AMATEURS WHO ARE INTERESTED in predicting the positions in orbit of the earth satellites that broadcast pictures of cloud cover can now make a device that will perform the calculations automatically. The device is designed for satellites in orbit within 2,000 miles of the earth, and a separate device must be made for each satellite. Details of the apparatus are described by William K. Widger, Jr., who was a pioneer in the development of weather-satellite technology and is now director of geophysical services of Biospherics Consultants International, Inc. (P.O. Box 529, Laconia, N.H. 03246).

"The device consists of a map in the form of a polar projection of the Northern or Southern Hemisphere. A covering sheet of clear plastic is loosely pinned to the center of the map. The pin is fixed at the North or the South Pole.

"The plastic overlay rotates around the pin. The track on the earth's surface over which the satellite passes during any orbit is plotted as a curved line on the overlay [left]. By rotating the overlay with respect to the polar projection the observer can determine the geographical position of the track, that is, the subpoints on the earth's surface directly under the satellite, at any instant.

"A pattern of concentric circles can be drawn around any point on the earth's surface to represent the local horizon and to serve in determining the series of points directly under a satellite that correspond to the several angles of elevation above the horizon at which the satellite would be seen. When the overlay is rotated to the position at which the track crosses the pattern of concentric circles, the approximate direction and elevation of the satellite are depicted by the points where the track intercepts the circles. The track can be calibrated in terms of the number of minutes that have elapsed since the satellite passed over the Equator on its most recent northbound crossing.

"Normally the observer is interested in the position of the satellite in relation to the observing station. For this reason the map need not be cluttered with outlines of the earth's landmasses, although they can be included if the experimenter wants. They are shown on polar projections of the earth in most world atlases.

"For the background map use polar-coordinate graph paper No. 464417, a product of the Keuffel & Esser Company, or an equivalent. The track of the satellite is drawn with india ink on the type of plastic sheet known as heavyweight clear acetate paper, which is available from dealers in artists' supplies. Mount the graph paper on a stiff backing of cardboard or plywood.

"Begin the project by numbering the major intervals of the graph paper to serve as the geographical coordinates: latitude () and longitude (). Note that longitude increases both west and east from 0 degrees at the bottom of the map to 180 degrees at the top. Latitude in-creases from 0 degrees at the Equator to 90 degrees at the Pole. By convention positive numbers designate northern latitudes, negative numbers southern latitudes. In making calculations it is convenient to express longitude in terms of a full circle of 360 degrees, beginning at the bottom of the map and proceeding clockwise. Conversion from one system of designating longitude to the other is easy. Subtract longitudes greater than 180 degrees from 360. Label the difference longitude east. For example, longitude 210 degrees is equivalent to 360- 210 = 150°E.

"Next select and identify (by placing a dot on the graph paper) the location where observations will be made. In the accompanying example [Figure 1] I placed the dot at latitude 40°N and longitude 80°W. I then determined the radii of the concentric circles to be in scribed around this point for indicating the horizon and the elevation angles. To simplify the calculations I assume the earth to be spherical and smooth.

"The accompanying diagram [Figure 2] depicts the earth as it might appear in cross section if it were cut through the meridians 80° W and 100°E. Note that a straight line tangent to the earth's surface at 40°N defines the horizon and is intercepted by the orbit of the satellite at distances from the observer of 35.6 degrees of arc. The points of interception mark the limits beyond which the receiving station could not pick up line-of-sight signals from a weather satellite that orbits at a height (h) of 1,464 kilometers (910 miles) above the earth's surface. The diagram is based on the U.S. weather satellite NOAA-2, which derives its acronym from the National Oceanic and Atmospheric Administration.

"Observe that an oblique triangle can be drawn by connecting with straight lines the center of the earth, the observing station and the satellite. In this triangle the lengths of two sides are known. One side is the earth's radius, R. In the calculations that follow R is assigned the value of 6,378 kilometers. The other known side is equal to the sum of R + h, 7,842 kilometers. The largest angle of the triangle, the one that faces the large known side, is also known. This angle is the sum of the assumed angle of elevation and the right angle made by the vertical line through the station and the tangent to the earth's surface.

"With this information one can easily determine the size of the angle (B) between the known sides and therefore the length of the arc at the earth's surface that is subtended by the known sides. This arc comprises the radius of the circle of points that could lie directly under the satellite when the satellite is at the assumed angle of elevation above the horizon. For example, the triangle that has been drawn with the darkest lines in the diagram corresponds to an elevation angle of 20 degrees.

