Manual / Guide · 1966
Jeppesen E6B Flight Computer Manual (1966)
Jeppesen's manual for the mechanical E6B: slide-rule side, wind side and worked problems.
Manual / Guide · 1966
Jeppesen's manual for the mechanical E6B: slide-rule side, wind side and worked problems.
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VARIATION WIND a ra 1, le SCALE DIRECTION ||@|<< ies 0 10; 2 LINE 4,0 ° es . ‘ fi x = (CENTER 2 So x: BT pal Fy ats) j SIRCLE: 3 - e e- 14 a “ Fett} LS) (2) 1 NAVIGATION f° > Hit 2 FORMULAS Q : A WIND y yy ae BS CIRCLES % il 7 ee 0 ie ose OE Ofc o C ove 08 er ge * % = as Hany os ah i - aN mre 2 ass 2 # Sevan mH= it -N® < Xo TEMPERATURE 20 Higy te onc 19 20. alt CONVERSION Ee oH AS : SCALE s z a NAVIGATION — WAC SCAL 7 i : a FORMULAS — = ART A-—-CALCULATOR SIDE Bieter rere Distance eo, 2 BPE UE sOpOUON a hk Se ae 3 or Er ee 6 MEMOIR Tose tr reson kyo yn)! 7 (a 8 SRN ac 10 Meemevep tite sr 3 Se i SS eT 11 Rep re ter ie a 12 Smeaie Speedie 2 RS ew ae 13 Bemesecn Number os. ee a 15 _ True Air Speed Computation at Higher Speeds... .. 18 3 A SOE 2a Sree erence Se aaa 20 _ Time and Distance to Station.................... 22 Drift and Off— Course Correction. ib ee 24 g _ PART B—-WIND SIDE _ Wind Solution on the Bg i ee 27 Flight Planning with Berecast Winds! ene 82 35 E- True Heading & Ground Speed ae eS 36 True Heading and Prue Air: Speeds Sy. eyes SSS 38 True Course (Track) & Ground Speed ati teas: 02) 41 _ Finding Winds in era, ue 45 Hints on the use of Pie WindsSide: oye wee tt 48 _ Additional Uses of the Jeppesen E6B 3— Way Slide ...50 PART C—-ANSWERS, DEFINITIONS AND TIPS Sine Oe Notion ei einei siieny Yo. or tweiec co 51 Seer Ue een cMter PE tien) Ueda ss Ie os ete a ee 53 PPmeseol- the EGB <0 fs: sks. oe 56 eppesen & Co, Jeppesen & Co., G.m.b.H. East 40th Avenue Frankfurt Am Main r, Colorado 80207 Kaiserstrasse 77 : Germany Part A CALCULATOR SIDE DISTANCE ” vU m m 2 TIME Time, speed and distance problems are solved with the E6B Computer in the conventional manner. . .using the outside scales on the calculator side. For the benefit of those ‘knot so inclined,’ the E6B is ‘knot knecessarily knautical' and you can get perfectly good answersin MPH and statute. First a word about reading the scales on the E6B. Each figure on the outer scales of the computer can stand for any number containing the given digits. The point marked '40' can stand for .4, 4, 40, 400, ete. You must determine, from the given problem, which value is correct. SIMPLE PROPORTION The slide rule face ofthe E6B computer is so constructed that any relationship between two numbers, one on the stationary scale and one on the movable scale, will hold true for all other numbers on thetwo scales. If the two 10's are placed opposite each other, all other numbers will be identical around the whole circle. If 20 onthe inner scale is placed opposite 70 on the outer scale, all numbers on the inner scale will be double those on the outer scale. If 12 on the outer scale is placed opposite 16 on the inner scale, all numbers will be in a three-to-four relationship. (See Fig. 1, below.) This feature of the slide rule permits you to supply the fourth term of any mathematical problem in proportion. The simplest and most useful application of the feature of the computer is in the solution of time-speed-distance problems. = as =n ALL NUMBERS IN s: i 3:4 RELATIONSHIP s= ate EN Ey ZP 2 TIME, SPEED, AND DISTANCE A plane has traveled 24 miles in 8 minutes. How many minutes will be required to travel 150 miles? The answer is the missing number in a simple proportion: 24:8::150: ? or 24 — 150 8 ? In the latter form, the proportion may be set on the com- puter exactly as it dppears in Fig. 2, with 24 on the outer MILES scale opposite 8 on the inner MINUTES scale; one then reads the number opposite 150 miles, and finds it to be 50 minutes. This is the basis of all time-speed-dis. tance computations on the computer. In most of these problems the arrowhead "speed index" is used for setting 9, reading the rate. In addition to the minutes scale, an inne, HOURS scale is used for reading times in excess of ono hour. 5,00 HOURS 6:00 p// » "Ai 2. Distance to fly, 150 miles. 2. w ~ 4y a h %, (e) -) "Ge. — ANSWER: Time to fly,}.\ 50 minutes. To solve for distance when speed and time are given: Example Given: Speed, 174 mph Time, 40 minutes Find: Distance Fig. 3 Dela ree To solve for time when speed and distance are known: Example Given: Speed, 113 knots Distance, 180 nm (nautical miles ) Find: Time ap ay wn scale read on inner Scales, 96 (min.) or 1 hour and ©, \ od8espeno 36 minutes. - 3 As “yo eantse: aay BO 40S .MPMUNIP ~ Fig. 4 ee To solve for speed when time and distance are known: Example Given: Distance, 266 miles Time, 1:25 Find: Speed \arweew ave > ‘83AN30 “05 y 301? Mi oo 1, Set the time, 1:25 x opposite 266 (miles) SS ale. 9 uf \y oe PROBLEMS 1 Find the missing quantities; check your answers agains; the answers in Part C. Ground Time Speed Distance 1 see 145 128 i ui, bi f 150 Wee 3. 