Hi Richard and Bob

Just wondering if you could show me an example on how to work out heading and ground speed using the CR-3 calculator as I am pretty sure the example shown in the PPl book in page 3.3.6 is not the CR-3

Kind Regards
Mark

Hi Mark,

As a start, your best bet would be to have a go at it using the instruction in the CR3 manual. The CR3 is a different beast to the E6B but still quite straightforward to use. Here's a link to the manual if you didn't get one with your CR3:

www.jeppesen.com/download/misc/crinstructions.pdf

The bit you need is on page 37 of the guide but it also includes information on how to solve all sorts of different problems using the CR3.

If you get stuck just shout.

Cheers,

Rich

Thanks Rich, will check it out now and if i get stuck i will let you know

Kind Regards
Mark

PPL 3.3.6

Given

FPT: 120 M

WIND: 180 T at 25kt

Variation: 10 E

TAS: 140

1. Using the wind side of the wheel. The first place I start off is rotating the TAS marker on the middle wheel to the given speed to the numbers on the outer wheel. The TAS marker should line up with 140 on the outer ring.

2. Using the directional wheel, turn to the given WIND (180 T) direction and align it with TC (True Course). Apply the variation 10 E. (10 E on the wheel is displayed as 170M. E & W are displayed under the directional wheel) Align the 170 with TC, drop down to 25 (Large numbers) and mark the speed on the wheel with a pencil.

3. Now that you have your mark, turn the directional disc to 120M so that it is aligned with TC.

4. Reading off the large numbers. Headwind = roughly 16-17kt and Crosswind = 20kts

5. GS = TAS - Headwind (As stated on the wheel. Headwind (-) Tailwind (+) 140 - 17 = 123kts GS

6. To get the heading. We found 20kts crosswind to the right. Go to the 20 on the outer wheel and match it up with the closest degree below it. In this case it is 8 degrees.

The crosswind is to the right so we add (As stated on the wheel) to 120M and we get 128M.

It's important to work using everything in magnetic or it won't work.

Hope this helped.

P.S. The Jefferson manual always confused me. If you don't know how to do IAS to TAS. There are two ways of doing it on the wheel.

Hi Rich,

Could you please help me to explain how in the PPL book, practice exam question #51, using the above technique, you get the answer of an amended heading of 087deg and 123kts?

I have tried it using the above technique and cannot get 123kts. Unfortunately the PPL book doesn't have a chapter on doing this with a CR3 (for future books maybe?) and the actual CR3 instructions are, well kinda weird.....There instructions also don't give the answer of 123 nor the amended heading of 087 because the instructions are asking me to add 10deg rather than subtract 10 (east is least) so now I'm doubly confused on how you get that answer.

Any help would be most appreciated. Exam on Wednesday.

Hi Matt,

When you do this problem, ensure you keep track of whether you are working in degrees magnetic or degrees true and don't mix them up. We are given wind from an area forecast. If wind is written down (e.g. such as in an area forecast, a TAF or a METAR) it will be given in degrees TRUE. Compare this to winds which are spoken by ATC, such as in an ATIS. These are given in degrees MAGNETIC.

So, we have a true wind direction of 150^o at 45kt and we also have a true heading of 080^o. We might as well keep things simple and work the answer in degrees TRUE and convert to magnetic at the end.

1) Set up the CR3 for the TAS of 145kt by aligning the TAS marker with the 145 line on the outer scale.

2) Set the true course of 080 (on the green disc) against the TC marker on the inner scale.

3) Find the radial line for 150 degrees and put a dot on that line which intersects the arc for 45kt. You can find the start of the arc by looking for 45kt on the vertical axis and following the arc lines around until they intersect the radial line for the wind of 150^o.

4) From the dot, read straight down to the horizontal scale for a right crosswind of about 42kt and a headwind of about 15kt.

5) Now, take a look at the heading corrections on the inner scale where they align with crosswind speeds on the outer scale. Find 42kt on the outer scale and you can see we need a heading correction of about 17^o. It is a right crosswind so we need to turn right to counteract it. Therefore our new heading should be (080^o + 17^o = 097^o T).

6) Since our crab angle is more than 10 degrees we need to work with what Jepp calls the "Effective TAS" and not the actual TAS when we calculate the ground speed. Looking at the effective TAS scale next to the TAS marker, we see a TAS of 145kt with a crab angle of 17^o corresponds to an effective TAS of about 138kt. We can now subtract the 15kt headwind from the effective TAS to get our groundspeed of: 138 - 15 = 123 kt.

All well and good so far but the question wants the answer in degrees magnetic. As you already said, "Variation East, Magnetic Least" applies. We are told there is 10^oE magnetic variation, so our magnetic heading will be 10^o less than the calculated true heading. 097^o - 10^o = 087^o.

The answer I get is 087^oM with a groundspeed of 123kt which is option (a).

Cheers,

Rich

P.S. Good luck with the exam on Wednesday!

That was the million dollar dot point, effective TAS! Thanks for explaining it clearly, I missed understanding of it in the CR manual.

Done this question so many times I know the values off the top of my head. Hope I get that question in the exam cause I won't even have to work it out haha

Many thanks again, you men should get medals.

M

Just to let you all know I passed the PPL exam!

Many thanks for your help and hope you guys have a productive new year.

Matt

That's fantastic news Matt. You certainly deserve it. Now you really can have a happy Christmas!!! All the best for the new year.

Bob

Well done Matt! Congratulations All the best for a merry Christmas and fantastic 2013 to you as well!

Cheers,

Rich