Manual / Guide · 1958

Kane Mark 6 Dead Reckoning Computer Instructions (1958)

Instructions for the Kane Mark 6 dead reckoning computer, an E-6B-type slide-rule computer.

Author: Kane
Year: 1958
Type: Manual / Guide
Credit: Scan via the Internet Archive.

KANE MARK VI DEAD RECKONING COMPUTER ILLUSTRATED INSTRUCTIONS AND PRACTICE PROBLEMS FOR MARK VI AND SKYKING COMPUTERS KANE AERO COMPANY MANUFACTURERS AND PUBLISHERS OF AERONAUTICAL AIDS 1845 FORD PARKWAY SAINT PAUL 16, MINNESOTA THE KANE COMPUTER TYPE MARK VI . AND "“SKYKING" ILLUSTRATED INSTRUCTIONS AND PRACTICE PROBLEMS COPYRIGHT 1958 BY KANE AERO COMPANY 1845 FORD PARKWAY ST. PAUL, MINN PRINTED IN USA PRICE $3.00 le SECTION I - THE COMPUTER SLIDE RULE SIDE DESCRIPTION The Mark VI is an all metal computer with a metal slide, The graduations are precis— ton engraved on a high grade aluminum alloy. All parts are anodized to provide a hard satin finish surface. The circular slide rule side of the computer enables the pilot to solve problems involving time, speed, distance and fuel consumption. Indexes are pro- vided on the inner circular scale for rapid conversion of gallons of fuel or oil to pounds, statute miles to nautical miles and miles to kilometers. Five logarithm sets of scales are used-one main and five aux- iliary. The auxiliary scales are positioned on cut-outs or win- dows. This side of the computer consists of two parts; namely, a stationary body plate known as the miles scale and a rotat- ing disc known as the minute scale. ee KANE AERO CO- It is convenient to consider the stationary and rotating scales establish arithmetical proportions, Note that the outer scales in Fig. 2 are positioned to establish a 2 FIG, 1 to 1 proportion. Knowing three factors of the proportion, the fourth can be determined. Example: 90 miles in 45 minutes is 120 miles in 60 minutes (120 MPH.). When using these two scales the decimal point will be positioned as necessary. For example, 12 may be con- sidered 1.2, 12, 120, 1200. A large triangular index has been positioned at the 60 minute graduation and is known as the miles per hour or gallons per hour index. TIME, GROUND SPEED, DISTANCE PROBLEMS On all distance, time, ground speed problems, the large triangular index is either set opposite ground speed or will position itself opposite ground speed. Distance and time are always opposite each other. 3. READ OPPOSITE GROUND SPEED| INDEX 120 MPH| T. ALIGN DIST| KONI) The rotating disc has two (2) scales, The outer rotating scale is used for minute and the inner rotating scale for hours and minutes. Note that 90 minutes is opposite 1 hour 30 min. ws 2. WITH TIME 45 SECs Ye BS ara FIG, 2 Given: -. Distance Flown-90 miles Solution: See Fig. 2. Time-45 minutes Align 90 on the fixed miles scale with 45 on the rotating minute scale. Opposite the ground speed Find: BZ fo) index read 120 on the fixed miles scale. Ground Speed ( -J- CONT, TIME, GROUND SPEED, DISTANCE PROBLEMS Exercise 1. Distance Ground Distance Ground No. (Miles) Time Speed (MPH) No. (Miles) Time Speed (MPH) ney 90 345 ( /20 ) 6. 65 230 ( 170) 2 80 2:00 ( go ) Ws 12 308 ( 90) 3. 60 230 ( “/20) g 140 1:55 ( 73) =F 4, 35 nll (yer) vs. 11s 2471/2 ( G6 IXIGS ! 5. 41000 20 (ge) XHF 10. 90 1:40 (5g ) Given: Solution: See Fig. 3. Distance-120 miles Ground Speed-80 mph 1. Rotate minute scale until Triangular Ground Speed Index is opposite Ground Speed (80 mph) on stationary miles scale. 2. Opposite distance (120 miles) on stationary miles scale read Pind: > time (90 min.) on rotating minute scale. The hour scale will Time ; (TO. }'32 indicate 1 hour 30 minutes. Exercise 2. Distance Ground Distance Ground No. (Miles) Time Speed (MPH) No. (Miles) Time Speed (MPH) 1. 80 X ( FOZ 60 6 110 ( }3s) 220 2. 120 ($lz0dz2i0e 60 7. 170 ( ‘//8O), 8522 85 3. 140 Cfypo)ete® 70 8. lis ( w77 y 7 30 4, 90 (40) 135 a. 300 ( ') SC) 22615 5. 95 ( 350) 190 10. 45 (3398) 78 Given: Solution: See Fig. 4. Ground Speed-110mph 1, Rotate minute scale until Triangular Ground Speed Index is Time~1 hour, 20 min opposite Ground Speed (110 mph) on miles scale, 2. Opposite Minute Scale (1:20) read Distance (147 miles) on * stationary miles scale. Find: vy Distance 1, PLAGE GROUND] SPEED INDEX ON 110 MPH 2. OPPOSITE DIST, 120 MI, ay, / FIG. 3 FIG. 4 Exercise 3. Distance Ground Distance Ground No. (Miles) Time Speed (MPH) No. (Miles) Time Speed (MPH) i. ( fo) 390 40 6. (297) 2:10 134 2 ( 45) ~ 330 90 7. (3p 1:50 175 So (170) 2:00 85 8. ( 37) 325 88 4, X( 23 BH 9:120 115 9. ( Sd) 1:25 59 5. ( Yo) 337 146 lo. ( Y) 307 78 =3- Exercise 4, (Mixed Problems) Distance Ground No. (Miles) Time Speed (MPH) No, ls 140 2:00 ?77oO 11. 2K 156 2120 ? 756-D iW, ae 98 337 2159 13. 4. 75 2136 125 14, 5. 36 2276.6 130 15. 6. 42 P47,.6 143 16, ihe 249O 3:00 60 uz 8. ? ¥2 1:05 76 18, oS 24St 1:37 112 19, 10, isé 1:32 210% 20. Distance Ground (Miles) Time Speed (MPH) 82 DG 115 275 246 98 81 PFS 107, 113 380 ? 8 210% 120 a PAP 207. 