164 196 4. 0:33 Bisse 78 5. 0:36 139 Ree 6. 0:18 arse 47 FUEL CONSUMPTION Problems in fuel consumption are solved in exactly the same manner as problems in speed, except that the rate is expressed as gallons per hour (gph) instead of miles per hour. Gallons are therefore treated as miles, and read on the outer scale. The arrow is used as a rate index, in the usual manner. To find the fuel consumed per hour (rate of consump- tion) when the total fuel consumed and the flying time are known: Example Given: Fuel consumed, 135 gals. Time in flight, 1:28 Find: Rate of fuel consumption Tr os <9 coe’ E6 ie Cc On, 3 <e® 90 MILES I U> @ S 1:10 2 s oa ; % R ov “<0 2 AS} ° “i 2. Opposite index, read rate 1. Set the time 1:28 (88 min.) of consumption, (92 GPH). opposite fuel consumed ; =o 135 gal. Dt RADE) oP Rs ee —s foi 2 ae Pe %, sth \ a~ % = eeln\ = | = ei s3/5/ © NF. é ev § : RY F /. CONVERSIONS "Things aren't always what you want them to be— But the E6B will help change them." For instance, if you want to change: _ Statute miles to nautical miles Statute miles to kilometers Minutes to seconds Fahrenheit to centigrade Or vice versa — Here's how: Note the following labeled arrows on the inside and outside scales of the calculator side of the computer: (Refer to Fig. 7) NAUTICAL miles............ near 66 on outer scale STATUTE miles,.......0000-. near 76 on outer scale KM. (kilometers).............. near 12 on outer scale BOON DS tiie ick en caeve's at 36 on inside scale STATUTE TO NAUTICAL TO KILOMETERS The arithmetical factor for converting nautical Miles to statute miles is 1.15; for converting statute miles to nautica) miles, .87. It will be found that 76 (the STAT. mark ) divided by 66 (the NAUT. mark) gives 1.15, and that 66 divided by 76 gives .87. In the same way, the conversion factor for statute miles to kilometers is .62, and 76 (the STAT. mark) divided by 122 (the Km. mark) gives the same conversion factor. The conversion on the computer is thus the familiar simple proportion: 66 ee 76 Nautical miles Statute miles ! To convert statute miles to nautical miles: Example Given: 136 miles | Find: Number of nautical miles | Number of kilometers CAL ALT. 12. Opposite the naut. arrow, e read distance in nautical : miles, 118 NM. a e ( m p a NS ras 0 ANSWER: 118 NM 1. Set 136 (statute miles) (-\< posite the stat. trow. Sore Als JON read distance in kilometers (218 Km). Y 9.09 A "J ANSWER: 218 Km | & z “e" { innndy Fig. 7 Conversions of any of the other units are made in the same way. The known figure is placed under the arrow for the unitin which that figure was expressed and the unknown quantity is read under the appropriate arrow. q 8 PROBLEMS 2 nd missing quantities; check your answers against the swers in Part C. Statute Nautical Miles Miles Kilometers 1: 240 2. 35 3. 15 4, 160 5. 47 6. 8 MINUTES TO SECONDS as t 36 on the inside scale is an arrow marked SEC. To nvert minutes to seconds, place the time index Aopposite e number of minutes and read seconds opposite SEC Example ae Find number of seconds in 13'/2 minutes. Place time index J opposite 13'2 . Opposite SEC arrow (near 360n inside scale) read 81. e of the E6B computer (see Fig. 8). Read tempera- nversions directly from this scale. ALTITUDE “Altitude comes in assorted varieties. Ever wonder how high is ‘up'?” No need for confusion if you remember the following points: Indicated Altitude is the altitude reading on the altimeter when it is set to nearest station altimeter setting. _ Calibrated Altitude is indicated altitude corrected for instrument and position error. Pressure Altitude is the reading on the altimeter when it is set at 29.92. Density Altitude is a measure of the weight or density 2. the air. Aircraft performance, such as take-off run, rate 0 climb and engine output, are affected by density altitude ‘True Altitude is calibrated altitude corrected for tem- perature and pressure. DENSITY ALTITUDE - Near the center of the computer is the density altitude window, entitled "For Air Speed and Density Altitude Computations". Example S Given: Pressure altitude.......... 3000' ‘B Air temperature........... 25°C Find: Density altitude \\ For 2. Spoesite density altitude oon read density altitude, | 1 Tevselaanal PROBLEMS 3 | Find density altitudes; check your answers against the ) ~—s answers in Part C. Pressure Air Density Altitude Temperature Altitude 1. 1500' 35°C 40° —10° 11 TRUE ALTITUDE e altitude it is necessary to use cali}, ate ae teesindicated if calibrated is not available), pe sure altitude and true air Rae oie cng Greater acount can be obtained if you also know t Ag titude of the grou,” station giving your altimeter setting. Example +4 . BTILULES PETERS Oe veo scceccecccs 10 000' ven: Pressure altitude ; =) Calibrated altitude.................00., 9,000: True air temperature................. ~20°C Ground station ihn as (io 5,000" Find: True altitude is unknown, f ai inside scale and oppo true alt. (8500'). pobiceele ANSWER: Add tru above ground, 3780' to station altitude 5000' to get , 4 8780' true altitude above @ % 5% sea level. 