"The sequence of arithmetical operations required to determine the corresponding radius (B) in degrees of arc for a circle of subpoints at 20-degree elevation is specified by the first of the accompanying formulas [Figure 5]. Divide the earth's radius by the sum of the radius and the height of the satellite: 6,378 / (6,378 + 1,464) = .8133. From a table of trigonometric functions determine the cosine of each desired angle of elevation. In this example the desired elevation is 20 degrees. According to the table, cos 20° = .9397. The formula requires the product of [R / (R + h)] x cos E, which in this example is .8133 x .9397 =.7642.

"Next the formula requires the angle (arc cosine) that corresponds to this cosine. According to the trigonometric table, the cosine .7642 corresponds to the angle 40.2 degrees. Finally, the formula states that the desired radius (B) in degrees of arc is found by subtracting the angle E (20 degrees) from the arc cosine angle: B = 40.2 - 20 = 20.2°.

"To draw on the map a circle that represents points in all directions that are 20 degrees above the local horizon, place the point of a pen compass on the equator of the graph paper at 0 degrees longitude and open the instrument until the nib of the pen corresponds to latitude 20.2 degrees. With this distance as the radius, transfer the point of the compass to the location of the observing station on the map and draw the circle. Similarly determine the radii and inscribe circles at other angles of elevation as desired. In the case of NOAA-2 I have tabulated radii for elevation angles at intervals of 10 degrees from 0 to 90 degrees [see Figure 6]

"Place the clear plastic overlay over the map and pin it at the Pole so that the overlay can be rotated around the pin. The path of the satellite will be plotted on the overlay in terms of the object's changing latitude and longitude. These coordinates can be calculated for a track of any length in increments of, say, 10 arc degrees, beginning on the Equator at 0 degrees longitude. Each length will be the hypotenuse of a right triangle as drawn on the surface of a sphere [see below left]. The lengths of the corresponding legs of the triangle are measured in degrees of latitude and longitude.

"The fact that members of the solar system rotate in step somewhat like the meshed gears of a machine enables one to plot the course of an earth satellite with a surprisingly small amount of initial information. For example, all that is needed to plot the course of NOAA-2 is the fact that it travels 1,464 kilometers above the earth's surface in a sun-synchronous orbit. The plane of a sun-synchronous orbit turns one full revolution with respect to the sun in a year of 365.2422 days. The angle made between the orbital plane and the sun remains constant. Moreover, the ascending node (the point at which a northward-bound earth satellite crosses the plane of the celestial equator-not, as in the usual case, the plane of the ecliptic) occurs at the same local time on each orbit.

"Offhand one might suppose that the orbit of any earth satellite would remain stationary in relation to the fixed stars. It probably would if the universe were symmetrical. Actually the orbits of earth satellites are strongly influenced by the aspherical shape of the earth, which bulges slightly at the Equator. Satellites that orbit in planes making an angle other than 0 or 90 degrees with respect to the earth's Equator respond to the gravitational pull of the bulge much as a gyroscope behaves when it is supported at one end of its shaft. Instead of falling the gyroscope rotates around the point of its support. In the same way the gravitational force between the bulge of the earth and the satellite causes the orbit of the satellite to precess in relation to the fixed stars. This effect is observed in the wobbling motion of an inclined top. Indeed, the rate at which the orbital plane of an earth satellite rotates varies with both the altitude of orbit and the angle at which the plane of the orbit is inclined with respect to the Equator.

"To establish the sun-synchronous condition, weather satellites such as NOAA-2 are injected into orbit at an extremely precise angle of inclination, which is based on mathematical relations worked out by specialists in astrodynamics. If the height of the orbit is known, the sun-synchronous angle of inclination (i) can be calculated, as shown in Formulas 2 and 3 [Figure 5]. In the case of NOAA-2 the inclination of the orbit (i) is equal to the angle (arc cosine) that corresponds to the cosine of -4.7349 x 10^-15 X (R + h)^7/2. To find the 7/2 power of R + h for NOAA-2, multiply by itself seven times the square root of the sum of 6,378 + 1,464. The square root of the sum is 88.56, and 88.56⁷ is 4.2723 x 10^l3. The product of 4.2723 x 10¹³ x -4.7349 x 10^-l5 is -.2022.

"The angle that is equivalent to the negative cosine-.2022 is found by subtracting from 180 degrees the angle that corresponds to (positive) cosine .2022: arc cos .2022 = 78.33; arc cos -.2022 = 180° - 78.33° = 101.67° = i, which is the desired angle of inclination of the satellite NOAA-2. Formulas that determine the track (L) on- the surface of the earth over which satellites such as NOAA-2 orbit make use of the supplement (I) of the angle of inclination: 180° - 101.67° = I.

