Manual / Guide
Jeppesen E6B Flight Computer Manual (undated)
A later Jeppesen manual for the mechanical E6B.
Manual / Guide
A later Jeppesen manual for the mechanical E6B.
Slide Graphic Computer Manual JS314305B CAUTION: Plastic computers are designed for a long service life. Since all plastic materials are sensitive to extreme heat, do not store the computer in direct sun- light or in an area subject to high temperatures. ©1981 Jeppesen Sanderson, Inc. All Rights Reserved Published Simultaneously in Germany TABLE OF CONTENTS Page Calculator Side of Flight Computer................ 1 Time and Distance .. 06... 00.0.0. ccc cece eee ace 4 Finding Time... 1. eee eee eee eee eee 4 FRRCHIFIG DISEAGE, Rese: oa sed ex wee Bette racnant ees oy 5 Fitiding Sp6e@ tcc wus ns er anmmaraieas Qa BR 6 Short Time:and Distance): 2. 5 oa eis 6866 ess coe ese ene one ; Finding Short Time. ............... 000000 0s 8 Fuel Consumption ........... 0.000 cc cee eeeeeaee 9 Finding Time . 11.0... 0... ccc c eee cee eens 9 Piel Burnes GAA oe cwanesacs xn us oe Ha RR BE 10 Fuel Consumption: .(6.0:H.) assess ws ea os ou oy we 11 THCIATRACRG Siti scia on ca acme oa Wh a ER OS RE OS 12 Converting Mach Number to True Airspeed .......... 14 Changing Nautical Values to Statute Equivalents, Statute to Nautical, and Statute to Kilometers...... 14 Multi-Part Problems ............ 00. cece ee ue cues 16 Finding Drift Angle (Off-Course Problem) ........... 17 Time and Distance toa VOR Station............... 20 True Altitude Density Altitude.... Multiplication and Division. .................0005 26 Multiplication. 2... eee eee ee ee ee 26 DIVO tence ede Ule ere is eee oe comeesecmeems wa § 27 Converting Feet Per Nautical Mile to Fest ReriniinGtem ees citi cx we memoria oe ve | 28 Converting Fahrenheit to Celsius (centigrade) ........ 29 Wind Sideiot Computers ys va as ae Hee RES he oe | 31 Determining Ground Speed and True Heading........ 33 Finding Unknown Wind ................00ee cues 35 Finding Altitude for Most Favorable Winds.......... 38 Radiusrof AGHOM sls Barts ax waremamscan an on ew we td 40 MGIC COnNGISIONS icc ws ec oe os wormage ba Oe OEE 46 OEMS CONVESIONS ws se He HE ENS NERT SD dae exe ene eos eed 48 Appendix | — Answers to Practice Problems........ A-l Appendix Il — Glossary of Terms..............20. A5 CALCULATOR SIDE OF FLIGHT COMPUTER ye A] 4 og aurituor COMPUTATIONS eying ony wit 308 “B" Scale (time in min- utes, calibrated altitude, of Climb/Descent. calibrated airspeed). Unit Index for Multipli- cation, Division, and Rate © “C" Scale (time in hours Imperial Gallon Conver e and minutes). sion Arrows. Temperature Conversion Kilometer Conversion Seale: Arrows. , Pounds Conversion U.S. Gallon Conversion Arrow: Arrows. Seconds Arrow. Foot Conversion Arrow. NMeélers: Conversion Pressure Altitude Arrow. Window. Liters Conversion Kilogram Conversion Arrows: Arrow. Speed Index. Nautical Conversion Density Altitude Arrow. Window. Statute Conversion Air Temperature AriawS: Window. : Fuel/Pounds Conversion Arrow. QOOoqoooeu Gee © “A Scale (miles, miles per hour, gallons, gallons “ 3 per hour, true airspeed, Oil/Pounds Conversion true altitude). Arrow. Figure 1 ® © © ©© © ©B©O ©OO®O INTRODUCTION Congratulations, you have just purchased one of the finest computers on the market today. It is well designed to assist you in solving the planning and navigating problems associated with flying. It is simple to Operate and adequate instructions and formulas are printed on the computer so that the user need not worry about forgetting how to work the computer. FLIGHT COMPUTER The flight computer has two sides; the calculator side is used to work problems involving time, distance, speed, fuel consumption, true airspeed, nautical to statute conversions, off-course, altitude, and multiplication and division. The wind side is used to determine the effect of wind on the aircraft. Ground speed, true heading, and unknown wind are solved on this side of the com- puter, CALCULATOR SIDE OF THE FLIGHT COMPUTER The main portion of the calculator side of the flight computer consists of three separate scales. The outer scale (10,fig.1) is fixed to the computer. The second (11, fig.1) and third (12,fig.1) scales, inward, are printed on a disc that pivots in the center of the computer, thus permitting them to be rotated within the outer fixed scale. To simplify the explanation, the scales on the calculator side of the computer are referred to as “AN, “B", and “’C’’ scales. (See fig. 1.) The “A” scale is used to represent miles, gallons, or true airspeed. When used as miles, the scale provides the distance traveled or speed of the aircraft in miles per hour. Gallons on this scale can represent two values: fuel consumption in gallons per hour, or total quantity of fuel used by the aircraft. “T.A.S."" on the “A” scale is an abbreviation for true airspeed. The graduations on the ‘’B’ scale are used for time in minutes or for “I.A.S.’" (Indicated Air Speed) in either miles per hour or knots. The “’C’’ scale graduations re- present hours and minutes only. CHANGING VALUES In order to accurately read the scales on the calculator side of the computer, it is necessary that the chang- ing values of the graduations of these scales be under- stood. When using the “A”, “B", and “C” scales to solve prob- lems, common sense must be used to determine the value of the numbers. If a short distance is involved, “25” on the “A” scale might be read as “2.5 miles”. If a long distance is involved, zeros are added to the "25" to get the proper answer. For example, 25" might be “250” or "2500 miles”. First, let's examine the changing values of the “A” scale. (See fig. 2.) If the number “14” is used as 14, each gradu- ation between “14” and “15” is equal to .1. If these num- bers are used as 140 and 150, each graduation is equal to 1; if used as 1400 and 1500, they represent 10; and if used as 14000 and 15000, they are equal to 100. Between the numbers “15” and “16,” there are only five graduations as compared to the 10 graduations between “14” and “15 .(See fig. 2.) When these num- bers are used as 15 and 16 respectively, each gradu- ation is equal to .2; when used as 150 and 160, each unit represents 2; when visualized as 1500 and 1600, each unit is equal to 20; and when used as 15000 and 16000, each graduation equals 200. NOTE The graduations on the “B” scale are identical with those on the “A” scale. CHANGING VALUES ZS —VT %, ‘0 2:00 “ 239 (Scale ee, CHANGING VALUES 5 .2-—2-20-200 A—Saale é A é rs V en .) —— Oo, Figure 3 The changing values on the “C” scale are somewhat different than those on the “A” and “B” scales. (See fig. 3.) These graduations always represent minutes and are equal to five or ten minutes as shown by the arrows. For example: between 1:50 and 2:00, the gradu- ations are equal to five minutes; and between 2:00 and 2:30, the graduations represent ten minutes. The graduations on the “C” scale are very large in com- parison with those on the “B” scale. As a result, there are times when finer graduations are needed than are provided on the ‘’C”’ scale. When this happens, the smaller graduations on the “B” scale can be used to supplement the measurements on the “C” scale. For example: in figure 4, the graduations on the “C” scale are equal to ten minutes,and the graduations on the “B” scale, immediately above, are equal to two min- utes. Starting at the left, the values are 3:30 on the HOURS ON C—SCALE MINUTES ON B—SCALE 36 3:38 3:40 3 a a Es ‘ 2 minutes & n B x y ial ee % sa 330 Figure 2 Figure 4 ww swouvanewo? 403 anil! 15630 asav Figure 5 “C" scale; then, moving up to the “B” scale for 3:32, 3:34, 3:36, and 3:38; and then back to the “C” scale for 3:40. Notice that 3:40 on the “C” scale is equal to 220 minutes on the “B” scale. SPEED INDEX The speed index is a large triangular symbol on the “B” and “C” scales and is used as a reference in time and distance problems. (See fig. 5.) The speed index always represents 60 minutes or one hour. The graduations covered by this symbol are also used as.6, 60, 600, or 6000 on the “B” scale. TIME AND DISTANCE In time and distance problems, there are three items for which to solve: time, distance, and speed. Two of these items must be known to work the problem. FINDING TIME If an aircraft is flying at a speed of 120 miles per hour, how long will it take to fly 140 miles? The steps involved in solving this problem are as follows: (See fig. 6.) 1. Turn to the calculator side of the computer and rotate the computer disc until the speed index is located directly under “12” which represents 120 miles per hour. ia % Noll Figure 6 2. Move clockwise on the “A” scale to “14” which represents 140 miles in this problem. Look direct- ly below “14” and find the answer to the prob- lem. It takes 70 minutes on the “B” scale, or 1:10 on the “C” scale, to travel 140 miles at 120 miles per hour. PRACTICE PROBLEMS — FINDING TIME The following practice problems should be worked to gain experience in working time problems on the com- puter. Answers to these problems are given in the Ap- endix of this manual. 1. Speed — 180 m.p.h.; distance — 240 miles; how long will it take to make the trip? 2. Speed — 142 m.p.h.; distance — 370 miles; how long will it take to make the trip? 3. Speed — 110 m.p.h.; distance — 33 miles; what is the time? 4. Speed — 136 m:p.h.; distance — 86 miles; what is the time? FINDING DISTANCE If an aircraft flies at 110 m.p.h. for a two-hour period, how many miles will it fly? The steps involved in solving this problem are as follows: (See fig. 7.) 1. Place speed index under “11” which represents 110 miles per hour. 2. Move clockwise on the “C” scale to 2:00. Look directly above 2:00 and find “22” on the “A” scale. This represents 220 miles, the distance traveled, which is the answer to this problem. PRACTICE PROBLEMS — FINDING DISTANCE The answers to the following practice problems are given in the Appendix of this manual. 