Ultimate Proof against flat earth, geocentrism and heliocentrism combined.

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Yeah I forget that Moon is <1 degree in the sky but >.5 degrees usually. I can't edit the image right now though as it's already released but I'll keep that in mind next time to indicate additions. Because Eiffel tower would be actually around 1/20th of a degree from London it should take roughly 8-16x optical zoom. Yea I indeed had errors in my calculations because I missed a decimal in my Moon calculation which was 5 instead of .5, big difference although I measured the Moon the other day as approximately .75 degrees so I'm not sure why my calculation varied, I'm guessing because it was daytime and also the Moon was pretty close to the horizon which both factors make the Moon appear bigger for some reason.

 As for the diagram itself I am simply saying it is wrong. The diagram clearly shows that the Earth is moving west to east but it never tells you if it's a positive degree or negative degree and that is a very important distinction. Yea theory of why is very close but the fact they are missing is that the Sun would move slightly west in the image which would actually speed up the solar day because remember Earth would be rotating West to East so you start in the west and it moves slightly west throughout the day then each day get's earlier and earlier starting point, because East to West movement would only be an illusion it would not account for any daytime lengthening unless of course the Sun was physically moving and the Earth was stationary.

But of course you won't believe me even if I made a fancy computer model to illustrate this and would still write me off as a quack.

As for the diagram itself I am simply saying it is wrong. The diagram clearly shows that the Earth is moving west to east but it never tells you if it's a positive degree or negative degree and that is a very important distinction. Yea theory of why is very close but the fact they are missing is that the Sun would move slightly west in the image which would actually speed up the solar day because remember Earth would be rotating West to East so you start in the west and it moves slightly west throughout the day then each day get's earlier and earlier starting point, because East to West movement would only be an illusion it would not account for any daytime lengthening unless of course the Sun was physically moving and the Earth was stationary.

But of course you won't believe me even if I made a fancy computer model to illustrate this and would still write me off as a quack.
What is wrong with the diagram?  From the vantage point it's drawn (north of the Solar System, looking "down"), it shows the Earth rotating counterclockwise while it revolves around the Sun in a counterclockwise direction.  Isn't that what you've been saying all along? Do you think one or the other of the curved arrows should be reversed? Why? As illustrated, they show what we see.

On Day 2 in the diagram, from the Earth's new location the Sun appears further east than it did on Day 1, not further west. Why do you think it should be further west? Whether the angle is considered positive or negative is immaterial; what matters is that the Earth has to turn a little farther to bring the Sun back directly overhead than to bring a distant star back overhead.

Making a computer model to illustrate something is fine, but if the model is wrong, it's not going to mean anything useful.

What did you use to measure the apparent size of the Moon? At least you're out looking and taking measurements, unlike a lot of very adamant posters here.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

As for the diagram itself I am simply saying it is wrong. The diagram clearly shows that the Earth is moving west to east but it never tells you if it's a positive degree or negative degree and that is a very important distinction. Yea theory of why is very close but the fact they are missing is that the Sun would move slightly west in the image which would actually speed up the solar day because remember Earth would be rotating West to East so you start in the west and it moves slightly west throughout the day then each day get's earlier and earlier starting point, because East to West movement would only be an illusion it would not account for any daytime lengthening unless of course the Sun was physically moving and the Earth was stationary.

But of course you won't believe me even if I made a fancy computer model to illustrate this and would still write me off as a quack.
What is wrong with the diagram?  From the vantage point it's drawn (north of the Solar System, looking "down"), it shows the Earth rotating counterclockwise while it revolves around the Sun in a counterclockwise direction.  Isn't that what you've been saying all along? Do you think one or the other of the curved arrows should be reversed? Why? As illustrated, they show what we see.

On Day 2 in the diagram, from the Earth's new location the Sun appears further east than it did on Day 1, not further west. Why do you think it should be further west? Whether the angle is considered positive or negative is immaterial; what matters is that the Earth has to turn a little farther to bring the Sun back directly overhead than to bring a distant star back overhead.

Making a computer model to illustrate something is fine, but if the model is wrong, it's not going to mean anything useful.

What did you use to measure the apparent size of the Moon? At least you're out looking and taking measurements, unlike a lot of very adamant posters here.

