Rule 39C

Mental DR Calculation (One in 60 Rule)


MENTAL DR NAVIGATION

Quick mental methods are extensively used in navigation and can also be applied by aviators in checking flight plans or estimating alterations of heading. They are particularly useful in estimating final airspeed changes to make good a timing point, when delay might leave insufficient time for a change within the aircraft’s speed range to be effective.

The mental solution of problems ac can be conveniently dealt with under the following headings:

  • Estimation of distances on maps.
  • Estimation of direction on maps (tracks and track errors).
  • Estimation of TAS (from RAS and from Mach number).
  • Estimation of wind effect (on speed and direction).
  • Estimation of groundspeed between fixes.
  • Estimation of ETA.
  • Estimation of heading corrections (to regain a track or achieve a turning point).
  • Estimation of speed adjustments to achieve an ETA.

 

 

ONE IN SIXTY RULE

The method of estimating direction which has the most general application is use of the one in sixty rule. This is based on the fact that an angle of approximately 1˚ will be subtended by a distance of 1 nm (or distance unit) at a range of 60 nm (or distance unit). The practical application of this rule is in a right-angled triangle. If the length of the hypotenuse is 60 units, the number of the same units in the length of the side opposite the small angle will be approximately the same as the number of degrees in the small angle.


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PROOF OF One In Sixty Rule

For a circle, Arc / Radius = Angle subtended at the centre

l / r = θ radians (by definition of Radians)

θ radians = l / r

Multiplying both sides by 57.3,

θ radians x 57.3 = l / r x 57.3

θ˚ = l / r x 57.3 ≈ l / r x 60 ( because 2 Π radians = 360 deg)

Therefore,

l / r x 60   =  θ˚

 

Track Error

Track error is the angular difference between Track and TMG.

Applying one to sixty rule –

TE =       Distance off Track x 60

Distance Gone

 

Closing Angle (CA):  The angular difference between Track required presently to Track calculated.

CA    =    Distance off Track x 60

Distance to go

 

Alternation of Heading to Regain Track (Tracking out):

To regain track, turn double of the TE towards the track and fly for the same time in which aircraft has gone off the track.

Course Correction, Double Track, Langley Flying School

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Alternation of Heading to Destination (Tracking in):

      Calculate TE and CA from the present position. Alter heading towards the destination by the amount TE+CA.

 

Estimation of Range by Change of Bearing:

Turn heading such that radio station comes to about abeam position. Note the time and bearing once and then again after a small change of bearing. Time to station (t) may be calculated by

 

Time in seconds for change of bearing

t =                   ———————————-

Change of bearing

 

Distance to station may be approximately calculated by time ‘t’ and G/S assessed by perpendicular P/L.

 

Note:     Change of bearing should be kept small.

 

Example:  An aircraft flying on heading 005 (M) observes RB of a Radio Station 080 at 1002 hrs. The next RB observed is 088 at 1004 hrs. Find the distance and ETA to the radio station, if  G/S estimated is 150 K.

Solution:

2×60

Time to Station =                   —–              = 15 min

8

Dist to Station = 15×150/60=37.5nm

ETA to Station  = 1004+15=1019 hrs.

 

Estimation of Range by Doubling the Angle at the Bow:

 i) 45º Method: When radio station is expected to be crossed abeam, note the time when RB is 045 and again when it is 090. The distance travelled in the time elapsed is the range of the Station. RB will be 315 and 270 if the station is to the port of the track.

double

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Example:  An aircraft observes RB 045 of a radio station at 1115 hrs. At 1121 hrs the RB changes to 090. Find the distance to the station at 1121 hrs if the G/S estimated is 140K.

 

Solution:  This is case of doubling the angle at the bow.

Change of bearing 45º in 6 mts

Dist to Station = (6/60)x140 =14 nm at 1121hrs

 

ii) 60º Method:  Note the time when station is 060(R) and again at 120(R). The distance travelled during the elapsed time is the Range of the station from later position. RB will be 300 and 240 for the station to the port of the track.

Revision of ETA:

Calculate the G/S from a perpendicular P/L. Revise the ETA for the distance to go to destination or next Reporting point.

To Maintain ETA:  Time may be approximately adjusted by adjusting speed by the following formula.

 

G/S x time required to be gained or lost

—————————————–                                =           (Speed adjustment)K

Time to go                                                                                 (Increase/decrease)

 

Example:  An aircraft flying at a G/S of 180 K finds himself 100 nm from destination at 1151 hrs. How much speed adjustment is required to make good an ETA of 1220 hrs.

 

Solution:     ETA with 180K              = 1151+33 = 1224 hrs

Time required to be gained = 4 mts

Time to go                 = 1220-1151 = 29min

Increase speed by = (180 x 4)/29 = 25K

 

Mental Calculations for Revision of ETAs:

 

Assess the G/S by using a known feature or a perpendicular P/L. Drop last figure of the G/S to get distance travelled in 6 mts and then for 3,2 or 1 minute. Revise ETA mentally with the information available and distance to go to the next reporting point/destination.

Estimation of TAS:

TAS = RAS 1.75% of RAS per 1000′ ft. For mental calculations, take approximately 2% of RAS, multiply it with thousandth figure of altitude and add to RAS for sub-sonic aircraft upto Flight Level 200.