1. Speed — 100 m.p.h.; time — 1:30; what dis- tance will be flown? 2. Speed — 126 m.p.h.; time — 2:05; what is the distance traveled? 3. Speed — 175 m.p.h.; time — 4:00; what is the distance? 4. Speed — 133 m.p.h.; time — 3:32; what is the distance? FINDING SPEED If an aircraft flies 210 miles in 1:30, what is the speed? To solve this problem, proceed as follows: (See fig. 8.) 1. Position 1:30 on the “C” scale directly under 210 miles on the “A” scale. 2. Directly over the speed index is the answer, 140 miles per hour. Figure 8 PRACTICE PROBLEMS — FINDING SPEED The answers to the following practice problems are given in the Appendix of this manual. 1. Distance — 90 miles; time — 0:43; what is the speed? 2. Distance — 320 miles; time — 2:00; what is the speed? 3. Distance — 35 miles; time — 0:19; what is the speed? 4. Distance — 182 miles; time — 1:54; what is the speed? SHORT TIME AND DISTANCE A procedure called “short time and distance” is used on the computer to solve problems involving short dis- tances, such as 2.5 miles. In these problems, a very small amount of time is involved in checking speed. For the short time and distance procedure, "36" replaces the speed index. This “36” is equal to 3600 seconds in one hour. Figure 9 shows the “36” position on the rotating “B” scale of the computer. When using the short time and distance procedure, all minutes on the “B” scale represent seconds. For example, figure 9 shows that at 101 m.p.h., 40 sec- onds are required to travel 1.12 miles. Similarly, the "C" scale is changed from hours to minutes. Figure 9 shows that at 101 m.p.h., it will take 5 minutes to fly 8.4 miles. Figure 9 The speed of the aircraft is always placed on the “A” scale directly over the 3600 index on the “B” scale. Distance is always found on the “A” scale directly over time in seconds on the “B” scale, or over time in min- utes on the “C” scale. FINDING SHORT TIME Flying at 120 m.p.h., for a distance of 1.5 miles, how much time will it take? This problem is solved as follows: (See fig. 10.) 1. Rotate the “B” scale until the “36” is directly under 120 m.p.h., on the “A” scale. 2. Then locate the number “15” on the “A” scale. For this problem, this number is actually “1.5” rather than “15”. 3. Under “1.5”, find, on the “B” scale, that the time to fly 1.5 miles is 45 seconds. Figure 10 PRACTICE PROBLEMS — SHORT TIME The answers to the following practice problems are given in the Appendix of this manual. 1. Speed — 140 m.p.h.; time — 3 minutes; how far would an aircraft fly? 2. Distance — 2.5 miles; time — one minute and 30 seconds; how fast is the aircraft traveling? FUEL CONSUMPTION Fuel consumption problems are solved in the same man- ner as time and distance problems, except that gallons per hour and gallons are used in lieu of miles per hour and miles. FINDING TIME If an aircraft burns fuel at the rate of nine gallons per hour and has 45 gallons of useable fuel on board, how long can the aircraft fly? The steps involved in solving the problem are as follows: (See fig. 11.) 1. Rotate the computer disc until the speed index is directly under “90 ", which represents 9 gal- lons per hour in this problem. 2. Then, move clockwise on the “A” scale and locate “45 "", which represents the 45 gallons of useable fuel. Directly under this on the “C” scale is found 5:00. The answer to our prob- lem is 5 hours. FOR Tim: AN ie Me. lo OISTANCE Figure 11 PRACTICE FUEL CONSUMPTION PROBLEMS — PRACTICE PROBLEMS — FUEL BURNED FERRERS EI: The answer to the following problems are given in the Appendix of this manual. 1. Fuel consumption — 14 g.p.h.; time — 15 min- utes; how much fuel was burned? The answers to the following practice problems are given in the Appendix of this manual. 1. Fuel consumption — 12 g.p.h.; useable fuel \ Note: Don’t forget to use care in determining the — 30 gallons; how long can the aircraft stay decimal point. in the air? ; 2. Fuel consumption — 21 g.p.h.; time — 2:40; 2. Fuel consumption — 18 g.p.h.; fuel burned how much fuel was burned? — 68 gallons; what was the time? { 3. Fuel consumption — 12% g.p.h.; time — 1:35; 3. Fuel consumption — 11 g.p.h.; fuel burned how much fuel was burned? — 24 gallons; what was the time? 4. Fuel consumption — 11 g.p.h.; time — 0:28; how much fuel was burned? 4. Fuel consumption — 15 g.p.h.; useable fuel — 35 gallons; how long can the aircraft stay in the air? FUEL CONSUMPTION If an aircraft burns 80 gallons of fuel in 2:30, how many gallons is it burning per hour? FUEL BURNED : , The problem is solved as follows: (See fig. 13.) lf an aircraft burns 82 gallons per hour for a period of 1. Rotate the calculator disc until 2:30 on the 2:00, how many gallons of fuel were burned? “C" scale is directly under “80” on the “A” “ . scale. The problem is solved as follows: (See fig. 12.) 