I used a camera to measure the Moon by counting pixels in reference to a 120 degree frame. It was 16 pixels out of 3019 which was 120 degrees fov so that means you have about 1/188 or 120/188 = .64 degrees.

As for the Earth I understand what they are trying to say, they say because the Earth is traveling east and because the day appears to move East to West because of opposing rotation then your reference frame moves slightly west to make for a slightly longer day.

However I'm saying heliocentrism is wrong because if you move in orbit West to East then the Sun will move physically West in relation to you which will cause the day to become shorter because the real day would start in the West and End in the East so optically you would see a shortening of the day and not a lengthening.

See this is actually how you tell an optical illusion from reality. Because if you have an optical motion it will be equal but opposite of what it would be if the motion was instead real. If Earth was stationary and only the Sun was moving around it then it stands that the Sun is simply moving slower then the stars. However if Earth is spinning you have to say both Sun and Stars are pretty much stationary from our point of view, thus any motion is equal but opposite to what it should be. Thus if you have the Earth move East in one day then the Sun will appear to move East (remember equal but opposite), because the real day would start in the West, Westward motion would shorten the day and not lengthen the day. Because we see the day is actually longer then the Stars this means Earth can not be rotating and if Earth is not rotating then the whole schism of heliocentrism is flushed down the toilet.

However the above would only be if people would be willing to accept real scientific proofs that were right in front of their own eyes when logically examined from an unbiased perspective. Both the real truth is many people only see what they want to see even if the truth is shoved right in their face they will still choose to ignore it. You know, I would gladly believe heliocentrism if it actually lined up with reality but the flaws I've seen in it actually turned me away from it as I couldn't explain my own observations with a heliocentric mindset.
« Last Edit: October 05, 2014, 12:25:52 PM by Sculelos »

The earth rotates relative to the sun.

The earth rotates relative to the sun.
... and much faster than it revolves around it in its orbit.  That's why the Sun appears to progress a full 360 degrees westward across the sky (due to the Earth's rotation) in the time the Sun appears to progress almost 1 degree eastward relative to the background stars (due to the revolution about the Sun along earth's orbit). This is what the original illustration showed clearly (and still did after it was cluttered up with extraneous 'E's and 'W's, especially if you ignore the added text to the right).

But enough arguing about your misinterpretation of Heliocentric theory. Check it out for yourself. There's going to be a total Lunar Eclipse visible from western  North America Wednesday morning, Oct 8. This will be followed by a partial Solar Eclipse on Oct 23, also visible in North America. The times for each stage of these eclipses are predicted, to fractions of a second here and here. The ability to predict these to the precision offered - for centuries in advance - requires a thorough understanding of orbital mechanics and excellent data sets based on detailed observations. Those are hosted on a NASA website, so see if they're lying to you. I'm going out on (not much of) a limb here and state that these will occur exactly as predicted.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

I forgot to touch on this.

What did you use to measure the apparent size of the Moon?
I used a camera to measure the Moon by counting pixels in reference to a 120 degree frame. It was 16 pixels out of 3019 which was 120 degrees fov so that means you have about 1/188 or 120/188 = .64 degrees.

That's a good way to measure the apparent size or separation of objects in an image, but the scale of the Moon in your image(s) is too small to get a very precise result. The results should be more accurate if you used a system with a longer focal length, yielding a larger image of the moon.

Your object is only 16 pixels, so an error of just one pixel represents more than 6% in the measured size. Consider that a well-designed imaging system will spread the image of a point source across at least two pixels (but, ideally, no more). Point sources theoretically have zero width, but can never, even with a theoretically perfect optical system of finite size, be focused to a point with zero width; the nature of digital sampling says you want a minimum of two pixels, anyway, so this isn't entirely a bad thing. This one additional pixel on each edge of your moon image can make it two extra pixels wider. Since your camera lens is not perfect, and/or the focus likely isn't perfect, both of which will make the image larger by spreading the energy, another pixel or so gain in size is easy to see. Now, with a not-unexpected three-pixel "bloat" you are right in line with the known angular size of the moon.  If you had a much longer lens and the image of the Moon was 160 pixels instead of 16, a 3-pixel "bloat" is only 2% instead of about 18%.