Estimation of Cross Wind:

 

 

Example:  Track 150 (T) TAS 120 k W/V 105/20 k

Estimate G/S & drift

Solution: W/V is 45º. Headwind  effect on speed is 70%

 

70% of W/S = (70/100)x 20 = 14k :. G/S 120-14 = 106k

W/S X 60   20 x 60

Effect of drift: Max drift = ——– = ——- = 10º

TAS       120

At 45º H/W effect is 70% of Max drift (10º) ie. 7º

 

Estimation of TAS

The following methods are however, simple and sufficiently accurate for most mental DR purposes:

  • RAS + RAS/60 X Alt in 1000’s of feet = TAS.
  • RAS + 1.75% RAS X Alt in 1000’s of feet = TAS.

TAS from Mach Number.

This method, which assumes an International Standard Atmosphere, gives reasonably accurate results up to 40,000 feet and for speeds up to Mach 1.

  • Multiply the indicated Mach Number (to two decimal places) by 600.
  • If the indicated Mach No. is 0.60 or less, add 1 knot for each 1,000 feet below 25,000 feet, or subtract 1 knot for each 1,000 feet above 25,000 feet,

OR

If the indicated Mach no. is greater than 0.60, add 2 knots per 1,000 feet below 25,000 feet or subtract 2 knots per 1,000 feet above 25,000 feet.

 

Example           (i)         Ht = 13,000’, MN = 0.43, TAS = (0.43 x 600) + (12 x 1) = 270 knots

(ii)       Ht = 32,000’, MN = 0.88, TAS = (0.88 x 600) – (7 x 2) = 514 knots

Deviation from ISA.   In case ISA conditions do not exist then the deviation from ISA is added algebraically to the value outlined in para 12 above to cater for non standard atmospheric conditions, Thus, when prevalent conditions are warmer than ISA, the ISA deviation is added and under conditions colder than ISA the ISA deviation is subtracted.

 

Example            Ht = 32,000’, MN = 0.88, OAT = -32 deg

 

TAS (under ISA conditions) = (0.88 x 600) – (7 x 2) = 514 knots ISA deviation = + 17

Corrected TAS = 514 + 17 = 531 knots

 

TAS in Climb or Descent

For altitudes up to 20,000 feet, the following formulae may be used to determine the altitude at which TAS may be calculated from RAS to give a reasonably accurate mean TAS for the climb or descent.

(a) Descent or Climb at a Constant Rate. The altitude at the lower level + Half the change of altitude

Example: 

Constant ROC 2,000’ to 20,000’, OR, Descent from 20,000’ to 2,000’,

Mean altitude for TAS calculation = 2,000’ + ½ 18,000’ = 11,000’

Decreasing ROC.:   The altitude at commencement of climb + two-thirds the change in altitude

Example:

Decreasing ROC 2,000’ to 20,000’, OR, Descent from 20,000’ to 2,000’, mean altitude for TAS calculation  =  2,000 + 2/3 x 18,000 = 14,000’

–        For altitude above 20,000’ these approximations become very inaccurate, and mean TAS should be calculated directly from the data in the ODM for the ac, or from simplified tables based on the data.

Estimation of Groundspeed Between Fixes

It is often possible to avoid the necessity of calculating groundspeed between fixes by working in time and map distance terms rather than in terms of speed. There will however be times, as for example when changing map sheets, when it is more convenient to use the groundspeed to revise an ETA. Groundspeed can be calculated by fractional proportion, increasing or decreasing the time to a convenient whole number fraction of 60 minutes, and increasing the distance in proportion. For example:

  • 5 minutes to cover 23 nm is equivalent to 6 minutes to cover 23 – (23/13) nm or 6.0 mins to cover 21 nm approx, which gives G/S 210k approx.
  • 5 mins to cover 87nm is equivalent to 15 mins to cover (87- 1 ½ x87/16 ½) nm.

i.e. (87 – 1 ½ x 5), i.e. 79 ½, which gives G/S 4 x 79 ½ = 318 approx.

Estimation of ETA

ETA can sometimes be estimated without calculation of groundspeed. The two methods commonly used are:

Modifying Flight Plan Times by Proportion. For example, suppose an aircraft arrives at a checkpoint after 18 mins instead of the expected 20 mins, thereby gaining 2 mins in 20 or 1 in 10. If 50 mins of flight plan time remains, the aircraft can be expected to gain a further 50 x 1/10 = 5 mins. Thus the aircraft will arrive 7 mins ahead of the original ETA.

Estimation of Leg Time to Go by Proportion. For example, suppose about two inches (on the map) of a leg has been covered in 11 mins. This distance is stepped off along the remaining track to go, using a thumbnail against a pencil. If the distance to go is estimated to be three and a third times the distance gone, the time to go will be further 37 mins.

When the estimation of ETA by the use groundspeed is necessary, a fractional proportion method is used as in para 20, except that for ETA problems it is easiest to increase or decrease the groundspeed to a multiple of 60 knots, adjusting the distance in proportion. For example:

  • G/S 287 k to cover 98 nm, is equivalent to 300 k to cover {98 + (1/20) x 98} nm i.e. 103 nm, Time = 103/5 = 20.6 mins.
  • G/S 374 to cover 205 nm, is equivalent to 360k to cover (205 – 205/27) nm, say (205 -205 / 30) nm, i.e. 198 nm, therefore time 198/6 = 33 mins.

Quite rough approximations can be used in the method with only minor errors being introduced.

 

 

 

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