# Clock Code Drift Calculations

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Hi,

Any ex-RAAF PLTs|NAVs|QFIs who can put down some simple words on using the "Clock Code"?

I teach Mental Ded Reckoning (MDR) to my pilots (GA/RAAus) and need to refresh my memory

Ta,

Flight Instructor

--------------------------------

" From a Senior Aviator (40 years RAAF/Airlines)":

# NEVER run out of fuel.

# NEVER crash untintentionally.

# ALWAYS have an out..be creative..but ALWAYS have an out!

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Not ex-RAAF or any of the above but when I was taught we used the clock code. It is pretty simple really.

Converting winds into a crosswind component can be done by calculating the sine of the angle of the wind. It is easier to workout with an example.

If the wind is coming is coming straight at you - i.e. 0 degrees - then clearly there is no crosswind component:

sin(0) * windspeed = 0 * windspeed

If it is coming at you at say 15 degrees then the crosswind component is:

sin(15) * windspeed = 0.25 * windspeed

At 60 degrees:

sin(60) * windspeed = 0.86 * windspeed

Calculating the sine of a number in your head is a bit difficult on the fly however so using the clock code you can work out good approximations.

The clock code theory says that the sine of a number is approximately equal to that number divided by 60. For angles greater than 60 the errors begin to get too large and you just assume the crosswind component is equal to 1.

Using the same examples above we have:

0/60 = 0 -> crosswind component = 0 * windspeed

15/60 = 0.25 -> crosswind component = 0.25 * windspeed

60/60 = 1 -> crosswind component = 1 * windspeed

You can see the error getting larger.

The 1 in 60 rules are also handy for track errors and work on a similar concept of 'near enough is good enough'. The idea is that if you stray off track by 1 mile in 60 then you have to correct your course by 1 mile. Imagine a right angle triangle, your planned track is one straight edge that is 60 miles long beginning with your origin and ending at a point we will call your positive fix point. From this positive fix point extend another line at right angles to your planned track for 1 mile, the end of this line represents where you actually are. Draw a third line from your origin to your actual location, the angle between your planned track and your actual track is the error. Now, recall your trigonometry from school.

The error angle can be worked out as tan-1(1/60) = 0.95 - given that we strayed off track to begin with and can't hold a heading anyway, this is near enough to 1 for us mere pilots! So let's extend this a little further.

Say you have strayed from track by 5 miles over the course 30. You can work out your track correction as follows:

5/30 = x/60 -> x = 10

Which is a lot easier than doing tan-1(5/30) = 9.46 in your head!

Thus, to parallel your planned track you have to adjust course by 10 degrees. If you have 60 miles to go then you have to adjust a bit more.

5/60 = x/60 -> x=5

So you have to correct by a further 5 degrees making the total course correction 15 degrees.

This is why I find it handy to mark my halfway point on charts. Say 30 miles is the halfway point and you have strayed by 5 miles. The total correction will simply be twice the error so far.

5/30 = x/60 -> x=10

2 * x = 20 -> total course correction is 20 degrees.

Maybe not what you were after but I hope it helps. :)

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A'sick,

Ta and it has helped.

FI:big_grin:

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How about visualising the degreed of crosswind as a clock face or similar. Make a mental picture of a line at the relative wind angle and compare the length of the line to the distance off at the end of the line. Trigonometry in your head.

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Drift correction

Having found the xw component (I use the 0.7 multiplier at 45degrees varying by 0.2 for each 15degree change - easier than it sounds!) divide by TAS in nm/hr to get WCA. eg17kts across and 120Kts = 17/2 or 8.5degrees WCA:loopy: