseffect
End If
'Update location
x1 = x1 + xd
x2 = x2 + xd
y1 = y1 + yd
y2 = y2 + yd
Wend
Next
Next

Beep

putat = "26"
putat = InputBox("Input line", "Line?", "26")
Range("A" & putat).Select
ActiveCell.Value = m1x
Range("B" & putat).Select
ActiveCell.Value = m1y
Range("C" & putat).Select
ActiveCell.Value = m2x
Range("D" & putat).Select
ActiveCell.Value = m2y
Range("F" & putat).Select
ActiveCell.Value = dist

endit:
End Sub

Sub upd(place)
Range("A24").Select
ActiveCell.Value = place
DoEvents
End Sub

'---END---'

THIS is multi-direcitonal and multi-sourced. As I've explained MANY MANY MANY times, your COS calculation is not... and has nothing to do with your description of the UniKEF field. I've shown that the field as defined by you does NOT result in an inverse square law as seen by gravity, but results in a huge offset and other problems.