3 Rotate the ‘camplete ‘computer and Find! the answer on the “A” scale above the speed index. The answer to the problem is 32 gallons per hour. 1. The “80” and “90” represent “8” and “9” re- spectively for this problem. Halfway between these two numbers is 8¥ gallons. Place the speed index under this graduation. PRACTICE PROBLEMS — FUEL CONSUMPTION 2. Locate 2:00 on the “C” scale. Look directly over 2:00 and find the number of gallons burn- The answers to the following problems are given in ed on the “A” scale. The answer is 17 gallons. the Appendix of this manual. 1. Fuel burned — 7 gallons; time — 0:40; what \ is the fuel consumption? WHEE Figure 13 Figure 12 10 11 Figure 14 2. Fuel burned — 47 gallons; time — 2:10; what is the fuel consumption? 3. Fuel burned — 75 gallons; time — 3:15; what is the fuel consumption? 4. Fuel burned — 36 gallons; time — 4:11; what is the fuel consumption? TRUE AIRSPEED The true airspeed problem is solved on the calculator side of the computer as follows: 1. In the small window labeled “FOR TRUE AIR- SPEED AND DENSITY ALTITUDE”, place the altitude for the flight under the temperature for the flight on the scale labeled “AIR TEM- PERATURE ° C". (See fig. 14.) 2. Without moving any scales, locate the indicated airspeed on the “B” scale. The “B” scale is label- ed “C.A.S." (calibrated airspeed) which, for all practical purposes, is equivalent to “indicated airspeed”. (See fig. 15.) Figure 15 3. Find the true airspeed on the “A” scale imme- diately above the indicated airspeed value. For example: what is the T.A.S. under the following conditions: altitude, 10,000 feet; temperature, -10° C.: and 1.A.S., 130 m.p.h.? The problem is solved as follows: (See fig. 16.) 1. Rotate the computer disc until -10° C. is located directly over 10,000 feet. 2. Then refer to the “B” scale and locate “13” which represents 130 miles per hour indicated. air- speed for this problem. 3. Look directly over 130 m.p.h. on the “B” scale and find that the true airspeed is 150 miles per hour. This is the answer to this problem. PRACTICE PROBLEMS — FINDING T.A.S. The answers to the following problems are given in the Appendix of this manual. 1. Altitude — 5,000 feet; temperature — +15° C.: 1.A.S. — 125 m.p.h.; what is the T.A.S.? 2. Altitude — 8,000 feet; temperature — O° C.; 1.A.S. — 110 m.p.h.; what is the T.A.S.? 3. Altitude — 12,000 feet; temperature — minus 5° C.; |.A.S. — 149 m.p.h.; what is the T.A.S.? 4. Altitude — 3,000 feet; temperature — +15° C.: |.A.S. — 105 m.p.h.; what is the T.A.S.? 13 s OF FEEr™, Figure 17 CONVERTING MACH NUMBER TO TRUE AIRSPEED To convert mach number to true airspeed in knots, use the following procedure: 1. Align the outside air temperature, in degrees Celsius (centigrade), with the mach number index as shown in figure 17, item 1. 2. Read the true airspeed, in knots, on the “A” scale opposite the mach number on the “B” scale. Figure 17 points out several mach num- bers and the corresponding true airspeeds at an air temperature of +15° C. The readings. are: Mach Number True Airspeed 0.8 (item 2) 528 knots (item 5) 1.0 (item 3) 661 knots (item 6) 1.36 (item 4) 898 knots (item 7) CHANGING NAUTICAL VALUES TO STATUTE EQUIVALENTS, STATUTE TO NAUTICAL, AND STATUTE TO KILOMETERS Another use for the calculator side of the computer is for converting nautical miles to statute miles, or knots to miles per hour. This type of problem is made very simple by a small conversion scale consisting of two arrows labeled “NAUT” (nautical) and “STAT” (statute) 14 respectively. These arrows are located on the “A” scale and point toward the “B” scale. NOTE Knots (nautical miles per hour) are changed to statute miles per hour in exactly the same manner. For example: to change 20 nautical miles to statute miles, the problem is solved as follows: (See fig. 18.) 1. First, rotate the calculator disc until 20 nau- tical miles is lined up with the nautical arrow. 2. Then, read the equivalent value, in statute miles, directly under the statute arrow. The answer is 23 statute miles. The conversion scale can be used to change statute to nautical miles by placing the statute value under the statute arrow and reading the nautical equivalent under the nautical arrow. In addition, another statute index arrow is included on the “B” scale. It can be used to convert statute miles on the “B” scale to nautical miles or kilometers on the “A” scale. (See fig. 19 and 20.) Figure 19 To convert statute to nautical equivalents: (See fig. 19.) 1. Align statute index on the “B” scale directly under the nautical index on the “A” scale. 2. Directly over the statute value on “B” scale, read the equivalent nautical value on the “A” scale. For example, 90 statute miles is equal to 78.2 nautical miles. To convert statute to kilometer equivalents: (See fig. 20.) 1. Align statute index on the “B” scale directly under the kilometer index on the “A” scale. 