Originally posted by Astner
No it doesn't, that's only the case if φ₁ = -φ₂.
Plug it in, sequentially, with 500 being used for your vectors during each change.
Originally posted by Master Han
Since longitudinal lines get progressively smaller as you move towards the north or south pole, wouldn't there be specific starting locations where the net displacement actually would be zero?
potentially if you started 250km south and west of the point where the prime meridian and equator meet. Or I guess any meridian would work, the route would just need to be bisected by the equator.
Originally posted by dadudemon
I'm pretty sure it does if you do that 4 times and then add the results together.
Solve x₂ and z₂ from that, and then reset the equations with the new values.
In this case you won't get the same values for x₁, y₁, z₁ in the first equation that you will for the x₂, y₂, z₂ in the fourth equation, unless c = 0.
Originally posted by Symmetric Chaos
How can the point where you start possibly have an effect on the final answer? Its a sphere, I can rotate it any way I want without changing anything.
You can't rotate it at will once the "problem" actually starts, at which point having a starting point at the "north pole" is certainly going to be different from having one near the sphere's bottom. If we use standard geographic definitions of NSEW, for example, starting at the North Pole would mean that you couldn't travel North, West or East at all, so it clearly matters where relative to the (fixed) sphere you are.
(correct me if I'm wrong, of course)
Originally posted by Master Han
You can't rotate it at will once the "problem" actually starts, at which point having a starting point at the "north pole" is certainly going to be different from having one near the sphere's bottom. If we use standard geographic definitions of NSEW, for example, starting at the North Pole would mean that you couldn't travel North, West or East at all.(correct me if I'm wrong, of course)
Astner said: "Apply this formula four times and voila, and you'll get your displacement. Obviously it depends on where on the earth you are."
Why and how would my displacement vary depending on where I started?
Originally posted by Symmetric Chaos
Astner said: "Apply this formula four times and voila, and you'll get your displacement. Obviously it depends on where on the earth you are."Why and how would my displacement vary depending on where I started?
Because latitudinal lines aren't parallel, and longitudinal lines grow more jam packed at the poles, where an equivalent angular displacement equals a far more massive arc-displacement. Again, I think that's right. It's the same fundamental reason why you can't travel north, west or east from the north pole.
To be honest, I don't exactly understand Astner's math, so I'm just assuming that he's using geographical definitions of the cardinal directions.
Originally posted by Symmetric Chaos
How can the point where you start possibly have an effect on the final answer? Its a sphere, I can rotate it any way I want without changing anything.
In fact I drew you a picture.
Two trajectories staring in the bottom right corner moving north then west then south then east. With one being equator-symmetric hence leaving no displacement, and the other not being equator-symmetric hence leaving one.
Originally posted by Astner
No. For instance, set the equation equal to the distance, c, set φ₁ = θ₁ = y₁ = z₁ = 0, x₁ = r, move north to get x₂ = √(r² - z₂²) with y₁ = y₂ = 0.Solve x₂ and z₂ from that, and then reset the equations with the new values.
In this case you won't get the same values for x₁, y₁, z₁ in the first equation that you will for the x₂, y₂, z₂ in the fourth equation, unless c = 0.
How it is supposed to work in the x,y,z coordinate system, while moving across the surface of a sphere, is you end up at the same place your started. Only on the fourth iteration should it equal if you set your starting position at zero (which you should be doing). On steps 1, 2, and 3, you will not end up back at your starting position and those results will not reflect zero.
Also, you referenced north in your post...
So what you were doing with matematca was accounting for the distortions that meridians add to moving about the system?
I think something that is also interesting to consider is what 500 km east/west is at the pole (or north at the north pole and south at the south pole). Because it really doesn't make much sense there. I'd say at the pole it becomes undefined, cause you could really make an argument for going in any direction from there.
Originally posted by dadudemon
How it is supposed to work in the x,y,z coordinate system, while moving across the surface of a sphere, is you end up at the same place your started.
You have a relation between spherical coordinates and Euclidean coordinates.
Originally posted by dadudemon
Only on the fourth iteration should it equal if you set your starting position at zero (which you should be doing). On steps 1, 2, and 3, you will not end up back at your starting position and those results will not reflect zero.
Originally posted by dadudemon
So what you were doing with matematca was accounting for the distortions that meridians add to moving about the system?
Either way, what I did was to find the vectors pointing north and east—which was pretty simple since that's the direction the derivates are pointing in—I then projected any change in movement to follow those vectors.
Originally posted by Astner
Not quite. By projecting your movement on the normalized n- and e vectors you're forcing any movement to remain on the surface of the sphere rather than drawing a straight line between point A and point B.
Yes, exactly. Any "z" axis movement will be negated. Likewise, so will the y and x axes. That means if you're moving across the surface of a sphere and you account for your vectors in such a system, your net displacement, on that sphere, should be zero, if you moved as described, four times.
The distortions only become involved once you mix in meridians and parallels.
Originally posted by Astner
Actually it's written in LaTeX and the backgrounds are removed in Photoshop. The formula is deduced by hand though.Either way, what I did was to find the vectors pointing north and east—which was pretty simple since that's the direction the derivates are pointing in—I then projected any change in movement to follow those vectors.
If you're doing that, then, yes, your distortions will show up.
Originally posted by dadudemon
Yes, exactly. Any "z" axis movement will be negated. Likewise, so will the y and x axes. That means if you're moving across the surface of a sphere and you account for your vectors in such a system, your net displacement, on that sphere, should be zero, if you moved as described, four times.
Originally posted by dadudemon
The distortions only become involved once you mix in meridians and parallels. If you're doing that, then, yes, your distortions will show up.
Originally posted by Astner
No, because the trigonometric functions of φ₁ makes the geometry anisotropic. You're not moving in straight lines.
Then let me rephrase it for you, too: you'll be moving 500km in an arc in each vector, not in a straight line. As a matter of impossibility, you wouldn't be able to just cut straight through this hypothetical sphere...you must have some awesome tunneling equipment in China (or wherever you are).
Originally posted by Astner
Let me rephrase it for you then. North is longitude, and east is latitude.
Let me rephrase that for you, then: North-South Longitudes are meridians. East-West Latitudes are parallels.