110 120 2EIFO 72 80 41 1/2 2ae 125 2 7Lafi/e 104 192 2 Phan pled 137 FUEL CONSUMPTION PROBLEMS Fuel consumed (gallons), flying time (hours), and rate of fuel consumed (gallons per hour) are com- puted in a similar manner to distance, time and ground speed problems, Fuel consumed is com- parable to distance and is found on the stationary miles scale. Flying fuel hours used or left is the same as time in flight and is found on the rotating time scale, Rate xf fuel consumption (gallons per hour) is found opposite the triangular ground speed index on the stationary miles scale. Given: Fuel Available-360 gals. Time in Flight-8 hours we) Find: Rate of Fuel Consumption Solution: See Fig. 5. 1, Rotate time scale to position time (8 hrs) opposite fuel available (360 gals) on miles scale. 2. Opposite the triangular rate of fuel consump- tion index, read rate of fuel consumption (45 gals/hr) on the stationary miles scale. Exercise 5 T 1, ALIGN FUEL AVAIL, 360 Fuel Available’ Flying Time Rate of Fuel No, orConsuméd Used or Left Consumption (Gallons) > (Gals/Hr) ag 42y (70,5 ) 2. 36 (42) a 27 ( 98 ) 4. 6 (17-8 9 5 32 ( %/ Given: Fuel Available~250 gals. Rate of Consumption-41 1/2 gals/hr. Find: Fuel Hours Available Solution: See Fig. 6. 1. Rotate the minute scale so that the triangular rate index (gals. per hr.) is opposite the rate of Consumption (41 1/2 gals. per hr.) on the 2. OPPOSITE FUEL AVAILe stationary miles scale. 1. SET INDEX OPPOSITE 41 1/2 GPH 2. Opposite Fuel Available (250 gals.) on the miles stationary scale read on the Time scale (6 hours). =-4- Exercise 6. Fuel Available Flying Time Rate of Fuel = 3. READ No. or Consumed Used or Left Consumption av , FUEL (Gallons) (Gals/Hr) » Oo aN Wy CONSUMED le 4a~ (ees ) Ee 36 GALS. 2. 42 (gree) 9 3. 37 ($749) 61/2 A. 26 (2’3h) 10 iq 40 (2:92) 15 Given: Rate of Fuel Consumption-9 gals. per hr. Fuel Time Used-4 hrs. <1 WEA Find: Fuel Consumed —E, DEV Solution: See Fig. 7. 1, Rotate the Time scale so that the triangular rate of fuel consumption Index is opposite. 9 on the stationary miles scale. 2. Opposite Fuel Time Used (4 hrs.) on the time scale read Fuel Consumed (36 gals.) on the stationary miles scale. act. DEV. oe +R. WEA +W. VAR, —E. VAR = ALTIMETER CORRECTION The slide rule side of the computer is used for altimeter and airspeed correction computa- tions. When flying from a warm region to a cold, the true altitude is lower than the alti- meter indicates. This is a dangerous condition, The pilot should continuously calculate in flight the true altitude from the indicated in accordance with the temperature aloft, The altimeter is calibrated for the standard atmosphere. It reads correctly only under standard conditions of sea level temperature (15 deg. C. or 59 deg. F.) and normal temp- erature lapse rate. For all practical purposes, the pilot may consider PRESSURE ALTITUDE AND CALIBRATED ALTITUDE EQUAL TO INDICATED ALTITUDE, The error in computations will be neglegible, Pressure altitude plus or minus pressure altitude variation is indicated altitude. Calib- rated altitude is indicated altitude corrected for installation errors, Exercise 7. Fuel Available Flying Time Rate of Fuel [7 va No. or Consumed Used or Left Consumption |; ate My, oe (Gals/Hr) (Gals/Hr) © sw ‘6 T. SET INDEX 1. ( 3:00 6 TA RW OPPOSITE 2 tay) 2:20 9 oninoxoss OY] SPECS 3. «(ap ) 3245 12 1/2 a 4. (137 ) 1:50 71/2 FIG. 7 5. ( 99) 4:14 8 MIXED FUEL CONSUMPTION PROBLEMS Exercise 8, Fuel Available Flying Time Rate of Fuel No. or Consumed Used or Left Consumption (Gallons) (Gals/Hr) 1. 36 4:00 ? 2. 27 3:30 ? 3. 16 1/2 ? 6 A. 25.3 ? 8 Se G 3:00 7 6. ? 4:15 9 7. 42 ? 14 8. 37 2:18 ? 9. 22.7 ? 81/2 10. 2 3:14 10 ll. ? 2:50 545 12. 25 ? 729 13. 80 6:10 ? 14, il 235 ? 15. 58 ? 15.5 16. 60 4:48 ? 17. ? 0 6 18. ? 5 75 19. 23.5 2 fel 20. 8.5 153 ? On the extreme outer scales, note "True Alt." 4, READ L} wes (Corrected altitude) and "Cal.Alt," (Indicated TRUE ALT, altitude) scales. 8.700 FT. It is sufficiently accurate for Pilot-Navigator e type of computation to consider that: & 1. Pressure altitude equals indicated altitude. 3. OPPOSITE |> 2. Calibrated altitude equals indicated altitude. IND. ALT.|'_ jf 53;] 3. Corrected altitude equals true altitude. 207000 FT dls ano HEADIN > h+e. wor Reverse] Givens =Uwra=Th*W VAR w+] Indicated Altitude~10, 000 ft. 1. ALIGN EV. Temperature Aloft-(-40 deg. C. IND. ALT. }'v"C2| ° ’ Baie 10,000 FT. Find: , True Altitude Solution: See Fig. 8. 1. Align in the altimeter cut-out window Indicated Altitude (10,000 ft. Press. Alt.) with Temp- erature Aloft (-40 deg. C.). 2. Opposite indicated altitude (10,000 ft.) on the calibrated altitude scale read True Altitude (8700 ft.) on corrected altitude scale. Caution: Do not use the airspeed window for FIG. 8 altimeter correction, Exercise 9, —|1. SET TRUE ALT.) Indicated Temp. True No. Altitude Aloft Alt. (Feet) (Deg. C.) (Feet) dG 20,000 -20 ( ) 26 10,000 -10 f ) 3. 30,000 -30 ) 2 4, 10,000 +10 () 2. OPPOSITE 5. 