has my Confuser got it wrong?

[liam548]

I am working through the confuser and get to question 4 on nav and cannot get anywhere near the answers it gives..

I get the other similar questiones correct.

Given a Hdg of 138degreesT and a track of 141 degrees true, a TAS of 122Kts and a G/S of 101Kts what is the W/V?

A -16520Kts
B - 12030Kts
C -13121Kts
D -31121Kts


p.s how do I type a degree mark??

Liam

[Islander2]

Nope, C is the correct answer.

[RTN11]

Hmm, I'm with you, I don't get any of them. What answer is listed in the confuser?

PS, like this °

[Keygrip]

Ballpark figure in a rush, with no time (or interest, "lol") to calculate it, I would GUESS that a low cross wind component from the left of 138 degrees with a 20 knot headwind would give you C.

A is from the wrong side, D would be a tailwind.

To get a ° - on a desktop keyboard, hold down the ALT key and hit the numbers 0176 on the number keypad.

No idea why it only works for me on a number pad, not on the top row nor on a laptop.

Ask in the Computer forum?

[Deeday]

That's a good one, liam548, because it can be solved only if you are able to 'see' the vectors, over the wind computer, rather than applying a series of obscure steps (Keygrip's estimate makes a lot of practical sense, though).

You use the wind-down, right? So the HDG/TAS vector will lie on the centre line, starting from the origin of the fan lines and ending onto the centre dot; you can then set the HDG under the index and the TAS under the centre dot. The TR/GS vector will start from the same point, lying on a fan line 3° to the right of HDG, and end where it meets the circle equal to the GS. Clearly, the latter is also the point where the W/V vector ends - if the triangle is to be closed - i.e. it's the wind mark, which gives you wind direction and speed, by whizzing the wheel as usual.

By the way, the Confuser does get it wrong, sometimes, like when it says that the properties of a gyroscope depend on its rotational speed and centre of gravity. Yeah... that makes sense.

Deeday

[Islander2]

Some elementary trig actually shows the correct answer (to the nearest whole degree and knot) to be 124deg/22kts.

But these are very acute angles, and the wizz-wheel simply won't be that accurate.

To see why, using the confuser's answer for W/V 131deg/21kts to achieve a track of 141deg and a G/S of 101kts, the required heading and TAS are (to the nearest whole degree and knot): 139deg/122kts, i.e. only 1deg and 1kt difference from the data provided.

[India Four Two]

I agree with Islander 2 - 124.0º 21.8 kts.

In Firefox, you can use Edit, Special Characters for degree and other special symbols. I'm sure there would be something similar in other browsers.

[liam548]

thanks Deeday and all. Clearly the method im using to calculate wind on the whizz wheel does not work with the above example. I set heading the mark in using the track and g s then rotate the wheel until it lines with the centre line and read off the wind velocity. But it does not work with this example.

[miroc]

"To get a ° - on a desktop keyboard, hold down
the ALT key and hit the numbers 0176 on the number keypad.

No idea why it only works for me on a number pad, not on the top row nor on a laptop.

Ask in the Computer forum?"

The key-codes (as the operating system senses them) are different for
the numeric block keys and top row keys on an ordinary keyboard. The feature
works only for the key codes of the numeric keypad. Notebook's top row keys are
equivalent to top row keys o an ordinary keyboard - that's the point why it don't work.

[miroc]

>>By the way, the Confuser does get it wrong, sometimes, like when it says that the properties of a gyroscope depend on its rotational speed and centre of gravity. Yeah... that makes sense.<<

Actually, it is near. The properties of a gyro depend on:
- rotational speed
- the DISTANCE between the centre of gravity of the CROSS SECTION of the wheel and the axis
- mass of the wheel

[Deeday]

No, sorry, it's not near, it's plain wrong, and what you say doesn't make much sense either: the centre of gravity of any cross section of a gyroscope always lies on its axis, otherwise the vibrations would blow up the whole thing.

The properties of a gyroscope depend on two things:
- Rotational speed
- Moment of inertia (http://en.wikipedia.org/wiki/Moment_of_inertia) (aka angular mass)
It's as simple as that.

Deeday

[PompeyPaul]

I always wondered:

1. Why is it called a confuser when it's attempting to do the opposite ?
2. Why is it written and produced by some Thai character it's for the UK CAA exams ?

[IO540]

I imagine the author is a pseudonym because the questions are very close to those in the real exams.

It's an excellent book, essential for learning the silly word plays which the exams are known for.

[miroc]

Deeday, you are right, it has to be the center of gravity of the "upper half" of that cross section. By rotating this shape around the axis you get the "wheel".

[BigEndBob]

Don't worry its in keeping with the real exams, even the CAA get the answers wrong to some of their questions?

[Gertrude the Wombat]

No idea why it only works for me on a number pad
By design.