2. Directly over the statute value on the “B" scale, read the equivalent kilometer value on the “A” scale. For example, 90 statute miles is equal to 144.5 kilometers. PRACTICE PROBLEMS — NAUTICAL TO STATUTE, STATUTE TO NAUTICAL, STATUTE TO KILOMETERS 1. 140 nautical miles is equal to statute miles. 2. 25 knots, wind velocity, is equal to m.p.h. 3. 77 statute miles is equal to nautical miles. 4. 17 knots is equal to. m.p.h. 5. 133 statute miles is equal to. kilometers. 6. 40 kilometers is equal to. statute miles. MULTI-PART PROBLEMS In practice, the pilot solves a series of interrelated prob- lems leading to a final solution, which may be consider- ed one multi-part problem. A good example of a multi- part problem is finding fuel consumption for an antici- pated flight. The following multi-part problems are provided to em- phasize the order of information required to determine the total fuel consumption. In each case, the ground speed is determined first, then the time en route, and finally, the fuel consumed. In each case, assume that 1.A.S. is equal to C.A.S. The answers to the problems are given in the Appendix of this manual. 16 1. Wind — Om.p.h.; ground speed — 120 m.p.h.; distance — 320 miles; fuel consumption — 9 g.p.h.; how much fuel will be burned? 2. Altitude — 7,500 feet; I|LA.S. — 105 m.p.h.; temperature — +15° C.; distance — 256 miles; fuel consumption — 11.5 g.p.h.; wind — O m.p.h.; how much fuel will be burned? 3. Altitude — 9,000 feet; |.A.S. — 115 m.p.h.; temperature — -10° C.; distance — 335 miles; fuel consumption — 8.5g.p.h.; wind — Om.p.h.; how much fuel will be burned? 4. Ground speed — 135 m.p.h.; wind — O m.p.h.; temperature — -20° C.; altitude — 9,000 feet; distance — 425 miles; fuel consumption — 12 g.p.h.; how much fuel will be burned? FINDING DRIFT ANGLE (OFF-COURSE PROBLEM) The computer can be used to find the drift angle when the aircraft drifts off course due to a wind shift, errone- ous wind information, or navigation error. This is often known as an off-course problem. Figure 21 shows a typical problem: 120 miles from departure and 18 miles off course to the right; what is the drift angle? 1. On the calculator side, set the number of miles the aircraft has flown, 120, on the “B” scale Destination OFF COURSE Figure 21 17 > Ce que directly under the number of miles off course, 18, on the “A” scale. (See fig. 22.) 2. Above the speed index is the drift angle, 9 degrees, (See fig. 22), which is also the number of degrees to change heading to the left in order to parallel the intended true course between Point A and Point B, as shown in figure 23. 3. To find the additional angle needed to further cor- rect the course from Point X to Point B, set the 170 miles to go on the ‘'B” scale, under 18, the number of miles off course, on the "A" scale (See fig. 24). (The number of “miles to go" is measured on the chart.) 4. Above the speed index is the additional angle needed to arrive at the destination (See fig. 24.) Use the nearest whole number, 6, as the answer. 5. Add the 9 degrees needed to parallel the intend- ed true course and the additional 6 degrees 18 Figure 23 Figure 24 for a total of 15 degrees heading correction to the left, necessary to fly from Point X to the destination, Point B. (See fig. 25.) PRACTICE PROBLEMS — OFF-COURSE Refer to the Appendix for answers to these problems. 1. GIVEN: a. Distance out from departure point — 110 stat. miles; b. Distance to left of intended true course — 13 stat. miles; c. Distance to destination from off course posi- tion — 200 stat. miles. PROBLEM: a. How many degrees should the aircraft be turned to parallel the intended course?. at NS? left >’ 77, \ to converge 19 b. How many total degrees should the aircraft be turned to converge on the destination? c. Which direction should the aircraft be turn- ed (right or left)? 2. GIVEN: a. Distance out from departure point — 150 stat. miles; b. Distance to right of intended true course — 20 stat. miles; c. Distance to destination from off course posi- tion — 140 stat. miles. PROBLEM: a. How many degrees should the aircraft be turned to parallel the intended true course? b. How many degrees should the aircraft be turned to converge on the destination? c. Which direction should the aircraft be turn- ed (right or left)? 3. GIVEN: a. Distance out from departure point — 90 stat. miles; b. Distance to left of intended true course — 6 stat. miles; c. Distance to destination from off course posi- tion — 180 stat. miles. PROBLEM: a. How many degrees must the aircraft be turned to parallel the intended true course? b. How many degrees must the aircraft be turned to converge on the destination? c. Which direction should the aircraft be turn- ed (right or left)? TIME AND DISTANCE TO A VOR STATION The time and/or distance from the airplane to a VOR station can be computed easily on the calculator side of the computer. These problems are based on the pilot 20 taking a time check to make a measured bearing change with a VOR radio. In this procedure, the pilot flies per- pendicular to the VOR radials involved with the time check. TIME TO A VOR STATION To find the time to a VOR station: 1. Set the time to make the bearing change, in minutes, on the “C” scale directly under the number of degrees of bearing change on the “A” scale. 