Bjarne on: Dark Energy, only an illusion.

[Bjarne]

The asymmetry is caused by two things:
-The meter is slightly elevated off each floor
-I don't think I included the roof

The roof is a bit strange
1.) The height of 30th floor is only about 1,2 meter, - it is a service floor I believe.
2.) The building does also have a elevator tower.
3.) Above this I assume the roof concrete layer is a bit thicker.

 

[Reality Check]

Completely Crap
35.000.000 * 6.67E-11 /50^2 = 0.000000934 m/s²Or……………………………………………………….. 93.4 MicroGal
Or……………………………………………………….. 0.0934 mGal.

Completely Crap and completely not what you originally posted and have been repeating since.

Originally Posted by Bjarne
35000000 * 6,67E-11 / 50^2 = 0.100 MicroGal.

which is wrong (it is actually 0.9338 microGal).

Doing the math right and to the correct decimal places:
35.000.000 * 6.67E-11 /50^2 = 0.0000009338 m/s
Or……………………………………………………….. 0.9338 microGal
Or……………………………………………………….. 0.009338 milliGal

A hint: micro = multplying the number by 10^6 or shifting the decimal point 6 places to the right; milli = multplying the number by 10^4 or shifting the decimal point 4 places to the right

The fatal flaws and basic ignorance in Bjarne's ideas in his threads
Yet another fundemantally flawed idea from Bjarne​

 

[Bjarne]

How is it bad? If you don't know the gradient without the building there, how do you know what the contribution of the building is in your measurements?

The free air gradient is there
It MUST be there, and it must also be easy to show, prove, calculate…..
At the physical center of the building
ACCORDING TO THE PREVAILING THESIS.
(BUT the hole speculation is completely wrong therefore you can’t find this “merge point”)

 

[Bjarne]

Doing the math right and to the correct decimal places:
35.000.000 * 6.67E-11 /50^2 = 0.0000009338 m/s
Or……………………………………………………….. 0.9338 microGal
Or……………………………………………………….. 0.09338 milliGal

RC
I think you need 11 year more at the University
Start to learn what 1 Gal is

 

[Reality Check]

This is mostly for a bit of practise and fun, since I know Bjarne won't pay attention, but here goes...

I've had a go at calculating the ADG we'd expect to measure inside Domus Vista by modelling it as a cylindrical building and including the effects of all 30 floors and the walls. Since I've no idea about the elevation and local geology, it's just a model of the building in space.

Here are the results:

[URL]http://www.internationalskeptics.com/forums/imagehosting/thum_365534c2ca9f73c9d5.png[/URL]

That's a pretty tiny field, so the variation will probably get drowned out by other effects.

If anyone wants to run the code themselves, here's the Matlab script. If you spot a mistake (which isn't unlikely!), do let me know.

Great graph, ArcturusA.

According to the PDF the standard deviation of the instrument reading is 0.008 mGal at most

 

[MRC_Hans]

According to the prevailing speculation, - the free air gradient and a "zeroline"
MUST merge perfectly at the buildings physical mass center.​

-

They do. Of course they do. But we are not able to determine where that is. Therefore we cannot adjust our curves correctly. So the problem is that we cannot draw our curves correctly, because your experiment provides insufficient data.

Hans

 

[Reality Check]

RC
I think you need 11 year more at the University
Start to learn what 1 Gal is

1 Gal = 0.01 m/s²
You are still wrong: the string "9338" is not the same as "1000". You got the math wrong.

I got the decimal place wrong (see how a university education teaches you to acknowledge mistakes).

You though are persisting in believing obsessively in your mistakes
The fatal flaws and basic ignorance in Bjarne's ideas in his threads
Yet another fundemantally flawed idea from Bjarne

 

[MRC_Hans]

The free air gradient is there
It MUST be there, and it must also be easy to show, prove, calculate…..
At the physical center of the building
ACCORDING TO THE PREVAILING THESIS.
(BUT the hole speculation is completely wrong therefore you can’t find this “merge point”)

Bjarne, I calculated all that. Find my post with a list of numbers, cut and paste it to a spreadsheet, and make a curve. You will find it crosses zero just under the ceiling of the 11th floor. That, however, is not necessarily the correct COG, because the basic Earht gravity I have subtracted is an estimated value. We don't know the exact value. Inserting different (but probable) values, we can make the curve cross zero anywhere we like.

We cannot determine the gravity without the building, because the building is there.

I think this is the 8th time I explain it. I can only assume you dont WANT to understand. That, however, is entirely your loss.

Hans

 

[Bjarne]

They do. Of course they do. But we are not able to determine where that is. Therefore we cannot adjust our curves correctly. So the problem is that we cannot draw our curves correctly, because your experiment provides insufficient data.

Hans believe me
IT IS NOT THERE, - IT “””””SHOULD “”””” Be according to SPECULATION
STOP TRYING TO CACH THE WIND
The thesis is an illusion
If you don’t believe me PRETEND it is at 15 the floor and you will see you will end up on Jens Peters Jensen ploughed fields.
Then try on 14th floor, try on 16th floor , wherever you want, - you will again end up on Jensen fields.
DOSENT MATTER WHERE YOU PRETEND IT IS, THE RESULT IS THE SAME. > JENSENS PLOUGHED FIELDS

I mean off course the >> "zero merge point" is not there - and that is a P R O B L E M
But not my problem, - my challenge...