2. Subtract station alt. from cal. alt. to get cal. alt. above ground: 9000'-5000' = 4000’. e altitude | SS KE 1.Set press.alt.(10,000' opposite true air temp. (-20°C) in window for true alt. calculation. 3. Opposite cal. alt. above ground (4000') on inside scale read true alt. above ground (3780') on outside PROBLEMS 4 Find true altitude; check your answers against the answers in Part C. Pressure True Air Calibrated Station True Altitude Temp. Altitude Altitude Altitude 4° 20,000! 25°C 11,400'. 4,200' 2. 5,000' 0° 6,000' Sea Level 3. 7,000' 10° 7,400' 1,900' 4. 20,000' ~1 5° 21,000' Unknown 42 = TRUE AIR SPEED "Let's knock over this true air speed thing.” “In the old days pilots listened to the wind in the wires and were happy to be flying at any speed. Today we have the air speed indicator. It's a mighty fine gadget, but its reading is affected by various items such as tem perature, pressure, compressibility and accidental misreading by the pilot who's thinking of the blonde in the tower." The air speed indicator is constructed to operate correctly under standard atmospheric conditions, at a pressure of _ 29.92 inches of mercury and a temperature of 15°C. In ) flight, these conditions willrarely be found.The pilot will there- _ fore have to correct the readings of his air speed indicator _ for pressure and temperature in order to discover his true air speed. The slide rule face of the E6B computer has a Special scale which will enable him to make the necessary corrections; it is labeled FOR AIRSPEED AND DENSITY ALTITUDE COMPUTATIONS. The most accurate use of is scale requires the setting of pressure altitude (set 29.92 in. Hg. into the altimeter correction window—the hands of e altimeter will then read "pressure altitude"). Calibrated airspeed (the indicated airspeed corrected to mpensate for error in the airspeed indicator or the pitot static system) will be used throughout this discussion. While calibrated airspeed is the technically correct airspeed use for true airspeed computations, many light airplane use indicated airspeed for solving true airspeed :putations since the difference between calibrated indicat- Q eed is relatively insignificant for light aircraft. 13 i i ined i tting. Usin True air speed is obtained in one se Batis. labeled FOR AIR SPEED AND DENSITY ALTITYDS COMPUTATIONS, take the following steps: R Example i i ture,— 10°C ven: Outside Air Tempera ; “ Pressure altitude, 20,000 ft. Calibrated air speed, 200 mph Find: True air speed 1.Set air temp. (-4 window opposite pre.” 20,000). Pressure <1 TEMPERATURE °¢ Te ”» \3 ous: eo (This solution disregards Compressibility, and Temper- ature rise. They are relatively insignificant at lower altitudes and air speeds.) Two mistakes are frequently made: One is using the wrong "window" and the other is forgetting that the tem- perature scale is reversed. The plus temperatures are to the left. At times, especially in case of controlled ground speed, the true air speed will be known and the pilot will want to know at what calibrated air speed to fly in order to a good the given true air speed. This solution on the Ria computer is the same as the preceding one, except! ie after outside air temperature and pressure altitude a s been set in the proper relationship, the calibrated alr pe is read on the minutes scale opposite the known true speed on the miles scale. 14 Example Outside air temperature, 0°C Pressure altitude, 5,000 ft. True air speed, 200 mph Calibrated air speed 1. Set air temp. (0°C) in window opposite pressur altitude (5000'). 2. Opposite true air speed (200 MPH) read desired calibrated air speed (188 MPH). ; = nd ANSWER: 188 MPH}e-\ < Foe, e PROBLEMS 5 Find true air speed; check your answers against the answers in Part C. Pressure Outside Air Calibrated True Altitude Temp. Air Speed Air Speed 1) #2 3,000' 0°C 140 mph 2.) 8,000' =6* 160 knots & 35-7 12,000' -10° 190 mph _ 4. ~~ 10,000° 0° 200 mph eB 6. 5,000" 3¢ 140 knots ar, 6. % 9,000' —10° 150 knots MACH NUMBER Every airplane, even a 60 H.P. "cub" flies at a Mach Number. What's yours? OK here goes! Since Mach Num- ber is dependent upon the speed of sound, which varies _ only with temperature, the same Mach Number represents _ different true air speeds at different temperatures. 15 MACH NUMBER FROM TRUE AIR SPEED AND TEMPERATURE In PROBLEMS 5 problem 6 (Pg. 15), we found Our true air speed to be 170 knots. The Mach Number inde, ( | ) appears to the right of the Figure 80 on the Pressy,, Altitude Scale in the Pressure Altitude Window. NOTE: The Mach Number Index ( { ) is positioned to relate Mach Numbers to True Air Speed in Knots only MPH must be converted. Example Given: True air speed, 170k Outside air temperature, — 10°C Find: Mach Number liisitiradann § SLES (OR MASP 4 1, Set Mach No. index oppo- @ = site air temp.