6,000 415 ( ) IND, ALT. |e 6. 14,000 45 ( ) 10,000 FT, 7. 5,000 -10 () ie or Se 8. 18,000 5 ( 2 3. OPPOSITE |e 9. 8,000 +10 ( 3 IND, ALT. ose 10. 12,000 -15 @ J 10,000 FT.|"%, % © Pe, Sel Given: Ve. "of True Altitude-11,000 ft. co Indicated Altitude-10,000 ft. & Pind: Sin AiR TRAD TE Temperature at Altitude See Fig. 9. FIG, 9 Solution; See Fig. 9. 1. Opposite True Altitude (11,000 ft.) on cor- 3, OPPOSITE 0 RECKONING rected altitude scale, place Indicated Altitude . TRUE ALT. eh? penta Cc (10,000 ft.) on the calibrated altitude scale. 2. Opposite Indicated Altitude (10,000 ft.) on pressure altitude scale, read Air Temperature +22 deg. Ce Exercise 10. Ind. Temp, True No, Alt. Aloft Alt. y 1. 20,000 ( ) 19,000 /* I. SET PRESS. 2. 25,000 ( ) 26,000 ALT. 8,000 FT. 3. 18,000 ( ) 18,500 a Ese 4, 10,000 ( ) 9,600 Se: 11,500 ( ) 12,000 6. 8,000 ( ) 8,700 Given: Pressure Altitude-8,000 ft. Temperature Aloft-(+10 deg. C.) True Altitude-9, 000 ft, Find: Indicated Altitude Solution: See. Fig. 10. 1. Opposite Pressure Altitude (8,000 ft.) on press.| alt. scale, place Temperature (+10) in the altimeter window. 2. Opposite True Altitude (9,000 ft.) on the cor- rected altitude scale, read Indicated Altitude (8,680 ft.) on the calibrated altitude scale. FIG. 10 Exercise 11. No. True Temp. Press. Ind. Alt, Alt. Alt. 1. 20,000 -10 21,000 ( 2. 10,000 +20 8,800 ( 3. 9,000 +5 8,000 ( 4, 15,500 ~15 16,000 ( 5. 4,900 -5 5,000 ( 6. 5,500 -20 6,000 ( AIRSPEED CORRECTION PROBLEMS The airspeed indicator like the altimeter will read correctly only in the standard atmosphere. Variation in temperature and pressure from standard at a given altitude ; will result in error. True Airspeed is cal— iF 1 {l. SET TE ibrated airspeed corrected for temperature 5, ~30% and pressure. For the pilot navigator, it ge TRUE A.S. 237 MPH is sufficiently accurate to use indicated altitude for pressure altitude in the air- speed window. KANE AERO CO- Again align the index "10" on the miles FIG, 11 scale with the index "10" on the minute scale. Note that the true airspeed is on the outer stationary miles scale and calibrated air~ speed (indicated airspeed) is on the minute scale; that temperature and pressure aloft are on the airspeed window scales, The Kane Mark VI computer permits the solution of airspeed correction problems directly us— ing either Centigrade or Fahrenheit terms. No conversion from Fahrenheit to Centigrade is os 75 necessary. Note further that at standard conditions (sea level alt, and 15 deg. C. or 59 deg. F.) true airspeed is equal to indicated. No error exists under this condition. Given: Find: Indicated Airspeed~210 mph.~ True Airspeed Indicated Altitude-10, 000 ft. Temperature Aloft-(-20 deg. GC, or -4 deg. F.) Solution: See Fig. 11. / 1. Using "For Airspeed Computation Window” align Temperature Aloft (-20 deg. C. or -4 deg. F.) with Indicated Altitude (10,000 ft. Press, Alt.). 2. Opposite Indicated Airspeed (210 mph.) on the calibrated airspeed scale, read True Airspeed (237 mph,) on the stationary true airspeed scale (miles scale). Exercise 12. No, Ind. Indicated Temp. True No, Ind. Indicated Temp. True A.S. Altitude Aloft A. S. A.S, Altitude Aloft AS. 1. 180 10,000 -lo ( ) 6. 140 4,000 -5 ( ) 2. 200 20,000 -30 ( ) 7. 140 4,000 -30 ( ) 3. 150 5,000 -15 { ) 8. 210 7,000 -15 ( ) 4, 160 12,000 -10 ( ) Ey 165 15,000 -20 ( ) 5. 140 3,000 +5 ( ) 10, 175 12,000 +5 ( ) Given: Find: True Airspeed-200 mph. Indicated Airspeed Indicated Altitude-10,000 ft. Temperature Aloft-(-20 deg. C. or ~4 deg. F.) Solutton: See Fig, 12. - 1. Using "For Airspeed Computation" window, align Temperature Aloft (-20 deg. C. or -4 deg. F.) with Indicated Altitude (10,000 ft.). 2. Opposite True Airspeed (200 mph.) on outer stationary true A. S. Scale, read Indicated Airspeed (177 mph.) on calibrated airspeed scale. Exercise 13, No. Ind. Indicated Temp. True No. Ind. Indicated Tomp, True A.S. Altitude Aloft A.S. A.S, Altitude Aloft A.S. ne )) 12,000 -5 150 4. ( )»* 6,000 -20 «125 Zann ) 20,000 45 210 eae ( ) 5,000 415 135 3. ( ) 18,000 +10 165 6. ( ) 7,000 #8 «112 Tae Thaw 5 ek IND. A.Se | oe ee 177 MPH le OPPOSILE |.% IND, ALT. ee, 10,000 FT| %% / [# Veo AIRSPEED - conse / Tap TEP lg, f i oo es| 3- OPPOSITEY 3 £4 TRUE AsSal ; ; \|_200 MPH Ni ul A WS > - Sa < 3. OPPOS. TEMP. =15°C or +5°F ITE 1. SET TEMP. y, =20°C or -40F |P ey a FIG, 12 FIG. 13 Given: Find: True Airspeed-156 mph. Altitude to Fly Indicated Airspeed-150 mph. See Fig. 13 Temperature Aloft-(-15 deg. C. or +5 deg. FP.) -8- Solution: 1, Align True Airspeed (156 mph.) on the outer stationary true airspeed scale with Ind— icated Airspeed (150 mph.) on the calibrated airspeed scale. 2, Using "For Airspeed Computation" window, opposite Temperature (-15 deg. C. or +5 deg. F.) read Indicated Altitude {5,000 ft.) on the pressure altitude scale. Exercise 14, No, Ind. Indicated Temp. True No. Ind. Indicated Temp. True A.S. Altitude Aloft A.S. A.S. Altitude Aloft A.S. i. 175 ( ) -10 200 4, 145 ( ) +15 150 2. 