 

[MRC_Hans]

Hans believe me
IT IS NOT THERE, - IT “””””SHOULD “”””” Be according to SPECULATION
STOP TRYING TO CACH THE WIND
The thesis is an illusion
If you don’t believe me PRETEND it is at 15 the floor and you will see you will end up on Jens Peters Jensen ploughed fields.
Then try on 14th floor, try on 16th floor , wherever you want, - you will again end up on Jensen fields.
DOSENT MATTER WHERE YOU PRETEND IT IS, THE RESULT IS THE SAME. > JENSENS PLOUGHED FIELDS

I mean off course the >> "zero merge point" is not there - and that is a P R O B L E M
But not my problem, - my challenge...

Bjarne, tell me where you want the zero point, and I will put it there. All using reasonable and plausible values for Earth gravity. I'm sorry, Bjarne, but it is your own fault. You designed an experiment that could not yield the information you needed.. And I told you so in advance. You have wasted your time and money.

Hans

 

[Bjarne]

Bjarne, tell me where you want the zero point, and I will put it there. All using reasonable and plausible values for Earth gravity. I'm sorry, Bjarne, but it is your own fault. You designed an experiment that could not yield the information you needed.. And I told you so in advance. You have wasted your time and money.

Hans

Pretend the "merge point" is at floor 15th
Even thoug it is 2 meter wrong it will according to the prevailing speculation only mean max. 10 microgal.

Now try to get the 3 line to form a synthesis, - it is impossible.

 

[MRC_Hans]

Pretend the "merge point" is at floor 15th
Even thoug it is 2 meter wrong it will according to the prevailing speculation only mean max. 10 microgal.

http://www.science27.com/forum/3lines.jpg

Now try to get the 3 line to form a synthesis, - it is impossible.

No problem. Only a slight adjustment to the assumed Earth gravity.

Hans

 

[Paulhoff]

Completely Crap and completely not what you originally posted and have been repeating since.

Doing the math right and to the correct decimal places:
35.000.000 * 6.67E-11 /50^2 = 0.0000009338 m/s
Or……………………………………………………….. 0.9338 microGal
Or……………………………………………………….. 0.009338 milliGal

A hint: micro = multplying the number by 10^6 or shifting the decimal point 6 places to the right; milli = multplying the number by 10^4 or shifting the decimal point 4 places to the right

The fatal flaws and basic ignorance in Bjarne's ideas in his threads

Yet another fundemantally flawed idea from Bjarne​

I posted earlier that I got 0.9338 microGal.

He can't even to the simple math right.

Paul

 

[Bjarne]

1 Gal = 0.01 m/s²


I got the decimal place wrong (see how a university education teaches you to acknowledge mistakes).

OK, so let's now get it clear.
The building’s expected max ADG (according to the prevailing speculation) is roughly 100microgal, - the same as 0.100 milligal, and don't write Mgal when you mean mGal Bjarne..

 

[Paulhoff]

OK, so let's now get it clear.
The building’s expected max ADG (according to the prevailing speculation) is roughly 100microgal, - the same as 0.100 milligal, and don't write Mgal when you mean mGal Bjarne..

Someone please tell him not to post the math and then tell us it doesn't count.

Gravity without math, what a joke.

Paul

 

[Reality Check]

OK, so let's now get it clear.
The building’s expected max ADG (according to the prevailing speculation) is roughly 100microgal, - the same as 0.100 milligal, and don't write Mgal when you mean mGal Bjarne..

OK, so let's now get it clear - your math is wrong.
The building’s expected max ADG is not roughly 100microgal.

Do the "spherical cow" approximation (a physics joke, Bjarne - not your bovine obsession) and treat the building as a sphere.
We will trust Bjarne's mass estimate (but we retain our doubts about it because of his displayed incompetence as below).
We will use 50 meters as the radius of the sphere (half the buiding's height). Then the ADG at the surface of this sphere approximation is:
ADG = MG/r^2 = 35.000.000 * 6.67E-11 /50^2 = 0.0000009338 m/s^2

Round this to 1 significant digit: ADG = 0.0000009 m/s^2

You cannot round this number any further because you have not given any errors for the original numbers or any other evidence that the ADG is less precise than the above number.

0.0000009 m/s^2... = 0.00009 cm/s^2 (100 cm per meter)
0.00009 cm/s^2..... = 0.00009 Gal (1 Gal per cm/s^2)
0.00009 Gal............ = 0.9 milliGal (1000 milliGal per Gal)
0.00009 Gal............ = 90 microGal (100000 microGal per Gal)

The building’s expected max ADG is roughly 90 microgal.
There is no reason to lose any further precision and round this up to 100 microgal.

I may have made a mistake in this and if so I would appreciate the error being pointed out.

