MDR Question

[Cricket23]

Hi,

I was 'playing' with MDR and am happy with calculations of around 90 degrees off, but things start to go awry with bigger numbers.

For example, let's say the wind is 040/30 and I want to divert from a heading of 125 degrees to a track of 185 degrees. TAS is 95kts and I make the Max Drift 18 degrees.

My MDR calcs tell me that the wind is 145 degrees (185-040) and therefore I should apply max drift, but I know that this is incorrect.

What am I doing wrong?

[tmmorris]

I'm not sure if I can explain this but I'll have a go...

You need to assess whether the wind is a head or tail wind. After your diversion it's now a tailwind. The angle you need to know is therefore not 185-040 = 145 deg, rather 040-(185-180) = 035 deg. In other words you have a wind 035 degrees left of a full tailwind. (You can do this on the DI: turn on to 185, then check the angle between the reciprocal (the number at the bottom of the DI) and the wind. You'll find 005 is at the bottom of the DI so the angle is 035 deg. In fact, always start a diversion by turning onto the track so you can build a mental picture of your situation better.)

Therefore your drift angle will be 3.5 sixths of your 18deg max drift, i.e. 10.5deg. The wind is from your left butt-cheek, so it's pushing you right 10.5deg; so your heading will be 174.5deg (call it 175!).

On the other hand, you will have (9-3.5)=5.5 sixths of the tailwind adding to your groundspeed, or about 27 knots, so your groundspeed should be 122 knots.

I haven't checked these on a whizz-wheel (I'm at work) so I'd be interested to know how accurately those measure up - that's the method I use in the air!

Tim

[High Wing Drifter]

Just a minor point, your Max Drift should be 19 deg.

I find the MDR technique is within a couple of degrees and a couple of knots of the whizzwheel, and it is much quicker creating PLOGs mentally...unless you're drunk that is :E

[Miserlou]

My method is to find the crosswind component which in this case rounds to 15kts. (35 degrees multiplied by 0.5 (the sine of 30)).
Then divide this by the sixty factor of your airspeed. 95 is a smidge over 1.5 times 60. 15 divided by 1.5 makes a wind correction angle of 10 degrees.

Heading to make good the 185 track=175 degrees.

[Miserlou]

As a point of interest and occasionally useful, one can also go the other way round and work out the approximate wind from your airspeed, heading and track made good.

[EGTC]

The calculations shown so far seem quite accurate, and would perhaps be difficult to do mentaly while in the air, for a diversion. This is the way I would have done it in the air, not that any particular way is right or wrong.

With a TAS of 95 knots, using the 1 in 60 rule the max drift is approximately 2/3 of the wind speed, so 20 degrees. The wind is 35 degrees from your track, so using the clock rule, the drift will be approx half of the max drift, so 10 degrees. The drift will be to the right, so taking 10 degrees off your track gives a heading of 175 degrees.

Then for the timing. With a TAS of 95, if there was no wind the ground speed would be approximately 1.5 miles per minute. So first work out the time it would take if there were no wind, which is the distance divided by 1.5 (I do the distance X 2, then divide by 3). Now you have a time for 0 head/tail wind conditions. As the wind is 55 degrees from the perpendicular to your track, using the clock rule again you have approximately a full tail wind of 30 knots. 30 knots is approx 30% of your TAS, so decrease the 0 wind time you worked out by 30% and then you have your time.

I know theres alot of rounding and approximations involved in this, but for making unplanned diversions in the air without a wizz wheel and while flying the plane, it simplifies things alot, and it works.