(-10°C) in Biince- s., . | window. Sit “pressure oly Orow Sonaity chide” ope. . 2, Opposite true air G s rT speed (170K), read oe aad s s Mach No. (Mach .27). [ ANSWER: Mach 27 2 e = th yo- an se Fig. 13 Your Mach Number is somewhat less than the speed of sound, but there it is. . .Mach .27. —— Os ae | ees TRUE AIR SPEED FROM MACH NUMBER AND TEMPERATURE (TAS IN KNOTS ONLY) In aircraft having a Mach indicator it is possible to get true air speed from Mach Number and temperature. At higher speeds, indicated airtemperature must be"corrected" - to true air temperature for "anywhere near" accuracy. See "Page 18 for method of computing TAT (True Air Temper- ' ature). a Example. Given: Mach Number 1.16 True air temperature + 10°C Find: True air speed 3 ne er Fx Opposite Mach No. E6B-2P ¢ = | (Mach 1.16) read true air Ou, 3= 4 speed (760K). 9 TYTYTPFR U> E. iy iy 7 ty &p 2:30 ; & & eS ANSWER: 760K Si Oy R as ; PE s (| |. Set Mach No. index S a ~ opposite true air temp. bi = (+10°C). SS eo pp. e zt ‘a = hb = 3 PROBLEMS 6 _ Find Mach Number or true air speed; check your answers | against the answers in Part C. - Peete. | Number y .. Speed” - 1.) Wes5°0 70 - . . 0° 67 me, go 180° te 605 TRUE AIR SPEED COMPUTATION... AT HIGHER SPEEDS (TAS IN KNOTS ONLY) Although the method of computing true air speeq as described on Pages 13, 14 and 15 is satisfactory for lowe, speeds, significant errors are introduced at speeds aboy, 200 knots. The "Huber System" for TAS determination (a, used with Jeppesen Computer Models CR, CB, is the mos; direct and accurate method. However, satisfactory (high speed) TAS computations may be made with the E6p Computers using the following method. (Note: Basic Air Temperature—BAT-—is Indicated Temperature corrected for scale error.) Example Given: Pressure altitude, 25,000 ft. Basic air temperature, — 30°C Calibrated air speed, 240 knots Find: True air temperature True air speed Solution: 1. Compute a trial TAS by using BAT, PA, and CAS. This trial TAS (363 k) will be used to compute temperature rise (heat of compression error). 2. Usethe formula TAS _ Temperature Rise 950 TAS by placing 950 on the minute scale under trial TAS of 363 on the miles scale. Read the temperature rise of 14°C on the MILES scale opposite 363 on the MIN- UTES scale. True air temperature would be approxi- mately—44°C (BAT- Temperature Rise). Remember that TAT (True Air Temperature) is always colder than indicated or basic air temperature. 3, Using TAT, PA, and CAS, compute TAS uncorrected for compressibility. Your answer should be 351 knots. Determine the F factor on the slide of your computer. Interpolating by inspection, you find it to be .97. am © jd ~ a *s ba 2] nj 10 SOLVE FOR T.H.AND G.S. a COURSE AT TRUE INDEX. DIRECTION OW COMPASS ROSE AND SPEED ON GRD. PLACE TAS UNDER INTERSECTION OF WIND DIRECTION AND VELOCITY W.C.A.AT INTERSECTION OF TAS AND WIND LINE REAO G.S. UNDER GROMMET F CORRECTION FACTORS FOR TAS CALIBRATED AIRSPEED KNOTS 250 | 300 | 350 2700 400 | 450 | 500 | 550 10 | 10 | 98 | 99 | 98] 98] 971 97 | 20,000 | 99 | 98 | 97 | 87 | 96 | 951 94 33 30000 | 37 | 96 | 95 | 94] 92191 50 | 89 40,000 | 96 | 94 | 92] 90 | a8 | 87) 87 88 53 | 90 | 87 | 96 [84 [a4 | ea) pe DIRECTIONS USE CALIBRATED AIRSPEED AND PRESS ALT To OBTAIN F FACTOR MULTIPLY F FACTOR 8Y TAS OBTAINED WITH COMPUTER 10 OBTAIN TAS CORRECTED FoR COMPRESSIBILITY 3 5f0 Bey ra yer rs i ie 2 1 ae T vy J Ng t ae Ti) ae ze t 50}0 n~ Be 5 ‘ 4510 4 | eetaS Fig. 16 Multiply TAS uncorrected for compressibility by the F factor. TAS is therefore 340 knots. PROBLEMS 7 ind true air temperature and true air speed; check your wers against the answers in Part C. Basic Air Calibrated True Air _ True Temperature Air Speed Temperature Air Speed —25°C 250k —35° 300 —10° 280 —30° 210 —15° 325 —35° 200 _ You can convince yourself of the necessity of using this (or the Huber System) method of calculating high speed TAS by comparing the above answers with answers that you would have gotten with the method (low speed) describ- _ ed on Pages 13, 14 and 15, 19 SLIDE RULE USE Multiplication and division are performed on the cal- culator side of the E6B in the same manner as ona straight slide rule. Be careful not to confuse the time index , which stands for 60, with the unit index in these prob- lems. Example: 28x15 Y 3 | i at a B20 © ms Aha, Set 182 on outside scale posite 14 on inside scale. : ene, / why \ A Deng EN 2, Opposite unit index on inside scale read 13 on a Example: 18214 { \ ; | : { L/HR. .. TOTAL GALS” | TIME Fig. 18 | Successive multiplication and division may be done on : the E6B. 2 ; Example: ax Scale read 15.8 an outside scale. Scale opposite 19 on inside scale. Fig. 19 PROBLEMS 8 j Find answers; check your answers against the answers in Part C. l. 12.6 X 31 = 2. 267 + 156= Aas i TS, ne As ad ee Se ae ie y oP Peril 25, shod ae al meth Ghat oe re ) uF ae _ Hlaped time.) 60 _ 3 x 60 to Station = Degrees of change 1B 9 culator side, set 3 on outside scale opposite 9 on ale. Opposite 60 (1:00) on inside scale read a outside scale: ‘ x PROBLEMS 9 Solve the golleihe problems; check your answers against the answers in Part C. 1st bearing 280° at 8:26 2nd Sips 269° at 8:31 age: = 120 met * Ste Hane to station Find gaunce to. enon DRIFT AND OFF- COURSE CORRECTION "Unless you have the instinct of a homing pigeon, some day you'll find yourself off course. This need not be distressing if you continue on your original heading until you reach a recognizable check point. Measure your dis- tance off course, miles flown and miles to destination. Then two easy computer adjustments will give you the number of degrees to correct your heading to take the shortest route to your destination." One form of a simple proportion has been set up for quick usage in low level navigation where corrections must be immediate or in a case where the pilot is unable to determine his exact wind for precise corrections. This is the DRIFT CORR window. The proportion is simply: distance off course correction to parallel distance flown = 57.3 and distance off course correction to converge distance to fly z 57.3 The window on the E6B was derived from prior usage of the old rule of thumb that one mile off course in sixty re- quires a one degree correction. The angular corrections given in the window are derived from the actual value of 57.3. The settings illustrated on Pages 25 and 26 are based on the above principle. .. The illustrations show the procedure in using the DRIFT | CORR window with a sample problem. | Example | Given: Course distance traveled, 400 miles Distance off course, 30 miles Find: (A) Correction angle to parallel desired course. (B) Total correction angle to intercept desired course in 150 additional miles. sPecestecusdas _—s + ttttt —_ —_ _ — 2 2. Opposite "correction y &p to parallel” index read fe ® @® 7 correction angle (to 3) 4. S OF parallel) 4,3° day 2 am Soe a. Site Bh ot ‘> acne ONS = 9 Yak far ) oe x»? GLE kl» a > % 40 Coup Spy > % 2 e 4 CORR | 4, ole ti o| Fa/ehh SO | LA Answer es |S 4 =f £ ° =§ . Per ct. = se Ci) ye So, = Tt — a %, Ss | =) “ r) %,%0 0m = 2 2 —SAL/HR. .. ToT, te of oe ¥ 1. Set "distance off course" “ig & sO (30 miles) (outer scale) - ra) opposite"distance traveled” A (400 miles) (inner scale). #°S/ © Pa ‘OfPsponding to > . % Mach No. on "Min 4-Sarig A s Se 90:9 a oy, > _ ) & % 00°Z G \ J ° = —2 € CO, DENVER, © _- iar eat PEND ee z } = Fig. 20A Answer to (A): 25 GQNid iva zs p10? '83AN30 “09 y 1 43 ILE SJy of 5 LTS ann A ® 2: 3: @ “<“7\\d_ 1. Set "distance off course” Ry ast < (30 miles) opposite miles & > RS yy to interception" (150 miles), ~“ SS: 1. $/3 Ne Vay COMPUTA e LTS fK Fora (B) ANSWER: Add 0, = ° ee pero correction to parallel 43° te4 tocorrection tointercept 11.3 = to find total correction 2. Opposite "correction to to intercept 15.6 intercept"index, read jj < e dditional angle to wr ic addition J yotie ch a intercept (11.3°). a a a5 v yer 0 A by + se J , 5 “Te ode $§ § w Zr oct "A PP 20 ‘ oy Answer to (B): To find the total correction angle, add correction to intercept to the correction to parallel. The total correction is 15.6° (4.3° + 11.3°). NOTE: Notice that it is still necessary to determine the decimal point. If the miles flown or miles to fly is more than 10 times the miles off course, use the smaller angle. For example: 9 miles off course in 300 miles, correct 1.7° to parallel; for 90 miles off course in 300 miles, correct 17° to parallel. PROBLEMS 10 Find the correction angles; check your answers against the answers in Part C. Correction Angle To to to Total Flown Course Intercept Parallel Intercept Correction 1.200 mi. 15 mi. 75 mi. 2.375 25 200 3. 260 10 145 4.185 5 75 5. 250 13 90 6, 425 24 125 SOT, < PartB WIND SIDE WIND SOLUTION ON THE JEPPESEN E6B "All ‘windy’ problems depend upon the solution ofthe old eternal wind triangle. In its basic form, this triangle is made up of three simple vectors: (1) A "wind" vector, made up of the direction and speed of the wind. (2) A "ground" vector, representing the movement of the plane with respect to the ground, and made up of the true course (or track) and the ground speed. (3) An "air" vector, representing the movement of the plane with respect to the air mass, and made up of the true heading and the true air speed. 27 le has a dir sured as an angle from ured by some Standard scajo ection ang a true hetrianglecan be determin. €., corrected for any four of these ther two may be t graphical or by a tion is always in is a simple arrans® angles of the standard graduated in degre complete azimut! 