Otherwise it looks like someting to add to the list in the first link:

The fatal flaws and basic ignorance in Bjarne's ideas in his threads
Yet another fundemantally flawed idea from Bjarne​

​

 

[W.D.Clinger]

OK, so let's now get it clear - your math is wrong.
The building’s expected max ADG is not roughly 100microgal.

Mumble. 100 microGal isn't all that far off from 93 microGal, which is what you compute below.

Do the "spherical cow" approximation (a physics joke, Bjarne - not your bovine obsession) and treat the building as a sphere.
We will trust Bjarne's mass estimate (but we retain our doubts about it because of his displayed incompetence as below).
We will use 50 meters as the radius of the sphere (half the buiding's height). Then the ADG at the surface of this sphere approximation is:
ADG = MG/r^2 = 35.000.000 * 6.67E-11 /50^2 = 0.0000009338 m/s^2

Agreed. Paulhoff computed this earlier, but then made a mistake converting m/s^2 to microGal.

0.0000009 m/s^2... = 0.00009 cm/s^2 (100 cm per meter)
0.00009 cm/s^2..... = 0.00009 Gal (1 Gal per cm/s^2)
0.00009 Gal............ = 0.9 milliGal (1000 milliGal per Gal)
0.00009 Gal............ = 90 microGal (100000 microGal per Gal)

The building’s expected max ADG is roughly 90 microgal.
There is no reason to lose any further precision and round this up to 100 microgal.

I may have made a mistake in this and if so I would appreciate the error being pointed out.

The two highlighted values are incorrect. The first value should be 0.09 milliGal. Since there are 1000000 microGal/Gal (not 100000), 0.09 mGal is the same as your final answer, 90 microGal.

I have been unable to reproduce the numbers presented by MRC_Hans and by Bjarne. I am not surprised by the latter.

What bothers me is that Strykowski and Lorenzen (Bjarne) report the acceleration measured at the ground floor as 26.4 mGal greater than the acceleration measured at the 30th floor, with a claimed std [sic] deviation of .008 mGal. Ignoring the mass of the building, and treating the earth as a point mass at its center, that difference would be about 31.4 mGal. If the difference between 31.4 and 26.4 mGal is attributed to the building, then that 5 mGal difference is approximately 50 times the 90 microGal that we calculated as above. That's too much to attribute to error in the estimated mass of the building itself.

My best guess is that the bedrock and/or crust below Domus Vista are unusually massive, invalidating any calculation that treats the earth as a point mass at its center.

Another possibility is that my calculations are incorrect, as is suggested by my inability to reproduce MRC_Hans's numbers. Has anyone else run into this discrepancy?

 

[Reality Check]

Mumble. 100 microGal isn't all that far off from 93 microGal, which is what you compute below.

My point (which Bjarne does not seem to quite get yet) is that is is incorrect to round 93.38 microGal to 100 microGal. The proper rounding is to 90 microGal.


As for the discrepancy: A few more possibilites

• The calculations posted seem to be all through the center of the building.
  But the measurements are from a stairwell on the side of the building. That is in fact the reason why Bjarne rejects the results as a test of his "central gravity" idea. For some reason he thinks that the same reason does not invalidate the results as a test of Newtonian gravity.
• I wonder what effect the building foundations have (and the other nearby buiding foundations)?
• Another possibility is an effect from the other buildings around Domus Vista.
• And there is of course the "spherical cow" approximation of treating the rectangular building as a sphere.

This PDF has a derivation of the gravitational field from a solid cylinder which ArcturusA seems to be using in his graph.

 

[MRC_Hans]

I think a cylinder is a much better approximation than a spere, for a building like that.

Hans

 

[Bjarne]

My point (which Bjarne does not seem to quite get yet) is that is is incorrect to round 93.38 microGal to 100 microGal. The proper rounding is to 90 microGal. .

This is because I am lazy-large and it could be several (fat) cows in that building. We never know. http://www.kahleerose.com/wp-content/uploads/2008/12/fat_cow-270x300.jpg

As for the discrepancy: A few more possibilites

• The calculations posted seem to be all through the center of the building.
  But the measurements are from a stairwell on the side of the building. That is in fact the reason why Bjarne rejects the results as a test of his "central gravity" idea. For some reason he thinks that the same reason does not invalidate the results as a test of Newtonian gravity.
• I wonder what effect the building foundations have (and the other nearby buiding foundations)?
• Another possibility is an effect from the other buildings around Domus Vista.
• And there is of course the "spherical cow" approximation of treating the rectangular building as a sphere.

This PDF has a derivation of the gravitational field from a solid cylinder which ArcturusA seems to be using in his graph.

Right
I have asked myself the same question.
I have also talked to a building engineer some time ago.
He said; that many places in Denmark the underground is not strong, and require very strong foundation.
He was not sure about Frederiksberg, but is could easy be 5 to 7 meter deep concrete foundation under the carrying construction parts.
All the measurements at ground floor have been done the exact same place, 20 cm from a wall.
If that wall have a heavy concrete block below, it is not so strange we measure stronger gravity here.
There are not heavy building nearby.