2. Find the answer on the “B” scale directly under the unit index (10) on the “A” scale. For example: what is the time to the station when it takes 2 minutes 30 seconds to accomplish five degrees of bearing change? (See fig. 26.) This problem is solved as follows: 1. Place the time (two minutes and thirty seconds --- 2:30) on the “C” scale directly under the degrees of bearing change (5) on the “A” scale. 2. Read the time to the station on the “B” scale directly under the unit index (10) located on the “A” scale. The answer is 30 minutes to the station. DISTANCE TO A VOR STATION To find distance to a VOR station, ground speed must also be known in addition to the time to make a bear- ing change. The solution to this problem involves the following steps: 1. First, solve for time to the station. FLIGHT COMPUTER 2. Then, set the speed index under the ground speed. 3. Next, locate the time to the station, in minutes, on the “B” scale. Look directly over this value and find the distance to the station. For example: Using the time to the station found in the previous problem (ref. fig. 26) and a ground speed of 142 m.p.h., what is the distance to the station? The solution is determined as follows: (See fig. 27.) 1. Place speed index under ground speed of 142 m.p.h. on the “A” scale. 2. Locate the time to the station, 30 minutes on the “B” scale. 3. Read distance to station on the “A” scale direct- ly above 30 minutes on the “B” scale. The answer is 71 miles to the station. If the time to the station is less than 10 minutes, the problem is worked in the same way, except that the short time and distance method is used. The short time speed index “36” is used in lieu of the triangular speed index,and the time in minutes and seconds are placed on the “C” scale. For example: Time to station — 4 min- utes, 30 seconds; ground speed — 127 m.p.h.; what is the distance to the station? (See fig. 28.) 1. Set “36” under 127 m.p.h. 2. Locate 4:30" on “C” scale and read distance directly above on the “A” scale, 9.5 miles. FOR TIME Fe AND DISTANCE CONSUMHo SPEED ___oy,, ear a Figure 27 22 KM US. GAL 13 Figure 28 PRACTICE PROBLEMS — TIME AND DISTANCE TO A VOR STATION Refer to the Appendix for answers to these problems. Degrees of Time Aircraft Time Distance Bearing Between Ground To To Change Bearings Speed Station Station 1. 5° 3 min. 110 mph 2.. 10° 5 min. 95imph 3; 5° 20 sec. IS5mMpAi a 4. 15° 9 min. TOP es og 5. 20° 1min. 50sec. 140 mph TRUE ALTITUDE True altitude is obtained by using the window labeled “FOR ALTITUDE COMPUTATIONS” which incorpo- rates temperature and pressure altitude scales. Pressure altitude is indicated on the altimeter when the baromet- ric scale is set to the atmospheric standard pressure of 29.92 inches of mercury. True altitude problems can be solved by following these steps: If an aircraft is flying at 12,500 feet with an outside air temperature of -20°C. and the altimeter set on 30.42 inches of mercury, what is the true altitude? This problem is solved as follows: 1. Referring to the altimeter, note the altimeter barometric reading of 30.42 and record it ona work sheet. 2. Now, rotate the barometric scale on the alti- meter to 29.92. Note that the pressure altitude is 12,000 feet. 23 x %, pe fe: Figure 29 3. Next, refer to the altitude computation window and place -20°C. on the temperature scale directly over the pressure altitude of 12,000 feet. (See fig. 29.) 4. Move to the “B” scale of the computer and locate the original indicated or calibrated alti- tude, 12,500 feet. Look directly over this value and find the true altitude is 12,000 feet. (See fig. 30.) This is the answer to the problem. The aircraft is actually flying 500 feet lower than the altitude originally indicated on the altimeter. PRACTICE PROBLEMS — TRUE ALTITUDE Refer to the Appendix for answers to these problems: 1. Pressure altitude, 7,000 feet; indicated altitude, 6,500 feet; temperature, -10°C. What is the true altitude? 2. Pressure altitude, 9,000 feet; indicated altitude, 10,000 feet; temperature -20°C. What is the true altitude? 3. Pressure altitude, 6,000 feet; indicated altitude, 5,500 feet; temperature -10°C. What is the true altitude? NOTE Figure answers to closest 50-foot increment. DENSITY ALTITUDE Density altitude problems are solved on the calculator side of the computer through the use of the window labeled ‘FOR TRUE AIRSPEED AND DENSITY ALT.” The procedure for solving this type of problem is shown in the following example: Flying at 10,000 feet pressure altitude with outside air temperature at -20°C., what is the density altitude? This problem is solved as follows: 1. Place -20°C directly over 10,000 feet pressure altitude in the airspeed and density altitude window. (See fig. 31.) 