1 1 tions sc tions "are stated; these may be used for plotting the wind when the wind value is stated in terms ofthese points rather than in degrees. The center of the disc is a transparent material ‘on which a wind grid is printed to provide for ease in plotting the wind. The solid line wind speed circles on the wind grid are always utilized with the low speed side of the moveable slide (Figure 24). The dashed wind speed circles labeled 50 and 100 are always utilized with the high speed side of the moveable slide (Figure 23). TRUE INDEX TITTTTTTYITW1TTT1T re OBEY CE Ras | 4 oe ] I TTY TITTT “TTT TT TITTY I —y TITTTT1TT Tr T I — TTTYVTYT t T mec remem jure hes io tak 16 hha ot , Wfiee = 4 ped Zi LAT TT AL, Py, Vifglty 4 WIT Tira i00dg HATTA) ; 4 r Tr p ai Lett 1964) thagg pitt (9494 19664 , y in tH Pttisdd thi seit ite Ltt td YT ‘re The thts . iY 7 ] Pt they tied f HY. $ ve + r j Wn TT t+ Tht Mist Tt thids. Mie fj 1 StH tH} THT ied ee HITT Wh Ui HY tit = Tit tb Y Viti hatte ttt HAL titties if HTH ad aes teddy ead it Tititt? TH ! a A i am iE: LOW SPEED SIDE Fig. 24 30 lit bre eas ae ‘ eetas Freee Gaettied ye De T ra Aa. 7 442 T O41. Pan eee, iT ri ioe Oa pg 2- WAY SLE Y_ HIGH SPEED SIDE Fig. 23 i ; 3-way slide L onal uses of is AMOUNT OF DRIFT 10° 5° 0° 5° 19° 5° 15° oe as * 2 25° °e 30 220 — 30° 35° —210 35° 40° -200 - 40° e 190 50° 180 50° 55° 0 55° 160 II Ne LINES 7 (peel US es LE Sad Be ph erie el koe “~ SPEED CIRCLES 180° Fig. 25 The second part is a movable slide (see Fig. 23) on which are printed two sets of lines. One set is a series of concentric arcs, known as "speed circles". These are at equal distances, each representing two units of distance, and numbered from 20 to 300 on one side of the slide and from 200 to 700 on the other side. Since these are "speed circles", they are used for all measurements of speeds for the wind triangle: wind speed, ground speed, true air speed. The second set of lines is a series of converging straight lines, which do not actually meet on the slide itself but which converge toward a point outside the slide. These are Called "track lines”, since they show the number of degrees by which track (or true course) differs from true heading. 31 On the high speed side of the 3-way slide (see Fig. 23) the track, or drift lines, are one degree apart for all speeds above 200 mph or knots. On the low speed side of the 3-way slide (see Fig. 24) the track, or drift lines, are five degrees apart up to 30 mph or knots, two degrees apart between 30 and 150 mph or knots, and one degree apart above 150 mph or knots. arts of the wind face make it possible to set : aes hts directions from true north, and angular ditteretce between true course and true heading—al) the elements necessary for the solution of wind vector triangles. f the computer is a matter of placing the aoe ranean on the computer in the proper relation- ship and reading the unknown ones from the completed triangle. e wind triangle graphically, only one basic Peace oie since the vectors have both direction and quantity. The three sides of the triangle (see Fig. 25) are known as the wind vector (wind direction and velocity — W/V), the air vector (true heading andtrue air speed —TH and TAS), and the ground vector (truecourse or track and ground speed——TC or TR and GS). There are four basic wind triangle problems with which the pilot is concerned. The first solution is for TH and GS. This problem is most common to pre-flight planning. To solve it graphic- ally, the wind vector is plotted first, the TC plotted on a reciprocal bearing from the head of the wind vector, and the TAS swung in an arc across the TC vector. Point A is established. MEASURE TH Fig. 26 32 oblem may be encountered if the flight re | GS. Again the wind vector Is plotted first. _ om ‘or is established with its head tothehead r, The components of the air vector, TH a en measured between points AandB. > ‘ 1 type of problem involves solving for TC and This solution is basic to pure dead reckoning when no s aids are available. The wind vectoris plotted and the the air vector attached to the tail of the wind vector. point A is established and TC and GS can be een points A and C. TH & TAS This problem is asually eocountered while in Might. by establishing a departure point, A, and plotting the a'r a> gtouné vectors trom it, points B and C are located. The wind is measured from B to C sincetbe sircraf will alway, be blown from heading to track FLIGHT PLANNING WITH FORECAST WINDS The most common type wind problem usually encounter- ed by the pilot is the one in which he is required to calculate true heading and ground speed. This problem is usually solved in the flight planning stages when true course, true airspeed and forecast or reported winds are known quanti- ties. It is essential that the pilot remember that winds aloft reports and forecasts are always provided in a format which utilizes directions measured from true north, wind speed in knots and temperature in degrees centigrade. Either knots or miles per hour may be used in a wind problem provided the chosen unit of measure is used consistently throughout the problem. There are several ways of solving flight planning pro- blems however, the one outlined here is the most consist- ent with the other solutions presented in the booklet. 