"Also required to calculate the track (L), as we have seen, is the period of time in minutes (P) necessary for the satellite to complete one orbit of the earth beginning and ending at the ascending node. The period (P), as indicated by Formula 4, is equal to 1.6586 x 10^-4 multiplied by the cube of the square root of R + h. For NOAA-2 the square root of R + h is 88.56. The cube of 88.56 is 6.9456 x 105. Therefore the nodal period of NOAA-2 (P) is 1.6586 x 10^-4 X 6.9456 x 10⁵ = 115.19 minutes.

"With this information one can calculate intervals of latitude () and longitude () that comprise the legs of the spherical

triangle that has as its hypotenuse the track (L). When I design a prediction device, I normally begin by assuming that the track starts on the Equator at 0 degrees longitude. I then compute the corresponding latitude and longitude as the satellite advances along its track in increments of 10 degrees of arc. A somewhat more accurate curve could be plotted by reducing the increments to five degrees of arc, which would double the number of computations

"Latitude () is determined with the aid of Formula 5. For example, when NOAA-2 has orbited through a distance of 10 degrees of arc from its assumed starting point above the Equator, it arrives at the latitude: arc sin (sin 78.33° X sin 10°) = .9793 X .1736 = .17 = 9.8° = . After NOAA-2 has advanced 90 arc degrees along its track its latitude has increased to arc sin (sin 78.33° x sin 90°) = .9793 x 1 = .9793 = 78.33°, its closest approach in latitude to the North Pole. When NOAA-2 completes half of its orbit (180 degrees), its latitude falls to arc sin (sin 78.33° x sin 180°) = .9793 x 0 = 0°. Stated differently, the satellite has arrived above the Equator on the opposite side of the earth. From 180 to 360 degrees the trigonometric sign of all angles is negative, indicating that these latitudes lie in the Southern Hemisphere.

"Each increment of longitude () constitutes the second leg of the associated spherical triangle. The increments are calculated with the aid of Formula 6. Longitude is generated by two motions: the westward advance of the satellite on its retrograde orbit and the slower motion of the observer as the earth's rotation carries the station eastward. The component of the longitudinal motion for which the satellite is responsible is described by the formula as an arc that corresponds in degrees (arc tangent) to a trigonometric tangent expressed as the product of cosine l multiplied by tangent L. To this arc, according to the formula, must be added the second component of motion: an arc equal in degrees to the product of P x L / 1,440.

"Assume that NOAA-2 has advanced 10 degrees along the orbit (along the hypotenuse of the spherical triangle). It has reached the latitude 9.8°N, as has been previously calculated. Simultaneously the satellite has moved westward along the inclined orbit: L = 10°. As has been previously determined, the cosine of I (78.33 degrees) is .2022. The trigonometric table lists the tangent of 10 degrees as .1763. Substitute these values in the formula: arc tan (.2022 x .1763) = arc tan .0356 = 2°. This is the arc through which NOAA-2 would have orbited in longitude if the earth were not rotating.

"The motion of the earth generated an interval of longitude: P X L /1,440 = 115.19 x 10° / 1,440 = .8°. The total increase in longitude is therefore A = 2 + .8° = 2.8°. Near the Equator, where the meridians are widely spaced, the rate at which the satellite moves westward is low in relation to its motion northward. Over the first 10 degrees of NOAA-2's orbit from the Equator its longitude increased by only 2.8 degrees, an average of only .28 degree of longitude per degree along its track.

"This rate increases dramatically as the satellite approaches the poles, where the meridians meet. When L is 90 degrees, the tangent of L is infinity. The product of cosine I times infinity is infinity. The arc tangent of infinity is 90 degrees. At this point along the track the component of longitude generated by the earth's rotation is 115.19° x 90 / 1,440 = 7.2°. Therefore = 90° + 7.2° = 97.2°. When L has increased to 100 degrees, the equation is tan L = 180°- 100° = tan 80°. According to the laws of trigonometry, tangents of angles that lie between 90 and 180 degrees are negative numbers. Hence tan 100° = -tan 80° = - 5.6713. Substitute the appropriate values in the formula, cosine I and tangent L: .2022 x-5.6713 = -1.1467. Therefore at this point along the track the component of arc for which the satellite is responsible is 180° + arc tan -1.1467 = 180° - 48.9° = 131.1°.

"To this longitude must be added the increment contributed by the earth's rotation: 115.19 x 100° / 1,440 = 8°. Hence after NOAA-2 has orbited 100 degrees from the Equator its longitude has increased to = 131.1° + 8° = 139.1°. During the 10 degrees of arc along the track from 90 to 100 degrees the longitude of the satellite shifted westward from 97.2 to 139.1 degrees, an advance of 41.9 degrees compared with the difference of only 2.8 degrees for the same length of track at the Equator!