230 ( ) =35 225 Se 118 ( ) +5 125 3. 160 ( ) -5 185 6. 113 ( ) #10 (125 MIXED AIRSPEED CORRECTION PROBLEMS Exercise 15. No. Ind. Indicated Temp. True No. Ind. Indicated Temp, True A.S. Altitude Aloft AS. A.S. Altitude Aloft A.S. 1. 150 5,000 ? 156 6. 128 7,000 2 152 Ze 178 = =12,000 -20 ? 7. 158 12,000 -10 ? 3. 235 ? “35 270 8. ? 18,300 -18 165 4. ? 16,000 45 245 9. 165 18,500 2 210 Se 116 ? -5 125 10, 175 ? ~45 165 DENSITY ALTITUDE Density Altitude is pressure altitude corrected for true air temperature at the flight level. Engine power out- put, fuel consumption and wing lift or rate of climb is dependent on the density altitude. The temperature and pressure altitude of the airspeed auxiliary scales are used to determine density altitude. Gap RECKONING MARK VI Given: Temperature~(-20 deg. C. or -4 deg. F.) Pressure Altitude~10,000 ft. Find: ok Density Altitude » ALIGN <b, 6 PRESS» ALT, Ry Solution: See Fig, 14. 10,000 FT, 1, On the airspeed auxiliary scales set pressure altitude (10,000 ft.) opposite Temperature (-20 deg. C. or -4 deg. F.). Read on the density altitude scale 8,000 ft. PES GY i OPPOSITE NE ACKY ~ FIG, 14 -9- DRIFT CORRECTION FOR OFF COURSE PROBLEMS The drift correction angle is the correction which is applied to true course to find the true heading. The amount of drift correction applied should be enough to either parallel the intend track or bring the aircraft back to the intended track at some point, If the aircraft drifted to the left of the intended track it was caused by a stronger than anticipated left wind accordingly, a right wind correction angle must be applied to the true course. NOTE: The formulae on the center of the computer. The Kane Mark VI computer is especially designed to solve the above type problems simply by the use of an auxiliary scale, This scale is based on the trignometric sine function of the off course distance to the distance traveled. It will be helpful to remember that a drift of one mile off course in sixty miles flown results, in a drift of approximately one degree (.95 deg.). If the drift is ten miles in sixty the drift angle is approximately ten degree: 4-5 deg.); in one hundred twenty miles, five degrees (4 3/4 deg.). t Given: J Solution: Off Course-12 miles 5 1. SEE FIG. 15. Align 12 on the stationary miles Distance Flown-136 miles oo scale with 136 on the minute scale then read Distance remaining-68 miles in the auxiliary scale 5 deg. to parallel course. Find: “ 2, SEE FIG, 16. Align 12 on the outer stationary Correction angle to parallel course scale with 68 on the minute scale then read Correction angle to converge on destination on the auxiliary scale 10 deg. Total correction angle to converge on 3. To converge on course add 5 deg, and 10 deg. destination (15 deg.). 1, SET DIST, OFF COURSE 1. SETDIST. OFF COURSE 1 ? a 2. OPPOSITE DIST. FLOWN 136 MI, 2 Tpiv=ce +. wea — ° ss A =— ‘ = RK = » ‘RS. FIG, 15 FIG, 16 OFF COURSE PROBLEMS Exercise 16. Miles Miles Miles Angle to Angle to Total No. off Out Left Parallel Converge Correction Course Angle is 8R 100 150 ( ) ( ) ( ) 2. 10 R 125 175 ( ) ( ) ( } 3. 121 60 100 ( ) ( ) ( ) 4, is L 75 50 ( ) ( ) ( ) Sa 6R 96 66 ( ) ( ) ( ) Given: Ground Speed made good~150 mph. Time Flown-35 minutes Heading Altered~18 deg. Find: Off Course Distance Solution: 1. Place ground speed index on 150 mph, and read opposite 35 minutes 87 miles traveled. 2. SEE FIG. 17. Set 18 deg. un drift scale. Opposite 87 miles on the min- ute scale read 27.2 miles off course on the miles scale. -10-~ The drift scale is also useful in determining the off course dist danger areas or storm areas when the aircraft heading has been ance traveled to changed, evold) 3. READ OFF|s +7 COURSE 2. OPPOSITE DIST. FLOWN CONVERSION INDEXES OPPOSITE STAT MILES INDEX 3A, READ oppositz fECKONING co KM, INDEX| “env! ™, 96 1/2 KM 60 STAT MILES |w. rn 7 1A, SET NAUT. MILES INDEX OPPOSITE Indexes are provided on the inner circular scale for Tapid conversion of gallons’ of fuel or oil to pounds, statute miles to nautical miles and miles to kil= ometers. Given: 19 U.S. gallons in tank - Find: Corresponding no. of Imperial gallons, pounds oil and pounds fuel. Solution: See Fig. 18. Set U.S. gallons index to 19 on the stationary scale. Read 15.8 Imperial gallons 143 pounds of] and 114 pounds gas opposite corresponding indexes. Similarly, 52 nautical miles equals 60 statute miles and 96 1/2 kilometers= -ll- coMMON MULTIPLICATION AND DIVISION sa terms may be set up 4 Diviston Of jmilierly to distance ae time ; % \ Le cept that the answer *S to find nin ea ‘ao “ index instead of BE ACONVERSION inv GE, % will be oppos: a Siar. 2. OPPOSITE Ci ZB } the ground spec | DIVISER ie i ‘|__12 Anaad Example: 7% we,Be AS q 4 9 ' “ty, = : See Fig- 19- | % phase ern athe fixed miles scale with 12 e ea lige Boe on the rotating minute scale. Opposite the “Wy ‘S, g my “rotating "10" index find 20 on the miles / f le Se scale AIRSPEED = 4 £—Sottscrion J 3 3 ae ree Ly ef OS gs 1 PLN AOAE é FIG. 19 r% 20 See x Multplication of two terms is sim- 15 = a iliar to solving problems of ground Ave ag Bo OE i ent HE got wee ae instead of the ground speed index. me speed and time to find distance, XA except that the "10" index is used Example: Multiply 4 by 15 Solution: See Fig. 20. Align the "10" index of the rotating scale opposite 4.0 on the miles scale. Opposite 15 on the minute FIG. 20 RATE OF CLIMB AND DESCENT Problems involving rate of climb and om @re solved as ordinary mult~ Hication and division problems using the "10" index as explained above. { Given: fare Tate of climb-800 ft. per minute “Altitude to climb-From 2000 ft. to 14,000 Pind: Time to ascend to cruising level Solution: see Fig, 21 ‘lace "10" index of the rotating scale te 800 on the miles fixed scale. posite (14,000~2, 000) ; 12,000 on the fixed miles scale read 15 minutes. rotating scale read 60 on the fixed miles scale. 1. SET "10" —|z. OPPOSITE INDEX ON R/CIREY © Exeva, 800 FT, /M. i we 90. .% aayt ee x i! tit 2. WAS CH L.WoA a DEVece f bev“ FIG. 21 Given: An aircraft cruising at 9,000 ft. re~ quired 15 minutes to descend to 1,000 ft. 7 READ OPPOSITE |, "4Q" INDEX R/D 534 FT,/MINe Find: Rate of Descent Solution: See Fig. 22. Align 15 minutes of the rotating scale with (9,000-1,000) 8,000 on the fixed miles scale. Opposite the "10" index of the rotating scale read 534 ft. per minute on the fixed miles scale. ax oY i FIG, 22 FORMULAS FOR WIND TRIANGLE SOLUTION (SEE FIGURES 23 AND 24) Accurate dead reckoning navigation is dependent on the formulas engraved on the slide rule side of the computer. A detailed explanation is contained in Section II of this manual. ONING ¢ o REE Ome, a oe ep TN H H ser ve WIND << e152 COURSES AND HEAD! WHEN RECTIFYING REVERSE SIGNS S oe Se —— FIG. 24 FIG. 23 DEFINITIONS: TC-True Course WCA-Wind Correction Angle TH-True Heading VAR. +Variation DEV.=Deviation CH-Compass Heading MC=Magnetic Course CC=Compass Course To make good a 360 deg. True Course (Track) it is necessary to steer @ Com- pass Heading of 32 deg. (360-15-10-7) ~13- SECTION II - SOLUTION OF WIND VECTOR TRIANGLES Solution of wind vector triangles is accomplished on the face of the computer using the ro- tating compass ring and plotting disc. The wind triangle is drawn on the plotting disc. A vector wind triangle is composed of three vectors. A vector has magnitude and direction. The magnitude is usually expressed in miles per hour or knots. The direction of each vector is clockwise of true north. The three vectors dealt with in solution of wind triangles are wind direction and speed, true heading and airspeed, true course (track) and ground speed. Given any combination of four quantities, the other two can be determined, TRUE HEADING qaca 10s dmeash wea tear 5 INDEX TRUE HEADING- AIRSPEED LINE oe DRIFT ARC 1 A GROMMET (AIRSPEED INDEX) RADIATING LINES FOR TRUE COURSE (TRACK) AND GROUND SPEED COMPASS RING i TT FRAME AND aa j STRAP * i iy at 4,88 8 “Suge gat ar a y ae han, ONE ss LPs. “ldinte ee hl Ad 7 METAL SLIDE A A oe FRONT SIDE a 40 MPH TO is 300 MPH tt # BACK SIDE t 200 MPH TO 600 MPH WIND TRIANGLE SIDE OF COMPUTER RULES FOR SOLUTION OF WIND TRIANGLES General Information 1. True Heading and Airspeed is a vector quantity. It established one side of the wind tri- angle. True Heading is the direction in degrees from True North and Airspeed is the length of the line. 2. True Course or Track and Ground Speed is the second vector quantity. It establishes the second side of the triangle. 3. Wind Direction and Speed is a vector quantity and establishes the third side of the wind triangle. 4, Remember that True Heading and Airspeed is on the same line; True Course (Track) and Ground Speed is on the same line; Wind Direction and Speed is on the same line. \ -14= WW QA 5. Use the following order for plotting a wind triangle when possible: a. First, plot wind direction and speed. b. Secondly, plot True Heading and Airspeed. c. Thirdly, plot True Course and Ground Speed. 6. Always draw wind down from the grommet (center of plotting disc). Place an arrow on the outer extremity of the wind vector. 7. True Heading is always under the True Heading Index. True Airspeed is always under the Grommet. The center horizontal line of the slide (card) is always True Heading-Airspeed side of the wind triangle. The origin of the Airspeed line is on the extreme bottom of the slide center line and the other extremity 1s under the grommet. 8. True Course (Track) and Ground Speed are on the diverging (radiating) lines. The origin of the True Course line is on the extreme bottom center of the slide and the other extremity is along the diverging (radiating) line, terminating at the head of the wind arrow. The wind arrow is always on the Ground Speed. 