Measuring the Moon

Please note: This is a guest post by @rokro11. He uses info from timeanddate.com to calculate the moon’s size. I, Stacey, have not looked into this personally. If you would like to comment or ask any questions, you can do that below or to on Twitter.

How to Calculate the Moon’s Size

In this article, I will provide some insight on how to calculate the size of the moon.  The moon moves above the earth and all we need to know is the speed of the moon to do some simple math to calculate its size.

In order to obtain the speed of the moon, which varies throughout the duration of its travel while it moves above the earth, go to the site Timeanddate.com.  This company is located in Norway and it reports the moon speed relative to the ground speed of earth. You can contact them at webmaster@timeanddate.com 

Go here for the Day and Night World Map.

Under the heading called “Sun & Moon”, click on “Day and Night Map.” About half-way down the page, it will indicate, “Position of the Moon” and you will find the speed of the moon relative to the ground speed of earth as indicated below:   

moon_speed_flat_earth
From timeanddate.com

According to the site, the speed of the moon changes between about 933 mph to about 1005 mph.  The moon’s speed can change over 20 mph in a 24 hour period.

I captured a video of the full moon at 2:10 am on 05-30-2018. See how it moves:

moon_flat_earth_caption
First screenshot from my video.
moon_flat_earth 2.0
Second screenshot from video.

For this specific example, I will be using the Timeanddate speed of the moon relative to the surface of the earth of 952.5 mph because I captured the video of the full moon at 2:10 am that morning.  Now that we have the moon’s speed, along with video from a camera on the ground, we have to view how long it takes the trailing edge of the moon to reach the leading edge of the moon. The time it takes the moon to travel its width is 2:10, or two minutes and ten seconds.  

Because the moon is traveling at 952.5 mph, it travels 15.875 miles in 1 minute.  (952.5/60 minutes = 15.875 miles). The moon traveled across itself in 2 minutes and 10 seconds.  We add 15.875 to 15.875 and come up with 31.75 miles, over 2 minutes.

How do we take into account the additional 10 seconds?  Simply take the distance traveled per minute of 15.875 miles and divide that by 60 to arrive at the distance it travels per second.  That yields .2646 miles per second. (15.875/60 = .2646 miles per second).

To account for 10 addition seconds, take .2646 and multiply it by 10 seconds, which results in 2.646 miles.

For the total amount of distance traveled, we add the 2 minutes of travel distance of 31.75 (for the 2 minutes) and 2.646 (for the 10 seconds) and that yields 34.396 miles.

By this method, the width of the moon is 34.396 miles.

moon size.jpeg
My calculation of the moon’s size.

This is not a complicated method and this is how to determine an object’s size if its speed is known.  

Nasa claims the size of the moon is 2,158 miles wide.  It’s not. The moon is also not 238,900 miles from earth.  The moon’s size and distance is much less than we have been led to believe.  

We don’t need Nasa to continue lying to us.  If anyone finds errors in this method, provide your feedback.  Don’t Google search the speed of the sun because that Google search result is misleading and the results only back up the heliocentric theory.  It’s a diversion to keep you from finding the truth. Find a way to calculate the moon’s speed using information that makes sense, not based on unfounded and unreliable information from Nasa and other subjective and unverifiable sources.

Do this simple measurement and provide your own moon size calculations in the comments below.  If you have any questions, ask here or @rokro11 on Twitter.

36 Comments

  1. The following information comes from timeanddate.com:

    “On Friday, May 18, 2018 at 12:00:00 UTC the Moon is at its zenith at Latitude: 20° 38′ North, Longitude: 43° 09′ East
    The ground speed is currently 416.51 meters/second, 1499.4 kilometres/hour, 931.7 miles/hour or 809.6 nautical miles/hour (knots).”

    “On Friday, June 1, 2018 at 12:00:00 UTC the Moon is at its zenith at Latitude: 20° 43′ South, Longitude: 145° 48′ West
    The ground speed is currently 419.89 meters/second, 1511.6 kilometres/hour, 939.3 miles/hour or 816.2 nautical miles/hour (knots).”

    That is a 0.8% difference in the moon’s ground speed between 20° North and 20° South, and yet the moon takes the 25 hours to make one complete circle at 20° North but only 24 hours, 50 minutes at 20° South.

    How does that work on a flat Earth, when 20° South would have a 60% larger circumference than 20° North?

    1. Your reference to timeanddate for May 18th and June 1 has been reviewed. You indicate an .8% difference and then reference 20 degrees N and 20 degrees S. You then ask how does that work on a flat earth. The simple answer is it’s not understood how it works on the flat earth but since it actually does work and it works on our flat earth, we need to find out how more about it.

      When the truth is being hidden from us about where we live, we can’t rely on those withholding information from us. We need to take the initiative and find out for ourselves.

      In regard to the 60% larger circumference, that is an assumption and I don’t entertain assumptions.

      If you have any more questions concerning timeanddate, they are quick to respond; just send them an e-mail.

  2. You are using data from timeanddate.com. Do you think that data is correct?
    They base their predictions on the heliocentric model, like the exact sunrise and sunset times for any given day and place.

    Why should they be more reliable than NASA?

  3. Marius,
    The timeanddate website is located in Norway. Do you have any reason to suspect their information is not reliable?

    The information timeanddate provides is more than predictions. They are fairly accurate in what I’ve been able to cross-check. You indicated the site provides their information based on a certain model – that’s an assumption. Their information is provided based on what is observable and happening on our actual earth. What the site reports is what actually happens from our perspective and is not dependent on any model. What happens does not need a model attached to it to happen – it’s going to happen no matter what.

    Nasa is subjective, and anything they provide is not able to be objectively verified. Nasa is not reliable in the sense that anything they provide cannot be objectively verified. Nasa is a military branch of the United States government and the laws in the United States, by means of the Smith–Mundt Act, allow the government to lie to its citizens and the entire world using propaganda in any form and fashion. Since this law is in place allowing the US government to lie freely, Nasa is not considered a reliable or trustworthy source, and should not be relied upon for anything.

  4. Marius,
    Timeanddate provides information that is observed on the actual earth and is not dependent on any model and it is visually accurate by observation. The data they provide matches reasonably well to what I’ve independently measured myself. If you have concerns with what timeanddate provides as data, take up your concerns with them.

    As far as nasa, they are a military branch of the Unites States government. The Smith–Mundt Act is a law that was passed in the US that allows the government to lie to their citizens and the world using propaganda of any form or fashion. Due to this law that allows them to disseminate propaganda to the world, which is supported by law, they are not a source of information to be relied upon. That is unless you accept their propaganda as factual information. Their information cannot be objectively verified as anything that can be relied upon.

  5. TimeAndDate.com has the facility to show you the position of the moon at any point in the past or future. Therefore they are calculating that position and so are using a model. They are clearly not simply providing “information that is observed on the actual earth and is not dependent on any model” else they would not be able to provide future data. I have reached out to them for confirmation of this and of what model they are actually using.

  6. They are definitely using a model to generate their data since they are able to provide details for dates and times in the future. Can’t do that with just observations.

    I have reached out to them to ask what model they use and will let you know what they say.

    1. Alistair,
      The actual earth is not a model – it is the actual earth. There is a heliocentric theory model that you may think they are using but they are not. No model has to be used when the historically observed earth has shown what happens on our actual earth will repeat it self over and over with a high degree of accuracy in the future. If you want to call it “Actual Earth model”, that would seem reasonable.

      1. I do think they are using the heliocentric model. Have you actually asked them? I have mentioned your idea to them and they also seem to think they ware using the heliocentric model (and I think they would know). They say…

        “Basically there are some assumptions there which do not make sense, and the speeds listed on our site is not of the moon itself (like that person assumed), but the speed of the point on earth where the moon is in zenith.”

        So you are using numbers that are derived using a globe model. I don’t think you should be trusting those numbers in a flat earth environment.

      2. Alistair,
        The measurement method used the moon to measure itself, and its speed. The ground speed is the speed of the moon relative to a continuously changing location on the earth. The moon traveled about 34 miles while crossing itself and the zenith location on earth changed in location the exact same as the moon did. The zenith is the imaginary spot directly below the moon at all times and is the closest point between the moon and the earth at all times. The shape of the earth is not part of the moon’s measurement outlined in the article so heliocentric or flat earth is not even taken into consideration to determine this measurement.

    2. I explicitly asked TimeAndDate.com whether they used a model and if so what model they used. Here is their reply:

      “Hello Alistair,

      Yes, we use a globe-based model and not observations for all of our astronomy calculators and tools on the site.

      Best regards,
      Steffen”

    3. Then it’s settled. As rokro11 said “They are fairly accurate in what I’ve been able to cross-check”

  7. I like this method, it’s creative and would work to measure the moon, except that it’s fundamentally flawed. Your argument works on the initial assumption that the earth is flat, then uses this assumption to show that the earth is flat.

    You couldn’t measure the moon by these means on a globe. In the heliocentric model, the speed of a point over the ground varies with the diameter of the globe (obviously, since the period is fixed at around 24 hours and the circumference varies with the diameter of the globe, so if you vary one without the other, then the ground speed will change in sympathy).

    Similarly, an object twice the size of the moon would look the same if it was twice as far away, and had its speed altered to compensate. Your calculation depends on the assumption that the ground and the orbit of the moon are both parallel and two-dimensional (this is how you get to boil the maths of the reality of our world down to such simple terms). There’s actually an infinite set of solutions to the size of the moon observed by these means, and you land on one of them by introducing the assumption of ‘flatness’.

    1. Call,
      The time measured using the camera is 2:10 minutes. There were no assumptions taking that measurement. The speed of the moon of about 952 mph is the moon’s speed relative to the earth. There were no assumptions applied to that.

      I made no assumptions about the topography of the ground I was standing on to capture the video. The ground was my driveway and it is fairly level. The elevation of the tripod was 1247 feet above sea level. The recorded video of the moon would not change if the ground upon which I was standing was hilly or mounded. You can assume something would have been different but since you were not part of the measurement of the moon, I won’t introduce your assumptions into the framework of the actual size of the moon results.

      The heliocentric model claims the moon moves prograde to the earth. (The direction of the moon used to be retrograde to the earth but it had to be changed to make other assumptions applied to that model work.) Regardless, the heliocentric theory model suggests the earth allegedly spins from west to east. On this same day, timeanddate reports the earth speed relative to the sun as 964.1 mph.

      During the early morning while videotaping the moon, the moon travel was clearly from east to west. The east to west movement of the moon makes no sense when the alleged spin of the earth is west to east. The moon’s travel should be with the alleged spin of the earth, not opposite of it – because that’s what prograde means. If I applied your assumptions, I would have to take the difference between 964 mph and 952 mph (12 mph) because the earth is allegedly spinning faster than the moon’s speed, in that prograde movement, and consider the moons size less than a half mile wide if it crosses itself in 2:10 minutes. However, since that relies on your assumptions, I will not add your assumptions to my measurements.

      In order to keep the moon size measurement as simple as possible, I did it the way I presented it. If you have a different speed you think I should have used for the moons speed, without applying any assumptions to the calculations, let me know.

  8. There are inherent errors with any experiment, I wouldn’t try to claim the effect of the ones in yours, I don’t know your situation well enough to compare it to others, and either way the local topology of the ground you’re on doesn’t affect my point.

    To your point about prograde and retrograde, the moon can orbit prograde while appearing to move retrograde. As you appear to recognise, it’s just that the angular velocity (rate of rotation) of the earth is much faster. This is why moonrise is a bit later every day, because while the moon doesn’t keep up with the Earth, it does make some progress each day.

    My issue is the assumption that the speed of the moon observed on the ground is equal to the moon’s speed. The assumption is that the ground track projected by the moon onto the earth is equal in size to the path through which it moves in space. If this were true, your method would be sound enough for me to only quibble about little errors of methodology. But it depends on the Earth being flat and unmoving, because your method wouldn’t work in the heliocentric model:

    In the heliocentric model, if the moon was orbiting, prograde, at the same rate as the earth was rotating (a period of around 24 hours), the moon would not appear to move in the sky. By your calculation, the moon would have an infinite diameter because it would never cross itself(or it would have a diameter of 0 because its speed would be 0 too, either is clearly wrong). To claim your method to be without assumption is not true, because it’s invalid in cases where you don’t lock down an assumption you’re claiming not to make.

    The correct conclusion (based on the heliocentric model) would be that the angular velocity of the moon’s orbit was the same as the earth’s rotation on its axis. It would not mean that the moon’s velocity was the same as the observer on the ground, because the moon is moving through a greater arc (due to its higher altitude), and so would have a higher velocity. The data would be insufficient to determine the moon’s speed and/or diameter because its altitude is unknown

    Different altitudes give rise to different diameters, for a given angle subtended the distance between two lines scales with the distance from where they meet. If I could put my problem in a question it would be: “How would your video look different if the moon were twice as far away, with twice the diameter to the one you calculated (with 4x the brightness, for completeness)?”

    In the heliocentric model, it would look no different. My ‘infinite set’ point is that you can scale by any number and get the same results in a heliocentric model. The projected ground track would still move at around 950 mph from west to east because the rates of rotation would be unchanged (while the size, speed and altitude would be very different).

    1. 1. The alleged spin speed of the earth is 12 miles per hour faster than that of the moon. 12 mph is not “much faster”. That is subjective and arbitrary wordage. This article is not about moon rise so I won’t discuss that. If you want to talk about the earth’s spin, you should not just state it, you should prove that spin because if the moon was circling the earth it would have the same effect. The falsifiability of the alleged earth spin would then be considered if there are other possible reasons for what the moon is observed to do.

      2. The moon’s size is not dependent upon proof the earth is either flat or round. (See your first paragraph.) The measurement process of the moon is not dependent on a flat earth or a sphere earth. The moon’s measurement is dependent on its travel to cross itself with a known speed of the moon relative to the camera’s location.

      3. Your paragraph of the earth’s alleged movement would produce the results of the moon being less than a 1/2 mile in diameter and that is not the size my measurements concluded. The moon traveled its entire width in 2:10 minutes. That has been established. I’m not suggesting the moon is less than 1/2 mile wide – the measurement concludes a size of about 34 miles wide.

      4. The moon’s elevation isn’t needed in the measurement process I used – only its speed relative to the earth is needed. Only if the speed was unknown, then moon’s elevation would be required to determine its size. Since the size is now determined to be about 34 miles, the elevation of the moon can also be determined of about 4000 miles. If I reverse calculated using the 4000 mile elevation of the moon, the result would then produce a moon size of roughly 34 miles. There are checks and balances with this method that you seem to gloss over.

      5. The measurement I did was the moon’s size, not its elevation. I’ve already previously stated to others, if you want to move the moon further away, with the same 952 mph moon speed relative to earth, it would still cross itself in exactly 2:10 minutes. For the measurement of its size, its distance isn’t required – just its speed.

      6. I did not use a model to calculate the moon’s size. I used the moon to measure itself and applied its speed relative to earth. The moon speed relative to earth is the parameter obtained from timeanddate so if the speed of the moon relative to earth is not correct (whether the earth is moving or not), you should discuss your concerns with them.

  9. 1. Not sure what you mean here by ‘spin speed’. The alleged phase of the moon is around 28 days, while the earth’s rotation phase is one day. It’s the two of these being out of phase that gives rise to the moonrise changing from one day to the next. Not sufficient to conclude heliocintrism but is predicted by the theory. I don’t see 12mph to be relevant here. I’m not stating earth spin per-se, I’m saying your measurements only work for one model, and don’t work in the one where it spins. While I’m not personally agnostic on the spinning earth, my critique is. I’m not telling you how things are, I’m telling you that you can’t reach your conclusions without making unproven assumptions.

    2. I disagree. You explained your method well, you’re not telling me anything new about it here. Re-read my 4th paragraph. I’m demonstrating a case (regardless of the reality of the world, we could both conceive of observing such a phenomenon in the sky) where your method would clearly give a nonsensical answer, and it follows that for cases like the one we actually see in the sky, the number produced would be equally meaningless (if we live in a heliocentric world). Your method is not model-neutral because it doesn’t work in one of them (I accept it totally works for the FE one, but that’s why you have to assume FE to use it).

    3. I’m not following you here. Sure, we’ve established 2:10. I’ll even grant you that the moon has travelled a distance equal to its diameter in 2:10. I won’t grant you that that distance is equal to the distance travelled at 952.5mph for 2:10. This is the leap that I’m not giving you. Thus I dispute the concluded 34 miles.

    4. The elevation is needed, because this is circular motion, and distances scale with elevation from the centre of rotation. I haven’t granted you 34 miles, so you can’t use that to get 4000 miles. Even if I had, you certainly can’t then back-calculate from the 4000 miles to reach the 34 again and claim this proves it. Yes, the maths works, that’s just algebra. It’s also a blatantly circular argument.

    5. No, we’ve established you measured the time it takes the moon to move a given distance, it’s diameter, but you haven’t determined what that number is. But you’ve put your finger on the problem here: If you moved the moon away and it looked just as big in the sky and the time was 2:10, you would still conclude that the diameter was 34 miles, despite the fact that it must be much bigger to be further away and still appear just as big.

