Discussion below the figure

The spider diffraction pattern from a spider covering surface areas of 1 percent, 2 percent, and 4 percent of the aperture. These are normalized to the radiance of the thinnest spider at the center. The 1-percent pattern is the dimmest one at the center and the one with the fewest minima. The 4-percent spider has 16 times the radiance at the center of the pattern yet settles down to the same approximate radiance at points distant in the image.

The normalized integrated energies are 4 percent, 2 percent, and 1 percent of the aperture. This neglects the effect of the secondary mirror and supresses the contribution of the primary. It is important to not confuse this pattern with the primary's diffraction pattern. The primary diffraction pattern would be a spike very near the center of this diagram. It would be a good deal higher than the central brightness of all three spiders.

If the one-percent surface area spider were fitted to a 200 mm aperture, it would be 0.79 mm thick and the first diffraction minimum would occur at the indicated 2.4 arcminutes.

Even though the pattern may superficially go to the same values in distant parts of the image, it is almost unlit there. Following is a split of the energy residing in various lobes as determined from a numerical integration of [sin(x)/x]^2. An overexposed photograph might possible reach out beyond pi, where it would show the spikes truncated at about the same place. However, this would neglect the very important topic of ensquared energy at that point.

Central lobe	90.3 percent
First side-lobes	4.7 percent
Second side lobes	1.6 percent
Third side lobes	0.8 percent

And so forth.

It is clear that the behavior inside the central lobe of the spider diffraction pattern is the most important.

Let's calculated the ensquared energy at the edge of the first minimum of the narrowest vane, which in this case is found at pi. Evaluating them at the same real angle is a fairer way of comparing them.

1% spider... E = 0.01 * (0.903) = 0.0090

2% spider... E = 0.02 * (0.903 + 0.047) = 0.0190

4% spider... E = 0.04 * (0.903 + 0.047 + 0.016 + 0.008)= 0.0390

Note that each of these has 0.001 of the total diffracted energy remaining to be enclosed.

12/31/99, prepared from materials generated by Bill Zmek and Dick Suiter for the spider article.