9. The wind vector has its origin at the grommet (center of plotting disc) and the other ex- tremity of the line terminates at the True Course (Track) line. 10. Consider the airplane headed along the center line of the slide and flying towards the True Heading Index. If the wind arrow strikes the airplane on the right, it will drift to the left of the center line; if it strikes on the left, it will drift to the right of the center line. THE WIND BLOWS THE AIRPLANE FROM HEADING TO TRUE COURSE (TRACK). The drift angle is the angle between the Heading-Airspeed line to True Course-Ground Speed line. 11. True Course is the direction you want to make good. Track is direction you actually make good. 12. True Heading is the direction you point the airplane to compensate for wind drift. The angle the airplane is turned from True Course to True Heading is the Wind Correction Angle. The Wind blows the airplane from True Heading to True Course causing the drift angle. Both angles are equal but opposite in sign. 13. Aright wind correction angle (left drift) is added to the True Course to obtain True Head- ing. Conversely, a left wind correction angle (right drift) is subtracted. Right drift angles are found on the right side of the True Heading-Airspeed line of the slide and left drift angles on the left. 14, Similar corrections are made for Variation and Deviation as for wind. West variation and deviation are added to the True Course to obtain True Heading and conversely east variation and deviation are subtracted. ~~ SS ~15~ Formulae for correction of wind, variation and deviation to compute headings and courses: 1. To obtain Compass Heading (CH) from True Course (IC) by correcting for wind, variation and deviation apply the following rule: "EAST IS LEAST; WEST IS BEST". Note that the primary difference between the two formulae below is in the order of correcting for wind variation and deviation. + W.DEV. _ oy FWVAR om + WREY — EVAR. + W.DEV. W.VAR + RWCA _ ea = MC _ EDEV. = b. TC T EVAR. = co wea = 2._To obtain True Course (IC) from Compass Heading (CH) (Rectifying a Course) reverse the signs e.g., "EAST IS BEST; WEST IS LEAST". — w.pey. — WAR. _ — R.WCA b, o HY Wow = MH EAR = TH wea = TC — R.WCA — W.DEV. _ — WAR. b CHT twea = oC | epev. = MC 4 Evan” = TC c. Abbreviations: DEV.--Deviation CH--Compass Heading MC--Magnetic Course CC--Compass Course TC--True Course WCA--Wind Correction Angle TH--True Heading VAR. -=Variation Solution of Type I Wind Triangles P ig: Mt Find: yey True Course Xo Ground Speed ory é Given: Wind Direction-45 deg. Wind Speed=20 mph. True Heading-276 deg. True Airspeed-130 mph. \ Pe = Solution: See Fig. 1. 1. Set Wind Direction (45 deg.) opposite True Heading Index by rotating compass ring. 2. Move slide so as to place any convenient whole number under grommet (center of rotat~ ing disc). fe 3. Draw a line from the grommet 20 units (10 & ay spaces) placing an arrow at the end of the th, wind line. 4. Rotate Compass ring so that the True Head- ing (276 deg.) is opposite the True Heading Index. 5. Move slide so that the True Airspeed (130 oR pl = 30 Ps mph.) is under the grommet. Bae ve 2” xe * % or D 6. Ground Speed (144 mph.) is read at the end A HE Ka of the wind arrow along the speed circle. Hin % 7. Since the wind is from the right, the drift is Hifsas left (6 deg.). The end of the wind arrow is 6 ites spaces to the left of the True Heading-Alrspeed \] line. | 3 8. Opposite 6 divisions to the left of the True 7 J Heading Index, read True Course (270 deg.). | 7 a, att ty % ra 3 xo NN. ea la FIG. 1b -16- PRACTICE PROBLEMS FOR TYPE I. WIND TRIANGLES Exercise 1. Wind Wind = True True True Ground WIND No. Direction Speed Heading Airspeed Course Speed DRIFT {Deg.) (MPH) (Deg.) (MPH) (Deg.) (MPH) (Deg. ) 1. 90 30 350 120 ( ) x ) ) 2. 50 20 260 130 ( » ¢ ) ( ) 3. 270 25 180 125 { ) ) ( ) 4, 300 22 360 110 ( ) x ) ( ) K6 225 18 45 150 ( » ¢ ) ¢ ) 6. 360 30 100 135 ( ) x ) ( ) Wo 325 26 37 117 ( > ¢ ) ( ) 8. 180 27 90 128 ( > x ) ( ) 9. 360 12 299 130 ( ) « ) ( ) 10. 360 ll 86 92 ( » ) () Solution of Type II. Wind Triangles m 30 Given: Find: True Heading-100 deg. True Airspeed-135 mph. % True Course-112 deg. Ground Speed-144 mph. Wind Direction Wind Speed Solution: See Fig. 2. 1, Rotate compass ring so that True Heading (100 deg.) appears under True Heading Index. 2. Move slide so that True Airspeed (135 mph) appears under the grommet. & (True Course) from T.H. (True Heading). If & T.H. (True Heading) is greater than T.C. (True Course), the drift is left and if T.H. (True Heading) is less than T.C. (True Course) the drift is right, REMEMBER THAT YOU HEAD THE AIRPLANE INTO THE WIND FROM TRUE COURSE, IF YOU HEAD TO THE FIG, 2a RIGHT TO COMPENSATE FOR A RIGHT WIND, THE DRIFT WILL BE LEFT, AND THE TRUE HEADING WILL BE GREATER THAN THE TRUE COURSE, 3a. Alternate method of determining Drift. The T.C. (True Course) is 112 degrees appears on the right drift side of the compass ring with respect to the T.H. (True Heading Index) of 100 degrees. Accordingly, the drift is right. The wind arrow will appear on the Tight side of the drift arc. Determine the dif- ference between T.H. (True Heading) and T.C. (True Course). In this example it is 12 degrees. 