What factors may influence the surface accuracy of space antennas?

Question

And is antenna size one of such factors?
Thanks in advance.

Answer 1

I assume you're talking about how close all the points on the surface of the antenna are to being the same distance from the focal point?

Yes, size is one the factors. Materials can deform under various stresses such as heat, wind, etc. The longer or wider the material you're talking about, the greater the impact of even a small amount of deformity once you reach the edges.

The stiffness of the material is also a factor. The siffness can be calculated using the Young's modulus. The shear modulus and density also affect the tendency to deform.

You can reduce the deformity by shortening the length of the stress, which is why a large antenna has a signficant amount of structural framing behind it.

The difference in stress between different points on an antenna are also a factor. For example, if the antenna were pointed nearly at the Sun, all points on the antenna would be heated more or less evenly. If the antenna is at some other angle, the incidence angle of the Sun will be different on different parts of the antenna and will be heated at different rates.

One other factor is the frequency range you intend to use with the antenna. The frequency may not actually affect the surface area, but it does establish what the surface accuracy has to be. If you're talking very high frequencies, that means you're talking very short wavelengths. A short wavelength would be more affected by a small deformity in the antenna than a longer wavelength.

Here's one article about improving the surface accuracy of ATCA antenna.

Answer 2

In radio telescopes, the resolution is a factor of the size. For large arrays, the resolution is a function of the length (and width in the other direction) of the array from the perspective of the arriving wave front.
The array could be 1000 m wide, but if the observed wave front is coming from a 45 degree angle, the wave front 'sees' a width of 700 m instead of 1000. Then 700 m determines the resolution (the wavelength used also affects the resolution).

For a single dish (usually steerable so it can point in the direction of the wave source), then the full diameter determines the accuracy (resolution).

For interferometer, the distance between the receivers determine the accuracy along the direction of the 'baseline'. if you have two 10-m dishes separated by 100 km, the resolution in the orientation of the baseline (the 100 km line) will be fantastic but the resolution in the other direction (perpendicular) will only be the result of the 10-m diameter of each dish.

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The resolution in radians is calculated with this formula:

R = (Lambda) / D

Where lambda is the wavelength of the observed photons (could be measured in metres for radio astronomy) and D is the effective diameter (as 'observed' by the wavefront) or the effective distance between the two elements of an interferometer.

One radian = 180/pi degrees = 57.2957795... degrees =
3437.7... minutes = 206,265 seconds.

Since is is customary to express the resolution in arcseconds, the formula we use becomes:

R = 206,265 * (Lambda) / D (where lambda and D must be in the same units: metres over metres, cm over cm, cubits over cubits, whatever).

On very large baselines, we even use milli-arcseconds (mas), each mas is 1/1000 of an arcsecond.

Answer 3

rain, heat, cold, autumn etc
weather factors can influence.

Answer 4

The question is meaningless.