8-1: The answer is simply wavelength (cm) = 30/20 = 1.50. This falls into the K-band. BACK

8-2: The British had been developing radar since before the opening of WW II. They had functioning systems by the beginning of the Battle of Britain in 1940; the Germans did not have this technology. The ability to spot incoming Nazi aircraft, both day and night, gave the English a great advantage from this forewarning that allowed them to "scramble" their fighters before the raids began, and also allowed ample warning to get the civilian population into shelters. Later, radar helped Allied bombers sight targets on bomb runs at night or when cloud cover was heavy. Radar was also important in moving vast flotillas of troop and supply ships across the Atlantic, being installed on naval destroyer escorts.BACK

8-3: Vegetation. The Altiplano is a high desert; the Lowlands are flat grasslands. For the L Band wavelength used (23.5 cm), the desert surface was a good backscatterer whereas the low plains responded differently as a smooth surface (reflecting most of the radar beam away). BACK

8-4: Physically, a SAR system can be much smaller - in length as well in weight. Thus, it can be flown underneath smaller aircraft, a real savings in operational costs.BACK

8-5: At any given moment, the Sun's radiation travels to any small area along parallel lines (rays), i.e., at some single angle. Radar, as it operates, sends its radiation out over a range of angles, so that different parts of the target area will respond differently. BACK

8-6: The cosine of 50 degrees is 0.6428; with these values Rr = (0.1 x 10-6)(3 x 108)/2 x 0.6428 = 23.33 meters. To calculate Ra at near range, after converting units to meters, one gets for the near range case: Ra = 0.7(5000 x 0.05)/5 = 35 m; for the far range this become 15/5 x 35 = 105 meters. Thus, at the far range, objects separated by more than 6.25 meters will not be resolved. BACK

8-7: The sine of 60 degrees is 0.8660. The smooth surface has h < 15/25 x 0.8660 = 0.693 or 0.7 cm. Thus, any surface with 0.7 cm variations (small pebbles) or less will act as smooth. To find the lower limit for surface roughness, multiply this number by 25/4.4 = 3.97 cm. Objects at the surface whose height dimensions exceed 3.97 or about 4 cm will cause that surface to act as a rough reflector. These values vary with depression angle: as that angle decreases, the value of sine also decreases, making the smoothness and roughness limits increase; thus as the slant range increases (looking farther out), the degree of roughness enlarges (i.e., the values of the heights increase) so that surfaces at greater distances will cause more backscatter. BACK

8-8: If the crop has at least partly grown, the tones of circular fields will be light, compared to darker for the soil in between circles. Mature crops are strong reflectors, so the fields will be at their brightest. BACK

8-9: Clouds! Most radar bands effectively penetrate cloud cover - one of this remote sensing system's chief advantages. In the valley, small hills or rises occur that don't show up as topographic features in that valley in the X-band mosaic. BACK

8-10: The high point is an elongate (up-down) hill or mountain in the left center. The width of its shadow is greater than widths of other shadows, a fact controlled by its greater height. BACK

8-11: Illumination is coming from the right. The real mountains probably have slopes of nearly the same angles on each side of their crests. But, here the mountains look as though they are grossly leaning towards the right. Their right sides look like bright cliffs whereas their left sides look like the equivalent slopes are gentle and shaded. This is the classic foreshortening effect. BACK

8-12: There is a large, massive hill in the upper left. While it appears generally similar in both images, its orientation (directional trend) and overall shape is somewhat different when the two images are compared. The large fault in the lower center (running slightly right of up-down) is visible in both images. Some faults or fractures are much more evident in the right image relative to the left, but the left has several that are hard to see in the right. There are subtle topographic differences as well. BACK

8-13: The HV mode is "flatter", i.e., doesn't show as much contrast as the HH mode (look at the fields). The HV mode also doesn't bring out the small hills near the Juniata as sharply as the HH mode. In this example, the HV mode is less informative but in other cases it may show more information. BACK

8-14: You betcha! The instances are small and require close inspection. But there are several places where the pattern of the bright, radar-facing slope cuts across another bright pattern of a neighboring topographic feature. I can't be precise in locating these for you, without using a grid, but look carefully and you shall find. BACK

8-15: First year ice is normally thin and smooth, so it will reflect most of the radar beam away from the receiver, hence appearing dark. This type is near the right edge of the strip. It is crossed by thin bright lines, called "leads", which are ice that filled cracks, was squeezed and broken, and is thus rough. Multiyear ice, on the far right, has been broken up, jumbled, and refrozen a number of times; its surface is rough, thus it is bright. On the left half are ice floes, segments of the ice crust that broke from the main sheets, moved about, and then were refrozen (with leads) into a new continuous sheet. Their tones tend to be medium light to medium dark gray. BACK

8-16: If you have ever been to the Mojave, your impression would be that it is a fairly flat desert surface with mostly light soils and small rocks. At the Seasat wavelength (23.5 cm), this pavement would, like a street surface, act as smooth and reflect most of the radar pulse radiation away - thus dark. BACK

8-17: These volcanoes (the shield type) are moderately high (rising in excess of a 1000 meters from the outer slopes and inner crater floors down at least 500 meters from the rims). But they show almost no foreshortening of any of the slopes (interior ones can be as steep as 35°). This is explained in part by the relatively low depression angle (around 40°) of SIR-A. Several of the lava flows are dark, implying smooth surface basaltic rock. Several others are bright, implying that the lava there is chunky (aa or block type). BACK

8-18: The differences are due mainly to look directions and depression angles (from the SW; around 70° and from the south; about 40°, for Seasat and SIR-A respectively). Seasat's higher depression angle produces more slope foreshortening. The approximately 45° difference in look (illumination) direction causes different ridge alignments between the two scenes (note that N-S trends are emphasized in Seasat with the same ones much subdued in SIR-A). In the SIR-A image, bedding units generally parallel to the coast line are emphasized in the mountains close to the coast, and a major fault is present about 40% in from the left margin (it is curved and has tonal differences on either side). There is a similar but shorter fault seen in Seasat (about 1/6th in from the left). Seasat also brings out more wave roughness in the Pacific Ocean. BACK

8-19: Using an atlas, you should have found the Richmond-San Rafael bridge, the Golden Gate bridge, the San Francisco-Oakland bridge, and (harder to see, at the bottom) the San Mateo bridge. What appears to be a curved bridge in the upper left is actually a shoreline. BACK

8-20: The first step is to determine the directional orientation of the image. In its lower right corner is a small patch of black with a curved boundary inside the image. That is probably ocean. Since the Pacific Ocean is due south of Mt Fuji, in a bay called Suruga Wan, the image needs to be rotated about 90° clockwise to get the new top pointed towards north. In that position, the white patches coincide with Fujiyoshida. (Of course, I could have rotated this for you, but I decided to keep it in the position that it assumed on downloading from the Net, and then issue this challenge to re-orient.) BACK