4. Count off 12 deg. R.D. to the right of the True Heading -Airspeed line along ground speed circle of 144 mph. and place a cross. The line from grommet to cross represents the wind direction and speed. 5. Rotate the compass ring so that the cross appears on the True Heading-Airspeed line below the grommet. FIG, 2b & 3. Compute the drift angle by subtracting T.C. 7 6. Move the slide to any convenient whole number under the grommet. 7, Read Wind Direction (359 deg.) under the True Heading Index. 8. Read Wind Speed by counting the units (30 mph.) from the grommet to cross. PRACTICE PROBLEMS FOR TYPE Il. WIND TRIANGLES Exercise 2. Wind Wind Wind True True True Ground No. Direction Speed Drift Heading Airspeed Course Speed (Deg.) (MPH) (Deg.) (Deg.) {MPH) (Deg.) (MPH) 1. ») ¢ ) ¢ ) 80 120 90 100 2. i¢ ») ¢ ) ¢ ) 355 135 360 130 3. ») ¢ ) ¢ ) 45 118 38 128 4 ») ¢ ) ¢ ) 192 150 183 137 5. >) ¢( ) ¢ ) 95 165 103 178 6. ») ¢ ) ¢ ) 270 65 263 72 7 ») € ) ¢ ) 158 68 158 78 8. ( ) { 2 ¢ ) 87 78 87 63 9 > « df ) 285 85 270 94 10. ( ») «€ 12 ¢ ) 183 155 180 161 Solution of Type III. Wind Triangles Given: Pind: True Heading Ground Speed Wind Direction-260 deg. Wind Speed-50 mph. True Course (Track)-240 deg. True Airspeed-150 mph. Solution: See Fig. 3 1. Rotate compass ring to Wind Direction (260 deg.) opposite True Heading Index. 2. Move slide to any convenient whole number under the grommet (center of disc). 3. Draw a line down from the grommet 50 units in length to represent the Wind speed and place an arrow at its end (end furthest from grommet). TRUE LHEAD Ra FE aig amie’ 30 0 9 p> KP. ° ~ : % fa ve e He 3 %» NS = F 2 . ete ae % 4, Place the grommet on the True Air— speed (150 mph.) by moving the slide. 5. Place the True Course (240 deg.) opposite the True Heading Index and draw an arrow on the compass ring at this position. zazes 6. Obviously step No. 5 is incorrect y ! since the True Course has been placed tt E under the True Heading Index, but this y may be done to obtain a first approxima- tion of the True Heading (T.H.). Under this condition read the drift angle (9 1/2 deg. L.D.) at the end of the wind arrow. (See Fig. 3b). x ay fs he as % 2 F yy" Mahl 08 og Oh chy etatcalaatelulio’™ FIG, 3a -18- 7. A leit drift (9 1/2 deg.) is a right wind correction to the True Course (240 deg.) and is added. The second True Heading (T.H.-2) becomes 249 1/2 deg. Place the second True Heading (249 1/2 deg.) under the True Heading Index, which is the same as rotating the compass ring (9 1/2 deg.) to the left of the True Heading Index. 8. Obviously the second True Heading is incorrect since five degrees appears on the slide at the wind arrow. To be correct the arc drift value should be equal to the drift on the slide at the wind arrow. 9. Finally rotate the compass ring back and forth until the drift angle on the drift arc oppos— ite the True Course (240 deg.) is equal to the drift on the slide at wind arrow. This occurs at 6 1/2 deg. See Fig. 3c. 10. Read True Heading at the True Heading Index (246 1/2 deg.). 11. Read Ground Speed at the wind arrow (102 mph.). 7 10 0 N10 lg GE Sluts a> aay ae lyje RE a Ml op ei mM LAW go 20 ye o% age? ee My, \ x w ° ‘ : Soe My y RS % Se Ae! Py Se & # cs “Medlialiutuals ice 2, s. x a es M2 * by, _ ey « “eda ty eto FIG, 3b FIG, 3c PRACTICE PROBLEMS FOR TYPE III. WIND TRIANGLES Exercise 3. Wind Wind True True True Ground No. Direction Speed Heading Airspeed Course Speed (Deg.) (MPH) (Deg.) (MPH) {(Deg.) (MPH) L. 320 30 ( ) 140 260 ( ) 2 50 25 ( ) 120 250 ( ) 3 45 20 ( ) 130 270 ( ) 4 90 32 ( ) 150 180 ( ) 5 180 28 ( ) 125 90 ( ) 6 240 22 ( ) 160 300 ¢ ) a 50 30 ( ) 174 315 ( ) 8 270 27 ( ) 180 360 ( ) 9. 180 32 ( ) 92 45 ( ) 10 270 30 ( ) 120 90 ( ) -19- PRACTICE PROBLEMS FOR TYPE I, II, AND III, WIND TRIANGLES MIXED SERIES Exercise 4. Wind Wind True True True Ground Drift No. Direction Speed Heading Airspeed Course Speed Angle (Deg.) (MPH) (Deg.) (MPH) {Deg.) (MPH) — (Deg.) al, 270 32 ? 125 135 ? ? 2. 250 25 ? 140 90 ? g fo 92 22 183 135 ? ? ? 4, 184 18 350 105 ? ? ? $ ? a 285 150 270 165 ? 6 ? ? 180 90 188 78 2 Uo 155 15 ? 145 180 ? ? 8. 175 20 180 125 ? ? ? 9. ? ? 255 100 265 88 tf 10. 186 25 8 140 152 ? 2 ab tS 180 25 ? 192 270 ? ? 12. 350 ? 390 125 ? ? 8R 13. 270 ? 180 150 ? ? 10L 14, 270 25 ? ? 350 120 15. AS 20 ? 2 90 150 SPECIAL TYPE OF WIND TRIANGLES Given: Find: Wind Velocity-260 deg/50 mph ‘True Heading True Course-240 deg. True Airspeed Ground Speed~110 mph. Solution: See Fig. 4. 1. Plot Wind Velocity as in Fig. 3a. 2. Rotate compass ring so that the True Course (240 deg.) is opposite the True Heading Index ease and place an arrow on the compass ring at 240 deg. 3. Obviously step No. 2 is incorrect since the True Course has been placed under the True Heading Index but this may be done to obtain a first approximation of the True Heading (T.H.) dud FIG. 4a 4, Move the slide until the Ground Speed (110 mph.) is at the wind arrow. 