2. Refer to the density altitude window and find that the density altitude is 8,000 feet. PRACTICE PROBLEMS — DENSITY ALTITUDE Refer to the Appendix for answers to these problems. 1. Pressure altitude, 10,000 feet; temperature, -20°C.; what is the density altitude? 2. Pressure altitude, 15,000 feet; temperature, -30°C.; what is the density altitude? 3. Pressure altitude, 4,000 feet; temperature, -25°C.; what is the density altitude? 9 x ar eS e ‘ ROR POG Figure 31 25 MULTIPLICATION AND DIVISION The computer can also be used for multiplication and division. The index for these problems is the “10” in the small box. (1, fig. 1.) To Multiply: 1. Rotate the “B” scale until the index "10" is direct- ly under the number to be multiplied. 2. Without moving the scales, locate the multi- plier on the “B” scale. 3. Look directly over the multiplier and find the answer on the “A” scale. To Divide: 1. Locate the number to be divided on the “A” scale. 2. Rotate the “B” scale until the divisor is located directly under the number to be divided. 3. Without moving the scales, find the answer onthe “A” scale directly over the index “10 ". MULTIPLICATION Climbing at 450 feet per minute for 8 minutes, how much altitude would be gained? This problem is solved by multiplying 450 f.p.m. x 8 minutes. (See fig. 32.) 1. Place 450 on the “A” scale directly over “10” in the box on the “B” scale. 2. Find 8 minutes on the “B” scale and, looking directly over this, find that we would climb 3,600 feet in the 8 minutes. > DIVISION An aircraft has to lose 8,000 feet in 19 minutes. What is the rate-of-descent? (See fig. 33.) 1. Place 8,000 feet on the “A” scale directly over 19 minutes on the “B” scale. 2. Look directly over the “10” and find that the aircraft should descend at 420 f.p.m. to lose 8,000 feet in 19 minutes. PRACTICE PROBLEMS — MULTIPLICATION AND DIVISION Refer to the Appendix for answers to these problems: 1. Rate-of-climb, 450 f.p.m.; time in climb, 18 minutes; what is the altitude gained? 2. Rate-of-descent, 600 f.p.m.; time in descent, 4% minutes; what is the altitude lost? 3. Altitude to lose, 6,500 feet; time to lose alti- tude, 9 minutes; what is the rate-of-descent? 4. Altitude to gain, 9,000 feet; time to gain alti- tude, 21 minutes; what is the rate-of-climb? 5. Aircraft cargo arm, 55 inches; cargo weight, 226 pounds; what is the approximate cargo moment? 6. Total aircraft moment, 162,000 pound-inches; total aircraft weight, 3,000 pounds; what is the center-of-gravity arm? Ssunt AUTIN PaSS NDS OF FE Figure 33 27 CONVERTING FEET PER NAUTICAL MILE TO FEET PER MINUTE Certain IFR departure procedures require a minimum climb rate to assure proper obstruction clearance. How- ever, the minimum climb requirement is stated in terms of feet to be gained per nautical mile. The pilot can easily convert this “feet per nautical mile” figure to a “feet per minute” figure on the calculator side of the com- puter. The following example outlines the procedure: (See fig. 34.) 1. Set the speed index at the appropriate ground speed in knots. (Figure 34 shows the speed index set at a ground speed of 120 knots.) 2. All figures on the “B” scale will represent the minimum climb requirement in feet per nau- tical mile. All figures on the “A” scale will repre- sent the equivalent vertical velocity in feet per minute and are read directly over the minimum climb requirement figures. For example: Ground speed — 120 knots; climb require- ment — 250 feet per nautical mile; what is the rate of climb in feet per minute? To solve this problem: (See fig. 34.) 1. First set the speed index at 120 knots. 2. Then, directly above the 250 on the “B” scale, find the answer, 500 feet per minute, on the “A” scale. Pr ry 354 1] FOR TIME AND DISTANCE SPEED FOR Fugt CONSUMPTION G.P.H. Dis la Min; Figure 34 Fue si MIN, == NOTE At the same ground speed, the climb require- ment of 350 feet per nautical mile is equivalent to a vertical speed of 700 feet per minute. PRACTICE PROBLEMS — CONVERTING FEET PER NAUTICAL MILE TO FEET PER MINUTE Answers to these problems are found in the Appendix of this manual. Climb Requirement in Feet/Nautical Required Minimum Ground Speed Mile Vertical Speed 1. 100 knots 240 f.p.m. 2. 80 knots 300 f.p.m. 3. 140 knots 300 f.p.m. 4. 105 knots 400 f.p.m. CONVERTING FAHRENHEIT TO CELSIUS (CENTIGRADE) In some instances, the pilot may know the temperature in degrees Fahrenheit but not know its equivalent in the Celsius scale. Since the computer uses Celsius to obtain T.A.S., it is important that the pilot have a means of converting Fahrenheit to Celsius. The flight computer described in this manual incor- porates a temperature conversion scale on the calculator side of the computer. Temperature conversions can be read directly from this scale. For example: Figure 35 shows how 50° F. is converted to 10° C. Figure 35 29 eo lOROROKC) oe SLIDING GRID INSTRUCTIONS FOR FINDING GROUND SPEED AND TRUE HEADING INSTRUCTIONS FOR DETERMINING MAGNETIC COURSE, MAGNETIC HEADING, AND COMPASS HEADING. TRUE INDEX. WIND CORRECTION ANGLE SCALE. CENTER. AZIMUTH. Figure 36 30 WIND SIDE OF COMPUTER The wind side of the computer consists of a rotating azimuth and a rectangular grid that slides up and down through the azimuth. (See fig. 36.) SLIDING GRID The sliding grid is nothing more than a section taken from a large graduated circle. (See fig. 37.) The lines, projecting from the center of the grid and radiating outward, represent degrees right or left of the center line. The lines that form the arcs around the center of the circle represent distance from the center and are labeled in miles. The computer grid has two sides, a high speed side and a low speed side. (See fig. 38.) Since most private aircraft operate in the speed range below 250 m.p.h., the low speed side is generally used because of its smaller graduations and greater accuracy. The low speed side should be used for all wind problems given in this course. AZIMUTH The azimuth circle rotates freely and is graduated into 360°. The transparent portion is frosted so that it can be written on with a pencil. The center, a small circle, lies directly over the centerline of the sliding grid. ww F * F wb SLIDING GRID i 180 10 distance from center He XX ‘degrees leftm— su —amtlegrees sight : 31 SECTIONAL AND SECTIONAL AND -—WAC SCALES WAC SCALES (STATUTE) (NAUTICAL) i ~ [—{ Horo s anor Tren LAAs LAMAAALAE ss : i + T as — + sae + ct YMA MA UAAAMAANA AALALARALS LAKAALLAD 11 20) FU EVENUONETS CYUURUO NY CUCU OVEN ED CUOTEN OED CUOENNOUED CUNENORUNY CERRO lm NS HIGH SPEED SIDE LOW SPEED SIDE Figure 38 Figure 39 82 DETERMINING GROUND SPEED AND TRUE HEADING To maintain a specific true course, the pilot must deter- mine the wind correction angle and adjust the aircraft heading accordingly. The time en route, an important factor in any flight, is influenced by wind velocity be- cause it affects ground speed. To determine the total affect of wind on a flight, the true course, true airspeed, and wind velocity must be known. The operation of the wind side of the computer is des- cribed in the determination of true heading and ground speed in the following problem: GIVEN: True course —030° True airspeed— 170 m.p.h. Wind —080° at 20 m.p.h. DETERMINE: True Heading Ground Speed The solution is as follows: 1. Rotate the azimuth until the wind direction of 080° is located directly under the true index. (See fig. 40.) 2. Next, slide the grid through the computer un- til the center is on any one of the heavy lines extending from right to left on the grid. For this problem, the 160-mile grid line has been arbitrarily chosen. (See fig. 40.) ‘ AAT LLy ae cule LT) = \\ 70 80 1d ENE Figure 40 33 : ,] riot O 0 SUR OL 1 6lo-H A" 30 30 oe iin, T Oo wie L < So Ht ' aN seb. __S Figure 41 3. Then, place the wind velocity on the azimuth by moving up from the center 20 miles and placing a pencil mark on the grid centerline. (See fig. 40.) 4. Now, place the true course of 30° under the true index. (See fig. 41.) 5. With the true course under the true index, slide the grid through the computer until the pen- cil mark, or wind dot, is positioned on the true airspeed line of 170 miles per hour. (See fig. 41.) 6. Find the wind correction angle by checking the number of degrees to the right or left be- tween the grid centerline and the wind dot. In this problem, the wind correction angle is 5° to the right. (See fig. 42.) 7. To find true heading, the scales on either side of the true index are used. Since the wind correc- tion angle is 5° right, start at the true index symbol, count five units to the right, and direct- ly below this graduation, find the true head- Yn Er CT t Ms q ff se Ry A NE - WHE 34 Figure 42 ing of 35° on the azimuth scale. (See fig. 42.) If wind dot is to left, count left for true head- ing; if wind dot is to right, count right. NOTE Another method is: If the wind correction angle is to the right, add the wind correction angle to the true course value; if the wind correction angle is to the left, subtract it from the true course. . Without moving the computer setting, read the ground speed under the center. The ground speed for this problem is 156 miles per hour. (See fig. 42.) PRACTICE PROBLEMS — TRUE HEADING AND GROUND SPEED For answers, refer to the Appendix of this manual. 1. True course — 310°, T.A.S. — 120 m.p.h.; wind — 180° at 16 m.p.h.; what is the true heading and ground speed? 2. True course — 1789, T.A.S. — 135 m.p.h.; wind — 045° at 23 m.p.h.; what is the true heading and ground speed? 3. True course — 050°, T.A.S. — 155 m.p.h.; wind — 165° at 18 knots; what is the true heading and ground speed? 4. True course — 270°, T.A.S. — 130 knots; wind — 344° at 18 knots; what is the true heading and ground speed? 5. True course — 095°, 1.A.S. — 111 m.p.h.; tem- perature — +25° C.;