35 Example Given: True Course, 195° Wind direction, 240° Wind speed, 30 knots True airspeed, 180 knots Find: Groundspeed True heading Solution: 1. Set the true course (195°) under the TRUE INDEX. 2. Locate the wind point on the transparent disc. This is accomplished by finding the wind direction (240°) on the periphery of the computer disc, then proceeding to- ward the center of the disc on the 240° degree line until it intersects the appropriate wind circle (30). The wind point may either be mentally noted or marked with a pencil dot. 3. Adjust the moveable slide until the arc representing the true airspeed (180 knots) is read directly under the wind point as illustrated in Figure 30. = t 5+ 24 Ex So, EN — Eh =s TRUE INDEX oe = CN 29 m0 20 “y >> 1. True Course 195° Ve : 0 ¢ IN . nN 419 + Ay 2. Wind Point 240° oF 20 Si i : sheer eel acescee 5. True Heading 202 a Y, ° 140. r oy6 1" 2 1 nes y | Be END 3.. True Airspeed 180 kts. 8 Sx = \ 7 5 Se Br = = Us © 4. Groundspeed 158 kts. ; e © A Gs dl 3 STATUTE 0 oe ee ae Read the groundspeed (158 knots) on the speed arc * under the small circle at the center of the disc. 5 Read the wind correction angle (7° right) directly under the wind point on the vertical lines printed on the move- able slide. Note that the wind point is approximately over the vertical line 7th from the right of the center line. 6. Since the wind correction angle is 7° right, the true " peading is 7° to the right of the true course. Thus, read the true heading (202°) 7° right of the TRUE INDEX. NOTE: All of the wind problems presented in this booklet will be solved with the true course (track) under the TRUE INDEX, the true airspeed under the wind point and the ground speed under the small circle in the center of the transparent disc. PROBLEMS 11 Find groundspeed and true heading; check your answers against the answers in Part C. True Wind Wind True Ground True Course Direction Speed Airspeed Speed Heading 1. 080° 148° 32k 170k C507, e ete: | >» (ee Bie 2. 010° 300° 18 TSO Haead Pb sinrksaly 3. 160° 015° 29 WOO eres ecieceas ene ce 4. 275° 225° 15 140 aicwees vce cco 0, 8 '< 5. 050° 180° 35 200 ice ans seetanls 6. 190° 078° 30 LTB ae ire tte colic ie TRUE AIR SPEED “Have a date to meet some ing gal, an airplane to interc good, there are times when speed that will enable you to make good a given ground speed. Here's how:" Your ground speed will be determined foryouin advance. The problem of pre-flight calculations then becomes oneof : finding true heading and true air speed. This solution differs only very slightly from the one for true heading and ground speed. one? Whether it's a 800d look- eptor a flight schedule to make Given: True course, 240° Wind direction, 090° Wind speed, 20 knots Ground speed, 140 knots Find: True airspeed True heading ) Example Solution: 1. Set the true course (240°) under the TRUE INDEX. 2. Locate the wind point (090° at 20 knots) on the trans- parent disc. 3. Adjust the moveable slide until the arc representing the groundspeed (140 knots) is read directly under the sm all circle in the center of the transparent disc. 4. Read the true airspeed (123 knots) under the wind point. 5, Read the wind correction angle of 5° left on the vertical lines under the wind point. The true heading (235°) is found by reading 5° to the left of the TRUE INDEX as illustrated in Figure 31. —e _ oe a ap a ae cue ©) |! pee T =—8 Ch TRUE INDEX — —j + = e010 bee - 4 at uy , 2 W ~ AS Rw 240°3 1. 7 S . True C Woe o™. N90 2 5. True Heading 235° FY =SSsi= fo) N o— at 1310+ + (7) aS eee cee 4 ° ~ E * i == = oO 4, True Airspeed 123 kts. aes, = o vA = aoe — 2. Wind Point 096° o 4 - "he - -2 7 Se aes w 3. Groundsgeed 140 kts. rs S NOTE: Remember that for any problem worked on the E6B computer, the true course is always set atthe TRUE INDEX , the true airspeed is set at the wind point and the groundspeed is set at the small circle in the center of the transparent 39 SS Fe IE i PROBLEMS 12 Find true airspeed and true heading; check your answers against the answers in Part C. True Wind Wind Ground True True Course Direction Speed Speed Airspeed Heading 1. 340° = 220° SRY oe 3 Cacia KB 2. 274° 050° 15 DUR eats) aici ey sce 3, 183° 150° S5 ISB ee 4. 005° —- 230° Se PARA ee 112° = §=070° 20 BG ie a 6. 