5. Read the Drift Angle at the end of the wind = arrow (9 deg. left drift). 6. A left drift (right wind correction angle) is added to the True Course (240 deg.) to obtain the second True Heading (T.H.-2-249 deg.). 3 Alternate Method: Place T.C. 240 deg. oppos- ite 9 deg. on the left drift portion of the arc. 7. Move slide so that the Ground Speed (110 mph.) is at the wind arrow. 8. Obviously this triangle is still incorrect since a True Heading of 249 deg. minus 5 deg. drift on slide is not equal to the True Course of 240 deg. Alternately, the drift on the arc is not equal to the drift on the slide. sails FIG. 4b -20- 9. Finally rotate the compass ring back and forth until the drift angle on the drift arc opposite the True Course (240 deg.) is equal to the drift on the slide. This occurs at 6 1/2 deg. Note: At each trial setting move the slide so that the wind arrow is on the Ground Speed (110 mph.). 10. Read True Heading at the True Heading Index (246 1/2 deg.). 11. Read Alr Speed on the grommet (158 mph.) Note: See Page 23 for Practice Problems. DRIFT ON MULTIPLE HEADINGS Problems involving drift on multiple headings depend on the use of a driftmeter. Given: TH~1,-50; TAS-1,-150 MPH; DRIFT 6 DEG, L TH-2,-180; TAS-2,-140 MPH; DRIFT 5 DEG. R Find: 1. Wind Direction and Speed 2, Ground Speed on each Heading Solution: See Fig. 5 1. Place TH-1 (50 deg.) at the True Heading Index and TAS-1 (150 mph.) under the grommet. (See Fig. 5a) 2. Draw a track line on the plotting disc 6 deg. to the left of the card heading line. 3. Place TH~2 (180 deg.) at the True Heading Index and TAS-1 (140 mph.) under the grommet. 4, Draw a track line on the plotting disc 5 deg. to the right of the card heading line intersecting the first drift line. (See Fig. 5b) tit fe itt i & leads FIG, 4c 5. Read the Ground Speed (130 mph) on this heading at the intersection of the drift lines. 6. Rotate the compass disc so that the inter- section of the two drift lines are below the grommet on True Heading line of the slide. 7. Read Wind Direction (135 deg.) under the True Heading Index. (See Fig. 5c) 8. Read the wind speed (15 mph.) 9. Place TH-1 (50 deg.) at the True Heading Index and TAS-1 (150 mph.) under the grommet. (See Fig. 5d) 10, Read GS-1 (149 mph.) at the drift line intersection. ey io % Ke e in == z sai Seer. ra lige aoe = % IS = 4, a Hes & PIG. 5c ast ‘e 53 Se, sat lalate FIG, MULTIPLE DRIFT PRACTICE PROBLEMS Exercise 5. TAS TH-1 Drift-l1 TH-2 No. MPH DEG, Deg. DEG. 96 134 6R 46 145 259 1aR 169 10R 125 360 15R 30 10L 150 100 4R 200 tt) 140 360 7L 300 4L 120 270 3R 200 8L 160 85 2L 135 LOR + 120 45 0 90 10L Drift-2 Deg. 14L ONO hwonE HALF SCALE WIND TRIANGLES Find: Wind Direction Wind Speed Given: True Heading-180 deg. True Airspeed-240 mph. True Course-160 deg. Ground Speed-260 mph. Note: Solving this problem as a conventional type II problem will take you off the face of the plotting disc. To solve this problem with a Kane Mark VI computer, use half scale for all speed quantities. Caution-1; Do not half scale the angles. Caution-2: Since half scale is used, double the wind speed quantity when determined. Solution: See Fig. 6. 1. The difference between True Course (160 deg.) and True Heading (180 deg.) is +20 deg. 2. Since the airplane must be headed further to the right by 20 deg. from a True Course of 160 deg., the wind must be from the right. W.D. Deg. wW.S. MPH eee ee SE om G.s M ( ( ( ( ( ( ( ( Sd Wh oe pa ot an G.8.-2 MPH FIG. 6a See KE be =O) 3. A right wind correction angle (20 deg.) will cause a left drift. 4. Place True Heading (180 deg.) under True Heading Index. 5. The True Airspeed (240 mph.) half scale is 120 mph. 6. Place grommet over Airspeed (120 mph.). 7. The Ground Speed (260 mph.) half scale is 130 mph. 8. Where the speed circle 130 intersects the 20 deg. left drift Track line, place a cross. 9. Rotate the compass ring so that the cross coincides with True Heading-Airspeed line and is below the grommet. 10. Count wind speed units (45 mph.) down from the grommet. 11. Doubling the wind quantity (45 mph.) is 90 mph. 12. Read wind direction (273 deg.) at the True FIG, 6b Index. Exercise 6. Wind Wind Drift True True True Ground No. Direction Speed Angle Heading Airspeed Course Speed (Deg.) (MPH) — (Deg.) (Deg.) (MPH) (Deg.) (MPH) 1 ( ) ( ) ( ) 70 180 90 160 a @ ) ( ) ¢ ) 250 140 275 100 3 ( ) ( ) ¢ ) 18 185 360 200 4 ( ) ( ) ( ) 153 145 175 102 5 ( ) ( » ( ) 170 110 150 150 RADIUS OF ACTION WIND TRIANGLES Given: Wind Direction-310 deg. Wind Speed-20 mph. True Course out-90 deg. True Airspeed-120 mph. F. iy True Heading out 2. Ground Speed out 3. True Heading back 4, True Course back 5. Ground Speed back Solution: See Fig.7. een 8 ae R 7 8 2 ‘sig iB S i - 4 uty ulunlisds inte Pas TR ng lig =-23- 5. For the return ttip rotate compass ring