4 ms·

It’s hard for me to understand the arc seconds and degrees. Is there some kind of illustration available that shows the transmission from earth as a megaphone c

It’s hard for me to understand the arc seconds and degrees. Is there some kind of illustration available that shows the transmission from earth as a megaphone cone expanding out into to space?

mikewarot 2y ago
Draw an equilateral triangle. The internal angle at a corner is 1/6 of a circle, 1.0471975511965 radians. That can be divided by 60 into degrees, which are 0.0174532925199 radians Those can be divided by 60 into minutes, which are 0.0002908882086 radians Those can be divided by 60 into seconds, which are 0.00000484813681109 radians Those can be divided by 60 into thirds, which are 8.08022801849E−8 radians I think going any further is likely to be meaningless in a modern context, but fourths and fifths were used in the past. https://en.m.wikipedia.org/wiki/Degree_(angle) https://en.m.wikipedia.org/wiki/Degree_(angle)
- semireg 2y ago
  I understand the division of units, what I’d like to see is an illustration of the signal’s cone at the astronomical distances in the article.
  - marcosdumay 2y ago
    Well, the area taken by a cone with angle "a" at the distance "r" is pi * r^2 * sin(a)^2. For an arcsec, the square sin is ~85e-9. At the distance of the Moon, that means the signal occupies ~400km^2. At the distance of the Sun, ~5 billions of km^2 (5Mm^2). The signal intensity is inversely proportional to that area.
  - semireg 2y ago
    For example, if the screw/servo controlling one axis of the dish is turned by a minimum adjustment, how many hundreds or thousands of miles does it throw the cone’s target at the distance of the probe? What’s the remaining margin for error at these scales?
  - mikewarot 2y ago
    The angle is about 9x10^-5 radians. If you were to take a meter stick (or a yard stick) and stick a piece of paper under one end, that's the precision within which it can point the dish. (While the earth is spinning, and going around the sun!) The actual width of the beam is about 5 times that angle.
  - deleted 2y ago
    [deleted]
GuB-42 2y ago
1 degree = 60 arcminutes, 1 arcminute = 60 arcseconds One arcminute is about the resolution of the human eye. That is, if two points are less than 1 arcminute apart on the your field of view, they appear as one point. The moon is 30 arc minutes (1/2 degree) on your field of view, that's why one can see the details of the surface with the naked eye. 18 arcseconds is about the size of Saturn (without the rings) as seen from the earth. It means it is just precise enough to shoot through the rings of Saturn. You can't see this without magnification, like with a small telescope or a good pair of binoculars.

It’s hard for me to understand the arc seconds and degrees. Is there some kind of illustration available that shows the transmission from earth as a megaphone cone expanding out into to space?

mikewarot 2y ago

Draw an equilateral triangle. The internal angle at a corner is 1/6 of a circle, 1.0471975511965 radians. That can be divided by 60 into degrees, which are 0.0174532925199 radians Those can be divided by 60 into minutes, which are 0.0002908882086 radians Those can be divided by 60 into seconds, which are 0.00000484813681109 radians Those can be divided by 60 into thirds, which are 8.08022801849E−8 radians I think going any further is likely to be meaningless in a modern context, but fourths and fifths were used in the past. https://en.m.wikipedia.org/wiki/Degree_(angle) https://en.m.wikipedia.org/wiki/Degree_(angle)

semireg 2y ago
I understand the division of units, what I’d like to see is an illustration of the signal’s cone at the astronomical distances in the article.
- marcosdumay 2y ago
  Well, the area taken by a cone with angle "a" at the distance "r" is pi * r^2 * sin(a)^2. For an arcsec, the square sin is ~85e-9. At the distance of the Moon, that means the signal occupies ~400km^2. At the distance of the Sun, ~5 billions of km^2 (5Mm^2). The signal intensity is inversely proportional to that area.
- semireg 2y ago
  For example, if the screw/servo controlling one axis of the dish is turned by a minimum adjustment, how many hundreds or thousands of miles does it throw the cone’s target at the distance of the probe? What’s the remaining margin for error at these scales?
- mikewarot 2y ago
  The angle is about 9x10^-5 radians. If you were to take a meter stick (or a yard stick) and stick a piece of paper under one end, that's the precision within which it can point the dish. (While the earth is spinning, and going around the sun!) The actual width of the beam is about 5 times that angle.
- deleted 2y ago
  [deleted]

GuB-42 2y ago

1 degree = 60 arcminutes, 1 arcminute = 60 arcseconds One arcminute is about the resolution of the human eye. That is, if two points are less than 1 arcminute apart on the your field of view, they appear as one point. The moon is 30 arc minutes (1/2 degree) on your field of view, that's why one can see the details of the surface with the naked eye. 18 arcseconds is about the size of Saturn (without the rings) as seen from the earth. It means it is just precise enough to shoot through the rings of Saturn. You can't see this without magnification, like with a small telescope or a good pair of binoculars.