Also, how accurate is that 120-degree FOV reference angle? Where did it come from?

Still, that's the right approach. Given the the data, yours is a reasonable answer.

[Edit] Some rewording and formatting for (I hope) clarity.
« Last Edit: October 05, 2014, 07:16:46 PM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re-examining the data for the night the picture was taken indeed confirms the picture was taken at 120 degrees fov based on the Sun, Moon and illumination of the Moon at 75%. So Moon is only 12 pixels according to illumination but 16 if you extend that into a full Moon. However you could easily have 2 pixels of error on top of everything else making the full moon only 13 pixels instead. I guess this sort of error would be called a compound error with the base being slightly too big the corrected full size would be even bigger. Anyways, it's accurate enough but definitely not perfect. Accuracy of the FOV was certainly very close to 120 degrees even though it could have been of a few percentiles because it was extracted from a panorama and perspective shifts are sometimes a little hard to compensate for.

Anyways as far as predictions are concerned I have only limited knowledge of how they are calculated. Most of the calculations I've seen are only geocentric and they seem to work with only a very small error in them so I'm not going to say what model the calculations came from unless I see the calculations for myself. I know a lot of models have only the sky moving and the Earth completely stationary even NASA commonly uses a geocentric concave Earth model for their sky predictions so I'm likely to say that if NASA's models are predictions of anything, it's a prediction of a concave Earth not moving which of course they constantly have to backpedal on why their concave geocentric models are more correct then their heliocentric models but the answer of course is clearly obvious to me.

Of course regarding Earth's spin I just can't grasp how a Earth moving east would not have the Sun physically move west in relation to the Earth thus because your day starts when you are parallel with the Sun but effectively 180 degrees opposite your day would start in the west and move east thus making it appear like the Sun is moving East to West but because your day starts in the west and the sun would move west it seems to me like your day would constantly get shorter so that you would have one less solar day per year then stellar, but of course this doesn't happen.

Of course NASA's accuracy like I said could easily be the fact that they use a skycentric (concave) Earth model when calculating celestial mechanics of course they then say that movement is all relative but I say movement is clearly not relative when my calculations show that a non-moving concave Earth is different then a heliocentric universe.

I mean the actual difference between the two is pretty much polar opposites in almost every way possible. Yea they mirror each other so if you mirror everything you might still end up with a model that predicts things accurately but it definitely will be inverted to reality in every way possible so that it will still be accurate but still wrong at the same time if that makes sense.

Sorry If I'm not making sense but if I wrote in the way I usually write to myself other people might view it as almost complete garbled nonsense.

What about organisations other than NASA across the world?

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ausGeoff

  • 6091
London is 465 ft elevation where Paris is 400 ft about 213 miles across this should put you about 65 ft above horizon level, Eiffel tower is 986 ft tall so you have about 1051ft total before the Eiffel tower fades completely. This puts your viewing distance at 39.7 miles until the Eiffel tower fades into the horizon with plain eyesight. This is because if an object is higher then you the light reflected from it will be above your head so you won't be able to see it.

I'm afraid that this is totally unsupported by any of the accredited sciences, and defies even the basic laws of common logic.  It reads like a fairy story.  I can only assume you're making all this stuff up as some sort of bizarre joke.  But one that doesn't have a punchline.


I used a camera to measure the Moon by counting pixels in reference to a 120 degree frame. It was 16 pixels out of 3019 which was 120 degrees fov so that means you have about 1/188 or 120/188 = .64 degrees.


Please, can you share the camera's specs?

Re-examining the data for the night the picture was taken indeed confirms the picture was taken at 120 degrees fov based on the Sun, Moon and illumination of the Moon at 75%. So Moon is only 12 pixels according to illumination but 16 if you extend that into a full Moon. However you could easily have 2 pixels of error on top of everything else making the full moon only 13 pixels instead. I guess this sort of error would be called a compound error with the base being slightly too big the corrected full size would be even bigger. Anyways, it's accurate enough but definitely not perfect. Accuracy of the FOV was certainly very close to 120 degrees even though it could have been of a few percentiles because it was extracted from a panorama and perspective shifts are sometimes a little hard to compensate for.
Yeah, a panorama isn't the best choice for that because the image manipulation required to stitch it together could be a significant source of error. A better way to get the pixel size (in fractions of a degree) is to take a picture of something a known size from a known distance away. A garage door from enough distance so that it fills, say, the middle third or quarter of your frame may be a good bet. Carefully measure the width of the door, then set your camera up square with its center and carefully measure the distance from the door to the front of the camera lens. Calculate the angle subtended by the door at the camera:

A = 2*atan(w/(2*d))

A is the angle, '*' is multiplication, atan is the arctangent (inverse tangent) function, w is the width of the door, '/' is division, and d is distance from door to camera.

Note that if you use Excel, it works with angles measured in radians, so you'll have to convert A from radians to degrees; you can use Excel's DEGREES() function for this. Most scientific calculators give you the option of working in degrees directly.

Once you know the angular size of a known object, count the pixels it subtends on the image and you can calculate the Instantaneous Field of View (angular size) of each pixel (IFOV).

This will only work if the focal length of your lens doesn't change. If you have a zoom lens, the IFOV of the pixels will change with the zoom setting, but is probably repeatable at maximum or minimum zoom.

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Anyways as far as predictions are concerned I have only limited knowledge of how they are calculated. Most of the calculations I've seen are only geocentric and they seem to work with only a very small error in them so I'm not going to say what model the calculations came from unless I see the calculations for myself.
Well, you're in luck then! The US Naval Observatory publishes the source code for the NOVAS package that, along with an ephemeris (available from JPL) does those calculations. I've used this library and the DE405 and DE420 Ephemerides in several applications; they're very accurate. The linked pages in the earlier post reference the DE406 Ephemeris. I think it's the same as DE405, but valid for thousands of years instead of hundreds, which makes the files huge!

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I know a lot of models have only the sky moving and the Earth completely stationary even NASA commonly uses a geocentric concave Earth model for their sky predictions so I'm likely to say that if NASA's models are predictions of anything, it's a prediction of a concave Earth not moving which of course they constantly have to backpedal on why their concave geocentric models are more correct then their heliocentric models but the answer of course is clearly obvious to me.
When the predictions refer to geocentric locations, they mean the celestial coordinates where the Sun or Moon would appear if you were observing from the center of the Earth. Obviously you can't do that, but from that information, along with your geographic location and time, you can calculate their actual celestial coordinates from any location on or near the surface of the Earth. NOVAS, among a lot of other things, provides a function that does exactly that. The Moon is close enough at about 240,000 miles that the 4,000-mile radius of earth can cause a significant amount of parallax, meaning it will appear at somewhat different places relative to the background stars at the same time from different locations on earth.

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Of course regarding Earth's spin I just can't grasp how a Earth moving east would not have the Sun physically move west in relation to the Earth thus because your day starts when you are parallel with the Sun but effectively 180 degrees opposite your day would start in the west and move east thus making it appear like the Sun is moving East to West but because your day starts in the west and the sun would move west it seems to me like your day would constantly get shorter so that you would have one less solar day per year then stellar, but of course this doesn't happen.
Applying the terms 'east' and 'west' to the Earth's motion around the Sun is a mistake that's confusing you.

If you're at a point on the Earth facing directly away from the Sun, lying on your back with your feet pointed south, east is toward your left, and that's the direction the earth is moving in its orbit; west, obviously, is to your right. 12 hours later, if you're still lying in exactly the same position (on your back, feet still pointed south), the Earth has made half a rotation and you're now facing directly toward the Sun. West is still toward your right, but now that's the direction earth is moving in it's orbit (because you've turned around wrt the orbit and the rest of space, but still in the same position wrt earth). Facing the Sun from this position, each day the Earth has progressed a little further along its orbit to your right (that is, westward to you), which would make the Sun appear to be moving to your left (eastward to you) relative to the distant stars.
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Of course NASA's accuracy like I said could easily be the fact that they use a skycentric (concave) Earth model when calculating celestial mechanics of course they then say that movement is all relative but I say movement is clearly not relative when my calculations show that a non-moving concave Earth is different then a heliocentric universe.
Except they don't use a "skycentric" model. The Ephemerides published by JPL (a division of NASA) are based on Barycentric coordinates.  That is, they provide locations for solar-system objects (including the Earth and Sun) relative to the center of mass (barycenter) of the solar system.  Since the Sun is far more massive than everything else in the solar system combined, the Barycenter is usually very close to the center of the Sun, almost always well inside its Photosphere (the visible surface of the Sun); only occasionally do the planets line up so that the Barycenter is "outside" the Sun, and even then, not very far. So, strictly speaking, it's not a Heliocentric model, but rather a Barycentric model, but the difference is so small that in most cases that's a good enough description and everyone knows what you mean.

Earth's motion around the Barycenter can then used to determine parallax for "nearby" stars relative to more distant ones. If earth were fixed relative to the stars we wouldn't see stellar parallax.

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I mean the actual difference between the two is pretty much polar opposites in almost every way possible. Yea they mirror each other so if you mirror everything you might still end up with a model that predicts things accurately but it definitely will be inverted to reality in every way possible so that it will still be accurate but still wrong at the same time if that makes sense.
I think it's the "orbiting from west to east" thing that's causing you to think this.

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Sorry If I'm not making sense but if I wrote in the way I usually write to myself other people might view it as almost complete garbled nonsense.
Some of your writing isn't very clear, but I think I see what you're saying. It's wrong, but I believe I understand why you think it.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

"Of course regarding Earth's spin I just can't grasp how a Earth moving east "

I can't grasp how the earth could be moving east.
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I used a camera to measure the Moon by counting pixels in reference to a 120 degree frame. It was 16 pixels out of 3019 which was 120 degrees fov so that means you have about 1/188 or 120/188 = .64 degrees.


Please, can you share the camera's specs?

It was just a crappy 2mp camera phone stitched together with 3 composite images. No zoom whatsoever. Each 3 images were about 50 degrees fov to start with but they lost about 10 degrees each to make sure I had reference points to make sure the panarama was exactly as it should be.

Anyways back to East and West. The two are not like north and south, because if you go north far enough you will go past north and start going south, so we know north and south are fixed directions but East and West are not. There is no East pole or South pole. They are simply directions with East being different then West and you can go East and West both forever and they will never ever change directions. (Psalms 103:12)

I mean if East did turn to west throughout the day you would see objects move backwards half the day! That clearly doesn't happen!

Anyways because East and West are relative directions and because North and South are physical points you can cross there is technically no East or West hemispheres as you could have your middle point technically be anywhere on Earth and nowhere would be more correct then any other location as far as the middle of East and West would go because there is no middle of infinity.

Okay because now we know East is right when facing the North pole and West is left when facing the North pole, we know that it never matters what time of day or night it is. East will always be East and West will always be West.

Okay now back to my original point. If Earth is turning West to East and also orbiting in the same direction, then it must be orbiting West to East. If Earth is orbiting west to east then the Sun must be moving West in reference to Earth, however because the Earth is supposedly rotating West to East that means the day must start in the West 180 degrees from the East (Which is why the Sun appears in the East first). If the Sun moves West and the day starts in the west and ends in the East, then of course that means your solar day should indeed be shorter by 4 minutes every single day until a full year has passed.

Now listen carefully, this would happen because the Earth would be moving East, so the Sun relative to the Earth would move West. Because West is the starting point for a day on the heliocentric model if the Sun moves West it brings your starting point closer to you. Now on any model which does not have the Earth rotating, this would be the exact opposite because the Sun would be moving East to West instead so moving West would cause the cause the day to be slightly longer as your reference point would be moving away from you. However like I said this is a huge deal because it's NOT RELATIVE. If the Sun's movement is an illusion every day would start in the West and therefore bringing the Sun more West would only make your day start SOONER. If the Sun's movement is REAL then bringing the Sun more West would cause the day to start LATER.

So if we understand this. We also understand that NASA is not using a heliocentric model. They are using a concave model and simulating the Earth's movement with completely different calculations. All space flight works with geostatic cordinates using a celestial sphere that spins and all space flights have used this model ever since the first Apollo missions. NASA actually likes using this model because it is more simple for them to assume a static Earth and just map what the planets and stars are doing and also assume you will be turned with the celestial sphere (which why would you turn with the Earth on a heliocentric model in space? That sounds simply too bizarre to be real)

I actually do know a little bit about space flight, but every single (accurate) model I've seen shows a completely stationary Earth in relation to all the planets and stars. Stellar parallax can easily be explained by a rotating celestial sphere because if a star was seen from the opposite side of Earth 180 degrees later it would be on the opposite side of the celestial sphere which would of course make it equal yet opposite displaying parallax or negative parallax depending on the tilt angle of the celestial sphere which of course would be tilted at 23.5 degrees.

Some concave Earth theorist say space flight is impossible, but I've actually seen the models and know everything shrinks in space so more space physically fits inside the Earth then on the Earth even though the Earth is technically bigger then what we call the Universe. The only reason I say that is because there is also another side to Earth with also Air and Water and Land and ALIENS. Of course. You could say I'm writing this because I'm stoned right now. But it's true. I've seen some of them in mirages over here. Yea right they are just holographic reflections and only best seen in sharply contrasting areas. Areas with lots of metal and a lot of reflections seem the best way to see these holograms which I think most refer to as grey's. Of course you can also see lots of flying orbs as well. And if you use a water filled telescope with about 45 power and a camcorder set to over-expose for about 15 seconds you can see the stars in the sky and they look almost like humans inside of a robotic exoskeleton that shape-shifts.

All I know is reality indeed does seem stranger then fiction to most people, but of course, most people just think I'm a lunatic when I tell them these things which is why the police made me go to the fifth floor once...     


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ausGeoff

  • 6091
It's impossible to accurately compute the size of any planetary or stellar body using a simple digital camera, and "counting" pixels.  This is largely due to the body being way outside the limits of the effective focal plane of the camera's sensor/lens combo—it's effectively focussed (somewhere!) at infinity.

This is because of a process known as "pixel binning" which is performed automatically—as necessary—by the camera's firmware.  This diagram illustrates the process:


Although the dynamic range of the sensor resulting from the larger charge
capacity of the summing node (typically 1.65-2 times increase in well depth)
increases, the disadvantage is a loss of image resolution—proportional to the binning level.


You also have to take into consideration the anti-aliasing techniques used to smooth "jaggies", particularly on high contrast images (of which astrophotography is a prime candidate).  This further blurs the edge of any object you'd be trying to "measure" with a digital camera.

And further to your latest comment (which I've just seen after writing this) a 2MP phone camera is nothing more than a toy camera technically and photographically speaking.




I used a camera to measure the Moon by counting pixels in reference to a 120 degree frame. It was 16 pixels out of 3019 which was 120 degrees fov so that means you have about 1/188 or 120/188 = .64 degrees.


Please, can you share the camera's specs?

It was just a crappy 2mp camera phone stitched together with 3 composite images. No zoom whatsoever. Each 3 images were about 50 degrees fov to start with but they lost about 10 degrees each to make sure I had reference points to make sure the panarama was exactly as it should be. 

So you have absolutely no idea of the cam's real angle of view Got it.

All I know is reality indeed does seem stranger then fiction to most people, but of course, most people just think I'm a lunatic when I tell them these things which is why the police made me go to the fifth floor once...   

OK, I get it. You're just funnin' us...
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

It's impossible to accurately compute the size of any planetary or stellar body using a simple digital camera, and "counting" pixels.  This is largely due to the body being way outside the limits of the effective focal plane of the camera's sensor/lens combo—it's effectively focussed (somewhere!) at infinity.

This is because of a process known as "pixel binning" which is performed automatically—as necessary—by the camera's firmware.  This diagram illustrates the process:


Although the dynamic range of the sensor resulting from the larger charge
capacity of the summing node (typically 1.65-2 times increase in well depth)
increases, the disadvantage is a loss of image resolution—proportional to the binning level.


You also have to take into consideration the anti-aliasing techniques used to smooth "jaggies", particularly on high contrast images (of which astrophotography is a prime candidate).  This further blurs the edge of any object you'd be trying to "measure" with a digital camera.

And further to your latest comment (which I've just seen after writing this) a 2MP phone camera is nothing more than a toy camera technically and photographically speaking.