"When you do the arithmetic, keep in mind that the tangents of angles that lie between 180 and 270 degrees are positive. The calculation of these angles is similar to that of angles between 0 and 90 degrees. The accompanying table lists the latitudes () and longitudes () that determine points at each 10 degrees of arc along the track of NOAA-2. Beginning on the Equator, plot the points on the plastic overlay. (The overlay can be anchored temporarily at the edge of the map with bits of adhesive tape.) Connect the points by a graph to represent the track. The graph can be drawn smoothly in india ink with a ruling pen and a draftsman's French curve.

"The track can be calibrated in minutes of time since the satellite crossed the ascending node. Intervals of two minutes are convenient. I prefer to plot the intervals in terms of latitude (_t) with the aid of Formula 7. For example, two minutes after NOAA-2 crosses the Equator its latitude, according to the formula, is arc sin [sin 78.33 X sin (360 x 2/ 115.19)] = arc sin .1066 = 6.1° = _t. If P is known in minutes, the formula can be simplified. Divide 360 by the known period. Thereafter multiply the intervals of time (t) by the resulting constant. For NOAA-2 the constant is 360/ 115.19 = 3.125. The formula becomes arc sin (sin I sin 3.125 t). The accompanying table [Figure 6] lists latitudes for calibrating the track of NOAA-2 during the first hour of its orbit from the ascending node.

"If two or more tick marks are inscribed on the overlay at the Equator, the device will serve for predicting the location of the satellite for as many as about 12S orbits. The marks are spaced at intervals of longitude equal to the satellite's apparent westward movement during each full orbit, as measured from consecutive ascending nodes. This distance, in degrees of arc, is about .25 times P. For NOAA-2 the tick marks are spaced at intervals of .25 X 115.19 = 28.8°. In general predictions of more than 24 hours are not useful because small errors build up rapidly.

"To use the instrument rotate the overlay to the point at which the zero time of the track coincides with the longitude of the satellite at its next ascending node near the longitude of the observer. This information is broadcast worldwide by radio station W1AW, as Eugene F. Ruperto explains in his description of an amateur weather-satellite station in this department [SCIENTIFIC AMERICAN, January].

"The fact that the orbit of the satellite can penetrate the observer's horizon as plotted on the map, or even some of the inner circles of elevation, does not necessarily ensure that radio signals from the satellite can be picked up. The device is designed on the unrealistic assumption that the earth is a smooth sphere. Local features of the terrain may interfere with the line-of-sight path. Experience at a given location will soon determine for the observer the bands of ascending node longitudes and the corresponding interval of time after the ascending node when the satellite is within receiving range of the station.

"Incidentally, lest readers misinterpret Ruperto's discussion of the information picked up and broadcast by NOAA-2, the scan to which he refers lies only across the earth at a right angle with respect to the satellite's track. The scans are analogous to the horizontal lines of a television picture. The signal in the infrared scan is the sum of all energy in the infrared band from 10.2 to 12.5 micrometers (not nanometers). The area resolved in the infrared is about four miles across when the scan is directly under the satellite. The visible channel integrates as a single signal all reflected solar energy between .5 and .7 micrometer, with a resolution of about two miles when the radiometer is looking straight down. At a distance of 500 miles on either side of the subpoint these resolutions are degraded by a factor of approximately 1.5.

"Amateurs should keep in mind the fact that although the visible channel picks up a true picture of cloud cover, the infrared channel broadcasts a temperature map of the earth's surface and of the highest clouds except thin cirrus clouds. Low temperatures appear as shades ranging from white to light gray, higher temperatures are in shades from dark gray to black.

"The tops of clouds at high and intermediate altitudes are cold. They appear white. For this reason the images of cloud cover made in infrared, even when they are picked up on the dark side of the earth, frequently resemble pictures made in visible light. Thin cirrus clouds and clouds with very low (warm) top may not be visible in infrared against the background of the earth's surface."

Bibliography

METEOROLOGICAL SATELLITES. W. K. Widger, Jr. Holt, Rinehart and Winston, Inc., 1965.

AN INTRODUCTION TO ASTRODYNAMICS. Robert M. Baker, Jr., and Maud W. Makemson. Academic Press, 1967.

The Society for Amateur Scientists (SAS) is a nonprofit research and educational organization dedicated to helping people enrich their lives by following their passion to take part in scientific adventures of all kinds.