CPL(A) - Critical Point Question

[Infinity Land]

Just been trying to help a mate out with a critical point question. The examples given in his book are the simple A -> B type ones but how does one calculate a CP when given a situation like this:

http://users.tpg.com.au/adslsttq/CP.jpg

The Critical Point from Town A to Aerodrome 1 is point CP (Assume nil wind conditions here).

What happens when we have flown past this point and we now wish to calculate a new CP to our destination (Aero 1) but which takes into account Aero 2?? - Does this seem overboard.....do many Pro pilots out there worry about CP?

The formula that we use is (from the book):

D x H
--------
O + H

where:

D = Distance from A to B
H = Ground Speed to the departure point.
O = Ground Speed to the destination point.

Regards


IL

[john_tullamarine]

Many years ago, I endeavoured to find a simple, closed form solution to the off-track problem for both CP and PNR .. eventually giving up in frustration .. not to say they don't exist .. just that I wasn't able, at the time, to tie them down.

My suggestion to the customer at that stage was to use a simple prayer wheel iterative solution and then interpolate to find an acceptably accurate answer. Alternatively, and this is what we did, one can set up an equivalent graphical solution a bit like a performance chart carpet which makes the slave labour for the pilot a tad easier .. doubt that I have that original solution on file anywhere these days.

If the system now is sticking it into the exams (?), then I guess that there is a simpler solution somewhere ... have to have a revisit in the next day or two if I get some spare time.

[marreeman]

Infinity, you have to draw a line between aero1 & aero2, pick the midway point between these two aerodromes, then at 90 degrees to that line project a line to your original track & where these two lines meet is the cp in nil wind, no formula is needed because this is an off track cp question unless a wind is used.

Hope this helps:ok:

[Arm out the window]

Same as mareeman's answer for nil wind, and there is a pretty simple graphical addition to that to account for wind.

If some computer head could enlighten me as to how to post a 'Paint' picture on here, I'll put it on for you and all will be revealed!



You need to upload the image to any of a number of free hosting sites - perhaps here. (http://www.freeupload.net/)

You can then use the IMG function to display the image on PPRuNe.

Alternatively, email the graphic to [email protected] as a JPG and I'll display the image for you.

Woomera

[Woomera]

Posted on behalf of Arm out the Window:

http://img20.exs.cx/img20/785/OfftrackCP.jpg

Here’s a good graphical solution courtesy of our P3 mates with too much time on their hands – I think this is what they do after all the frozos are eaten. It looks tricky at first but is really dead-set simple.

Explanation: you are flying from A to B (naturally). C is an off-track alternate.

Draw a line AC, and also CB.

Using a set of dividers and a protractor, find the midpoint of both those lines and draw in lines running at right angles to them (perpendicular bisectors, for the mathematically inclined) – these cut your track at d and e. In nil wind, the job is done, because e, for example, is the equi-time point to go to either C or B. If you have passed e, then it’s quicker to go to B.

Wind is fairly easily included. Looking at point e again, all you do is work out the nil wind time from e to B (or C for that matter, same thing). Then, extend a line out into wind using wind speed, for that amount of time. For example, say it’s a 20 kt wind and the still air time at your cruise speed from e to B or C was 30 minutes, you would draw a line out from e into wind for 30 minutes worth of 20 kts, or 10 nautical miles.

This gives you point Q.

Now, project a line back to your track parallel to the original perpendicular bisector, which gives you point Y.

Y is the equi-time point for going to C or B taking into account the wind.

Easy once you’ve done it a couple of times.

[Infinity Land]

Hey Arm out the Window,

What a great response - thanks a lot for that......prolly apply that to my PNR calc's as well.

Cheers mate.....


:ok:

[Tinstaafl]

That was in the SCPL/ATPL Navigation syllabus & exams 15 years ago.

[4dogs]

AOTW,

I think point X is derived by paralleling the normal to AC - not as drawn where you have paralled the normal to CB.

Unfortunately, I have to go elsewhere for the moment, but perhaps one of you young mathematicians might want to consider if the equitime points X and Y as proposed are the geometric equivalent of calculating the two equitime points using the known wind for AC and BC respectively and then finding the intercepts of the normal projections onto AB.