    6. Again, I get your calculation. You don’t feed the numbers through a model but you can only draw the conclusions you do by eliminating one possible state of affairs in favour of another. Their number is correct. They’re not stating the speed of the moon, they’re stating the speed of the moon’s projection on the earth. In flat earth these are the same number, in globe earth they are not. That’s why to use their number as the speed of the moon is to assume flat earth.

    My question again: “In a heliocentric model, I’d the moon we’re twice as big and twice as far away and travelling twice as fast such that it would give the same ground track speed on timeanddate, how would your measurements give you the new value of the moon’s diameter?” Based on circular motion, it wouldn’t, you’d get the same result every time, despite the changing size of the moon.

    1. 1. My article is not about the phases of the moon, it’s about the moon’s size. I don’t rely on any model to confirm that the moon’s size is about 34 miles wide.

      2. The moon’s measurement is not depended upon what the earth does or does not do. It’s simply the moon’s measurement. The article is not about anything other than measuring the size of the moon.

      3. I’m not asking for you to grant me anything. The moon traveled 2:10 minutes and in that time it traveled about 34 miles from edge to edge at 952 mph.

      4. Your assertion of a circular motion is an assumption. I will not entertain assumptions. The 4000 mile height confirms the size I measured of 34 miles wide. I didn’t use the height to determine this but by cross-checking it, the size is accurate. When something is verified by another method, it’s considered reliable.

      5. I can’t believe I have to address this. If the moon was 10 times further away, it would still take exactly 2:10 minutes to travel across itself. Appearance is not a factor I used to determine the size of the moon is about 34 miles wide. I don’t care if it looked big or small – the fact remains that it crossed itself in 2:10 minutes and was traveling 952 mph. The looks of the moon have nothing to do with calculating its size.

      6. In this paragraph you talk about globe earth or ball earth. The moon’s size measurement is not dependent on another object (the earth).

      7. If the moon were twice as far away, I would simply apply more zoom and all other factors remained the same (moon speed of 952 mph, 2:10 to cross itself), the size of the moon would still be the same size of about 34 miles.

  10. For the sake of understanding, because I keep repeating the same argument in multiple points, I’ll re-make my case in a more concise way. You don’t need to answer all these points, just the ones where you think my reasoning doesn’t follow (or I’ve slipped in a sneaky assumption). Similarly, if there are any important points that I’ve omitted that refute the reasoning below, let me know.

    My thesis: “Your calculation depends on the assumption that the flat earth model is the reality of the world, since the measurements you make do not give you the size of the moon in a heliocentric model.”

    My reasoning:
    1. The moon travels a distance equal to its diameter in 2:10. I think we both agree on this.

    2. If the moon’s speed is known, you can use these data to calculate its diameter. I think we both agree on this too.

    3. To say that the speed quoted on timeanddate is the speed of the moon is to assume flat earth, because that is not the case in heliocentrism.

    3.1 In heliocentrism, to derive the speed of the moon from the timeanddate value, you must divide that number by the radius of the earth then multiply it by the radius of the moon’s orbit. This is just orbital mechanics and circular motion. Not assumptions I’m making, they are the implications of heliocentrism.

    3.2 In flat earth, no such conversion is needed because the moon simply moves in a plane parallel to the ground.

    3.3. Having not made the adjustment in 3.1, you assume flat earth. Similarly, to use the conversion in 3.1 you’d be assuming heliocentrism. My point is you need to pick one or the other to do the calculation.

    1. Call,
      The article is about measuring the moon and not about your thesis so I ‘m not going to discuss your thesis with you.

      You, or anybody, can measure the moon using the simple, easy steps presented in the article. I can’t stress this enough; the measurement methods don’t require you to believe you live on either a sphere or a flat plane. Measuring the moon is the subject of the article, not about the earth or your beliefs about the earth.

      1. What a cop out.

        I’m not asking you to read my degree, as you know full well. It’s a turn of phrase to represent a hypothesis and a point being made.

        Stress things all you like, it doesn’t make them true.

        Pedantry being used to discredit logical and physical reasoning, possibly the lowest form of specious argumentation.

        I despair.

    2. 3.1 is pseudoscience. Without it you have no argument. Like all heliocentric believers you add to confuse truth for the benefit of argument. . 3.1 is unnecessary in simple formulation. Simply nonsense and a waste of possible talent.

