I was thinking about how airplanes would approach an airport and how they should look if on a globe. I was looking at various sites that talk about when a commercial airline (flying at 30,000 feet) starts it’s decent. At about 300 miles out the plane will start descending at about 500 feet per minute. If you could look 200 miles out with a powerful enough telescope, the horizon would equal about 5.6 miles. In other words no airplane could be seen since at that distance since 5.6 miles is the highest a plane normally flies.
With that descent time, a plane would reach ground in just under 1 hour. It would be assumed that the cruising speed would also decrease since the average speed is about 500 miles per hour. Assuming that the average speed during decent is about 300 miles per hour and 500 feet is dropped per minute then at what distance should the plane be visible?
After 30 minutes the plane will have dropped 15,000 feet (2.84 miles) and be 30 minutes out and 150 miles from the airport. Again, if you could look out that far with a telescope the plane would be at or below the horizon. Over the next 15 minutes the plan would drop another 7500 feet and be 75 miles out. At that point the plan should still be at or below the horizon. Over the next 7.5 minutes the plane would drop 3750 feet and be 37.5 miles out and 937.5 feet above the ground. Again, the plan would not be visible. Over the proceeding 3.75 minutes, the plane would be 18.75 miles out and 234 feet above the ground and still not visible. However, it is mostly likely that the decent and speed would be throttle to match the runway so the final few miles one would finally allow the plane to appear at the horizon. So what we would see would be the plane shoot upward from the horizon and then rapidly drop down.
Of course this is an extreme example to demonstrate the idea. But the point it that if you watch airplanes approach airports they are always very high in the sky and slowly descend over time. They never shoot up from the horizon. What we do see is planes ascending or descending but never appear at the horizon.
If someone was to film airplanes as they approach the airport, I can pretty much guarantee that not a single plan will shoot up from the horizon and then curve downwards toward the airport.
On another note, whenever you see jet streams they are always straight. At plane at 30,000 feet will be visible for a shot period of time before it starts curving towards the horizon. They should have a curve in them. No ones sees that either.
How do we know if the world is spinning or stationary? Do current images from space or the material in text books prove the Earth is spinning? Only if you believe the source to be valid. If, however, you are a natural scientist you would want to verify the veracity of the statements by repeating the experiments.
Using these two examples, let’s look at what is experienced when a person walks upon the earth. If the earth is spinning it is comparable to the conveyor belt example except that the earth’s axis acts as the central wheel that “belt” of the earth’s surface or ground moves around. It cannot be the train example since the floor of the train is not in motion. We can see that additional energy is provided by the moving ground whereas no additional energy is provided to the person who is beside the belt. We could then create an experiment where a person would leap against the direction of the belt (both on the belt and off). We could place an object 5 feet from the person both on and off the belt and ask them to leap towards the object. Since the additional energy of the belt is continually compelling the object forward, we would theorize that the total distance between the person and the object would be less than then person leaping beside the moving belt.
Why is leaping important in this experiment? If the person simply walks on the belt, the forward momentum will be continually added to the individual since there is always contact with the belt. By disengaging with the belt in a direction against the motion, we are subtracting (for a brief moment) the forward momentum of the belt and allowing the belt to move underneath. It would be possible to land nearer or even surpass the object whereas the person leaping beside the belt would gain no advantage like that. The question is how quickly does the forward momentum of the belt dissipate once the person leaps? If the person can leap (towards the object like a long jump) with an acceleration of 1 m/s/s and can stay airborne for 2 seconds then a total distance of 2m can be achieved (with a final velocity of 2m/s). Since the belt is moving at 1 m/s then in the first second the momentum has been overcome and the object has moved 1m closer. In the 2nd second the object moves 2m closer if we add the motion of the belt and the acceleration of the person. The total distance covered would be 3 meters. In the case of the person beside the belt, they would only be able to cover 2m if they leaped with the same acceleration. (you can work out the acceleration equation if you want)
We could increase the acceleration and distance by using something like a canon. If we use the same arc and acceleration the ball should land at a greater distance from the canon if on a moving belt than if stationary. If we assume that the acceleration of the canon is 10/m/s/s, the tennis ball can reach 80 m/s velocity and has a total airtime of 10 seconds, we can calculate the total distance travelled (640m). If the velocity of the belt is 1 m/s then within 1 second the momentum of the belt has been overcome and we can add an additional 9 seconds of belt movement to the distance for a total of 649m. If we then turn around and fire the canon in the opposite direction then the distance travelled by the canon is subtracted from the total distance of the ball (1m/s x 10 seconds = 10m). This would mean that the total distance between the canon and the ball would be 630m.
If we take the above example and apply it to a spinning earth then a similar result must take place. Taking the spin as the same as the conveyor belt one need only launch an object (a tennis ball thrower would suffice) to the west and then to the east. If the objects land at similar distances from the thrower then we are not spinning. However, if they land at different distances then we must be spinning.