I suspect it must be very close and can more easily be done as a right angle triangle solution without the need to plot.

Stay Alive,

[Arm out the window]

4dogs, yes, definitely! Sorry about that, it's my stuff up in producing the picture.
The 'PX' line should go back to it's section of track parallel to the perpendicular bisector that runs through D.

D'oh!



If you want to change the drawing, email the new drawing to me and I'll replace it in the post.

Woomera

[john_tullamarine]

I'd be interested in seeing if anyone comes up with a closed form solution without needing any graphical intermediate process to determine a solution. Having wasted a few hours on the subject many years ago, if no-one else knows a solution, or wants to have a go, I'll revisit it over an evening's coffee sometime.

[Chimbu chuckles]

JT what about:

Safe endurance avail x ETI->desto/ ETI->Desto+ETI Desto->Altn

[Tinstaafl]

Considering that the wind effect will change as it moves 360 deg around the nil wind position, I suspect Pi will be involved in mathematical solution.

[Captahab]

A fairly simple (rough) rule is to find the mid-point and then move the CP into wind by the percentage that the actual wind would be/is of your TAS.

ie. TAS 300
H/W 30
= 10%

Lets say mid point is 200 nm from each landing point, 10% of 200 is 20, move the CP 20 nm along track into wind.


Hope that makes sense.

[Arm out the window]

That sounds like a good one Cap, easy to apply in flight which is the main thing, and probably fairly accurate too.
Trying to work these things out to the nth degree is a bit over the top, because by the time you've had a problem requiring a diversion or whatever, panicked a bit and thought about it, 10 nm is going to make stuff all difference anyway.

Having said that, it would be an interesting exercise as John says to come up with a hopefully simple and elegant solution so you could plug a few numbers in and get it exactly, if there is one.

I'm thinking there must be some fairly straightforward way to include head / tailwind component on both the 'keep going on track' option and the 'go to off track alternate' option that might result in a simple formula; must have a think on it. I guess given that the application of wind moves the CP along track, there will have to be some kind of iterative stuffing around to do it really accurately.

[Arm out the window]

Further to the last, I came up with a seemingly reasonable solution to working out the wind-adjusted CP by plugging numbers into a formula. I guess this would have been the re-invention of a wheel that's been done many a time before, but not sure. See what you think anyway.

1. Find the nil wind CP.

2. Find the components of head or tail wind along both legs from the nil wind CP (e, in the picture above) to B and C.

3. Assumption number 1 : adjusting the CP for wind won't significantly change the wind effect on the legs to destination or alternate.

4. By definition, the time from CP to B or C is the same.
Time = distance/velocity, so using TAS adjusted for wind along each leg and making dist/vel for each leg equal to one another, then doing some simplification, you get the ratio of distances for the legs; eg if you've called the distance from Y to B d1, and from Y to C d2, you get d1/d2 = some number.

5. Assumption number 2 : The sum of distances of the two legs to destination or alternate will be approximately equal to twice the distance from the nil wind CP to dest or alternate.

6. Now you have d1/d2 = some number, and d1 + d2 = some number. Two equations and two variables, so solving for d1, you get the distance from wind adjusted CP to destination.

I did a few test calculations with this using various leg distances, TAS's and wind directions / velocities, and it comes out to be pretty close to what it should; within 2 or 3 minutes of flying time each way which ain't bad.

Sorry, didn't get a picture done this time, but hopefully it makes some kind of sense!

[Counter-rotation]

Let me stick my beak in here and add something important. As has been said, working these things out to the nth degree is really an exam "thing" not a real life flying "thing".
If you're faced with a situation and are weighing up alternates, there are other important considerations (terrain, services, runways, etc.) Have a look at CAO 20.6 for some other practical considerations.
To CPL candidates I would say: Exams are an important test of your ability to determine a CP should you have to. Your descision in a real situation will rarely be just about distance A v's distance B. In the air, beware of getting caught up in the numbers too much.
Obvious I know, but I thought it worth a mention.

;) C.R.