It's entirely possible to estimate the size of celestial bodies using a digital camera and counting pixels if the object is resolved and you know the IFOV of a pixel (I described a method to determine that). Realistically, only the Sun and Moon are large enough to adequately resolve unless you're using a telescope or very long telephoto lens, but the Moon is the topic of discussion. Knowing the limitations and characteristics of the equipment such as whether it's binning or not, and pixel size of a point source, can improve the accuracy of the estimate, but larger images will be less subject to this sort of error. Whether it's "accurate" depends on what you're trying to accomplish.

I'm not sure what "This is largely due to the body being way outside the limits of the effective focal plane of the camera's sensor/lens combo—it's effectively focussed (somewhere!) at infinity" is supposed to mean. The Moon is distant enough to be "at infinity" for every optical system you can pick up and carry around (and most, if not all, all larger ones). Objects at infinity can easily be focused on a sensor; in fact, many optical systems are designed to perform best (minimize aberrations) at infinity.

There are lots of issues with Sculelos' technique :
 - The image size is small.
 - An overexposed object will 'bleed' into the adjacent pixels, making it appear larger (I'm not sure if this was a problem here).
 - How well focused is the image? It sounds like the Moon is close to the edge of the frame; images are typically sharpest near the center.
 - How much distortion is there? The IFOV of the center pixels may not be (probably aren't) the same as those near the edge of the frame.
 - How does he know his reference angle is really 120 degrees?

This notwithstanding, I think Sculelos is to be commended for trying this. His result is certainly in the ballpark given the above; maybe he was lucky, but, in principle, the overall approach is reasonable. All but the first issue could be mitigated by correctly exposing the Moon while centered in the frame and determining the IFOV of center pixels using a reliable technique. Using a better camera and longer-focus lens will help with the first.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

And if you use a water filled telescope with about 45 power and a camcorder set to over-expose for about 15 seconds you can see the stars in the sky and they look almost like humans inside of a robotic exoskeleton that shape-shifts.
So when will you be sharing this video with us?

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macrohard

  • 139
  • IQ over 180
This topic derailed fast.  The size of the moon doesn't mean squat when your "model" applies one dimensional east/west logic to a two dimensional orbit.

And if you use a water filled telescope with about 45 power and a camcorder set to over-expose for about 15 seconds you can see the stars in the sky and they look almost like humans inside of a robotic exoskeleton that shape-shifts.
So when will you be sharing this video with us?

There was already a video made on this and it was completely ignored: Interstellar - Mystery Space Machines on Vimeo

Perhaps someday I should make a video of my own on this but I have a feeling it would suffer the same fate as the above video which is being ignored.

As for the width of the Moon, I'm not even sure why it matters at this point. I think it was something to do with measuring f.o.v. because I think I said you could tell an approximation of how many degrees the Moon was from the Sun by it's illumination level (from Earth). I think this was of importance because I used the same method to measure Jupiter the other day and found Jupiter to be around 7 AU away from the Earth which would put it a full 1 AU farther away then heliocentric calculations predict. Objects in a mirror may appear closer then they appear though, which if there is glass in the sky, perhaps that might explain why others think it's 6 AU because they are not calculating for distortion? I don't really know though. I'm just a science nerd with math abilities that don't really exceed basic geometry. A lot of my calculations seem to work well but I have been pointed out that I seem to lack any units of real significance in my calculations so they don't mean much to professionals. However saying that I'm simply not writing stuff in the proper way doesn't really mean my results are null they just mean I might be missing important elements but it's really hard to say because I don't really do things the way normal people do.

As for the more important topic of East and West you really can't treat either as fixed points like I said. If you did treat them as poles your day would move the opposite way of night for example and also half of the year the Sun would go East to West and the other half it would go West to East.

However if you treat them as relative directions only to North and South and each other then you could easily have an orbit say the same direction the entire year but the problem then becomes if Earth is orbiting West to East it is also rotating the same way thus your day starts West and ends East so if you move the Sun West then that moves your starting point of your day earlier. Like I said before the way the Sun appears to move in the Sky doesn't matter because it would only start in the East and move West because of your West starting frame. Move your West starting frame more West and you effectively shorten your day because you are also at the same time moving your (apparent) Sun East because remember the Sun would be an illusion and therefore the exact polar opposite of what Earth would be doing. Therefore move Sun West (in relative position to Earth) and it would appear to move more East (In relative starting position to the prior sunset).

Both scenarios would mean your day would be shortened relative to your sidereal day.