266° 320° BO 163 8 TRUE COURSE (TRACK) AND GROUND SPEED "Sometimes it's mighty interesting to know where you're going and how fast. Your air speed and heading are ® usually available in flight. If, in addition, you have some wind information (either reported or forecast) you can easily find true course and ground speed." In a practical sense, this solutionis usefulto the airborne pilot when he knows the wind, true heading and true air- Speed but has no means by which he can determine the true course or groundspeed. A problem such asthis can be solved as follows: 41 Example Given: True heading, 175° Wind direction, 320° Wind speed, 30 knots True airspeed, 180 knots Find: True course (track) Groundspeed Solution: temporary m the problem progresses. ) ' 2. Locate the wind point (320° parent disc. 1. Set the true heading (175°) at the TRUE INDEX as a easure. (Since a correct solution cannotbe read from the computer with the true heading at the TRUE INDEX, a further adjustment must be made as at 30 knots) on the trans- 3, Adjust the moveable slide until the true airspeed arc (180) is directly under the wind point as illustrated in Figure 32. ss f i T . : SS 3 = TRUE INDEX Zz =e eS n0 10 | 1. True Heading 175° Bg a we % _— ~ s ao 170 l 90 4 - to 9 \ 2 es ane v "0 2 %0% \ ‘ t ny Yo S » a r 2 fo) & “we . Pe M : ° * +2. =) | Fz * Ic to 1 © F Sf s ‘ ao : 2. Wind Point 320° at 30 kts = Pt = = =s 2 z ' =~ ° NOH PAN | =) a & : xs i) oot . 3 26 : KO 4 7) + 2a "Ky 0 LG OH, ON . a OR tN Ose 3. True Airspeed 180 kts - 420 eye mde itt -N oP oN ~ = %, +) cae 0 = “2 "C-19 -20 =, = +60 e's) Ei. = 40 +20 oF -20 2 2 ae 7s - Fig. 32 42 > _ By noting the position of the wind point in relation to the vertical wind correction angle lines, one may notice approximately a 5° right wind correction angle. Since this value is the difference between the true heading and the true course, adjust the transparent disc 5° to the right placing 170° under the TRUE INDEX which corresponds to the 5° right wind correction angle noted at the wind point. This angular relationship must be maintained in order to reach a correct solution. . Adjust the moveable slide to keep the true airspeed arc (180) under the wind point. (If this action appreciably affects the angular relationship referred to in step 4, a second adjustment of the transparent disc and the move- able slide may be necessary). . Read the groundspeed (206 knots) under the small circle in the middle of the transparent disc. . Read the true course (170°) at the TRUE INDEX as illustrated in Figure 33. a 3010 ; te = } = TRUE INDEX 6. True Come 170 = - + 10 V4 = =S Pe ue : 20 MES fs ao 160 170 30 “> vo \ 9 te 199 Mp, we AN tt SL io yo” wwe & at 3 + VY iy % @ Cs + —4—-25]0-F = @ “ HI a ° L oo ise 24 Ls ~) SY f ; As 0 r ~ P Ss 5. Groundspeed 206 kts 4 LIE o ° NE @ s 3 o> = E « = ax 2 3 stot = No ’ ' IT s- 2 . = . O, i ar on eee 4. Wind Point 320 w Sn USSS at 30 kts. °°. ; @ a yy A c. % 30 St US IL 8 PROBLEMS 13 Find groundspeed and true course (track); check you, answers against the answers in Part C. True Wind Wind True Ground True Heading Direction Speed Airspeed Speed Course 1. 075° 350° 30k 140k .44t. .82%. 2. 240° 115° 28 166 cu Yee e448 ° 3. 352° 025° 36 146... 44: 300° PLL 185° 40 170 026° 230° 25 200 a 173° 190° 30 160 > ee ee 88 @ @ 44 FINDING WINDS IN FLIGHT "Gentlemen, often after you get ‘upstairs’ you find the wind is not behaving asthe 'weather-guesser’ said it would. You must hold a different heading from that originally est- imated, in order to make good your course, and your are crossing check points ahead of or behind planned times. You know how fast and where you're going (ground speed and true course). . . .also the heading that's getting you there (true heading). . . .. . .but without accurate wind information you can't re-estimate the legs ahead. Hence you need to determine the actual wind direction and speed.” Let's assume the following: Example Given: True course (track), 122° Groundspeed, 154 knots True heading, 118° True airspeed, 160 knots (45 Find: Wind direction Wind speed Solution: 1. Set the true course (122°) under the TRUE INDEx 2, Adjust the moveable slide until the groundspeed (154 knots) is directly under the small circle in the middle of the transparent disc. 3. Compare the figures for true heading and track. The difference between the true heading and track is the wind correction angle (WCA). Note on the compute; disc that the true heading (118°) is 4° left of the true course (122°), thus it is immediately apparent that the wind point will be on the left side of the transparent disc 4. Place the wind point at the intersection of the 4° lef; vertical wind correction line and the 160 knot true air speed arc. Read the wind direction as 059° and the wind speed as 13 knots as shown in Figure 34. = t | ar as Od cea HS a | = == tt = = “TRUE IND — = 4! lo y fais 1. True Course 122°5 A mH Zot > z ® ie] By; R4O a WwW, oO I (7) ay 9 o= ty Jo” r ® io — +t Oy. SS \ 711. gs is ay me : } > os @ eed he x . bay HINTS ON THE USE OF THE WIND SIDE 1. The wind side of the E6B Computer incorporate, | handy method for converting true course to Magnet, course or vice versa. Notice — VAR. te (Variati., and VAR. WEST (variation west) are printed Saavely on the left and right sides of the TRUE I