In response to my latest post, I received a comment that due to refraction, objects at certain distance would be visible even if they are below the horizon. He also talked about adjusting curvature every 8-10 miles. I’ve started reading his blog but the reason for the adjustment is not clear. However, I think he is talking about this. In the end, the height of the person and the object being observed are taken into account. Here is the original comment:
“You are using an improper formula for curvature that has to be constantly corrected for every 8-10 miles. The correct formula for curvature is found here. https://chizzlewit.wordpress.com/2015/05/13/working-with-the-curvaure-of-a-spherical-earth/ Also you are over water which tends to make refraction a bit of a problem. You will find however that with the height of you the observer on the deck as you pointed out and the height of the lands you put in that it does indeed fit in with the sphere earth model especially when you account for refraction. If you want to use a crude formula for distances between lighthouses and boats that includes for refraction you can use this one though its only an average correction for refraction. http://www.pajack.com/stories/pitts/viewdistance.html I suggest you try this with surveying equipment though if you are at sea the choppiness of the water will make it tough. Also being over water with refraction conditions makes it tough. You will find though that if you use surveying equipment you can clearly measure landmarks and see the curvature of the earth. https://www.youtube.com/watch?v=IOFmLHHwtic”
To be fair to the individual, I wanted to address the objections empirically and see if he had valid points. You can read my response here:
“I appreciate your comment. The main issue with the video is that it was taken with a crappy Samsung smart phone. It is possible to take a stable view from land (from either side) which is what I plan to do so any shakiness could be taken out. This will require a telescope with a camera to bring a sharper focus at a greater length. There are distances of 70 miles or more that can be observed from Vancouver. Specifically from Kits beach to Denman Island.
I would like to address your blog and the concept within them. If we wish to proceed using a scientific process (which must involve empirical evidence not just math), we must apply your math to observations.
If I misrepresent your ideas or math in anyway please forgive as I’m simply trying to apply them the best I can. If there are errors, please point them out and we can move forward from that. If we disagree, then the empirical evidence must be the final judge, not our opinion. Do you agree with that?
If you can apply your formula to the distances given in my post (29 and 33 miles) that would provide me with an actual representation of how your formula represents curvature. From there we can compare the two formulas and see the differences. Are you ok with doing that?
Refraction is defined as “the bending of a wave when it enters a medium where its speed is different. The refraction of light when it passes from a fast medium to a slow medium bends the light ray toward the normal to the boundary between the two media.”
As I’m looking towards Vancouver from the deck of the ferry, I would not be looking into water which would be a slower medium. As well the refraction requires the object be within the slower medium (or at least in between my eye and the object in question). When looking at this wiki article, the superior mirage is the only explanation that you are referring to that could bend light (if you have a different one, please explain). It would require that the object in question is within an area of higher temperature than the observer. As well, if this object (like a building) occupies various temperature strata then the object would appear quite distorted since different temperature regions would overlap and the image would not make much sense to the eye (https://en.wikipedia.org/wiki/Mirage#/media/File:Superior_and_inferior_mirage.svg). The video that I shot shows objects that are consistent in shape and size without any distortion. With both superior and inferior mirages the objects are inverted and/or distorted. The other issue is that the temperature differential required for a superior mirage is significant:
“The “resting” state of the Earth’s atmosphere has a vertical gradient of about -1° Celsius per 100 metres of altitude. (The value is negative because it gets colder as altitude increases.) For a mirage to happen, the temperature gradient has to be much greater than that. According to Minnaert,[1] the magnitude of the gradient needs to be at least 2°C per metre, and the mirage does not get strong until the magnitude reaches 4° or 5°C per metre. These conditions do occur with strong heating at ground level, for example when the sun has been shining on sand or asphalt, commonly generating an inferior image”
The temperature differential from the bottom of UBC to the peak would require a consistent temperature change over 81 meters. According to the above formula, the temperature from the bottom of UBC (let’s say 12 degrees C) to the peak would be impossible. It makes sense at small distances (like asphalt) since the temperature difference can be significant. However, the temperature from sea level to 81 meters will not be 4-5 degrees per meter. So unless you see an error in that formula, we can safely rule out a superior mirage. So once I can see your math using your equations then we can move forward with the next discussion. Hopefully I will be able to get a sharper image next time that will show more detail.
Thanks again for your comment.”
It occurred to me that the image of Chicago across lake Michigan was described as a superior mirage. As I wrote in my post above, the temperature differential between the lake surface and the highest object would require at least 2 degrees per meter. If you think about it, an object 30 meters high (~100 feet) would require 60 degrees difference for the affect to occur. And that is not a very strong affect. According to the above article, you would need 4-5 degrees for it really be visible. There simply is not that kind of temperature differential in such short distances. As well, the object would need to be inverted or distorted which is obviously not the case in this video. Near the end of the video, they show a “duct” between the hot air and the cold air. For the affect to work, the temperature difference must be so great as to bend the light. The distance between the hot air and cold air is greater than 1 meter. The important thing to remember is that the “temperature gradient” must change not just one portion but the entire length of the object. In the case of a small object on the horizon at a great distance (70 miles), atmospheric distortion would occur and you would see inverted images. But if the object(s) in question are visible at 70 miles at a greater distance above the horizon (like a tall building) then temperature must be that much greater. Finally, if the image in question was taken with a camera with a zoom lens it completely removes the possibility of a mirage since the focal point of the lens would bypass the affect as seen in this video:
As the lens sharpens the image, any distortion would be removed and any mirage would disappear.
You can read about Fata Morgana to see how these distance object are explained within the commonly accept ideas about the curvature of the earth. However, mirages are distinctive by their distorted image and changing conditions. In the above video of Chicago, there is no distortion.
Terra Firma 