      1. Bob,
        Thank you for being able to think and know what is real science and what is not. It all falls into place when the garbage can be taken out to simplify things. Adding in assumptions makes the results assumption-based. There is no reason to apply many of the assumptions that are relied upon.

      2. 3.1 is how you do the maths if we’re in a heliocentric world. The author isn’t doing this and so has assumed the world not to be heliocentric.

        I’m not stating the world is heliocentric. I’m saying that the maths in a heliocentric world has to be done a particular way, and to not do this is to assume the world is not heliocentric

      3. Callum,
        If you want to read the Measuring the Sun article, you will find that MIT, the US military, nasa, Princeton, Russian military, and many other engineers and engineering firms apply the criteria of a flat and non-rotating earth. When that is the multi-industry world standard, it would make no sense to assume a different parameter than what everybody else is using. If you can find a standard that the assumption that the earth should be assumed to be round and spinning by MIT, the US military, nasa, Princeton, Russian military, and many other engineers and engineering firms, send that information to me so I can review it. Short of that guidance, I will continue doing the measurements correctly based on the multi-industry world standards.

        Take the time to obtain video of the moon crossing itself tonight since it’s a full moon and share that video with me with your latitude. Since the earth is allegedly spinning about 1000 mph at the equator at 0 degrees latitude, zero mph at 90 degrees latitude, and 500 mph at 45 degrees latitude, and all linearly changing speeds in between, you can be part of my multi-latitude experiment to determine if the earth is spinning or stationary.

        Since the sun crossing itself situation applies, I am soliciting videos of the sun crossing itself at different latitudes to determine if the earth is spinning or stationary as well. The video of the sun crossing itself should be obtained today when it’s at the highest point overhead to eliminate refraction effects of the atmosplane. Send that video to me with your latitude to be included in this experiment.

  11. When you look through the telescope, you will also notice that the background stars are moving across the field of view, slightly faster than the moon is moving. Surely the time that you need to record is the difference between the time taken for the moon to move its own diameter and the time taken for the background of fixed stars to move the same distance? That will isolate the speed of the moon from the speed of the rotating dome of the sky itself.

    1. Colin,
      The stars are not part of the moon measurement calculations. The only thing needed is the time the moon crosses itself (2:10 minutes), and the moon’s speed (952 mph). Nothing else is needed.

  12. By using ‘ground speed’ you have measured how far it has moved in relation to the Earth’s surface, not how far it has moved along it’s own, much longer orbital path. Your measurement is about 0.50° of the Earth’s circumference – how far is 0.50° of the Moon’s orbital path?

    1. Drac,
      The moon takes 2:10 minutes to cross itself. That is the motion of the moon only. Since the moon’s speed from the carera’s perspective is roughly 952 mph, the moon traveled a little over 34 miles as the moon’s own width was used to measure itself.

      No assumptions have been used in this measurement method – only the moon’s movement to cross itself was used and that is not an assumption, it actually happened and it’s recorded on video doing exactly that.

      1. You have to make a number of assumptions for you calculations to work at all.

        1. That the Earth is a flat plane.
        2. That the moon circles above the Earth on a flat plane parallel to the plane of the Earth.
        3. That the moons actual orbital speed is the same as it’s apparent ‘ground speed’ over the Earth.

        Without at least those three assumptions your calculations simply don’t work. On a spherical model an apparent ground speed of 952.5mph equates to the moon moving about 2100 miles (about 1 diameter) along it’s own orbit in 130 seconds.

      2. Draco,
        The only things taken into account include the 2:10 minutes for the sun to travel across itself, and the moon’s speed relative to earth. One is an actual event captured on video, and the other is a reported moon speed relative to earth, neither of which are assumptions.

        These calculations did not take into account the shape of the earth since the shape of the earth has no bearing to calculate the speed of the moon. I don’t know how to make this any clearer – the moon was measured and nothing was injected into the measurements. The only thing related to the earth is that the camera recorded the sun crossing itself from earth.

        I cannot give any weight to your assumptions that assumptions were made to calculate the moon’s speed because no assumptions were made – just actual events based on the reported moon speed from earth.

        You mention spherical model – the moon measurement does not assume any model; it was based on actual observations and models were not part of the measurement. The result of the measurement is that the moon is a little over 34 miles wide. If you have concern that the moon did not cross itself in 2:10 seconds, you are welcome to replicate the measurement yourself.

        Thank you for your comment and your concerns.

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