Wind - air molecules in motion

From meteorology we know that:

forces which drive wind or affect it are the pressure gradient force, the Coriolis force, buoyancy forces, and friction forces. - Wikipedia

The good news is, we now know where wind comes from...As you prepare to write the PPL exam however, you will realize the theory is good, but TC will expect you to know the application of that knowledge. For good reason.

You can expect questions about wind dynamics including; direction it is expected to blow, how the time of day affects it, cross-wind components during takeoff and landings, change with altitude, hazards it poses and more...

Here are a couple exam questions you can expect....

For a take-off on runway 31 with the wind from 270° at 20 kt, the aircraft would be
subject to head wind and cross-wind components respectively of
(1) 20 and 15 kt.
(2) 15 and 13 kt.
(3) 15 and 20 kt.
(4) 13 and 15 kt.

To solve this question you will need a wind component chart (provided below - it will be provided to you during the exam) and some simple math and most importantly common sense...

Let's try it then.

If the runway is 310 and wind is coming from 270 we know that the angle of the wind direction to the runway or your flight path is the difference of the two, hence 310 minus 270 or 40°. We also know the wind speed is 20kt, so that is all we need to figure out the answer...Looking at the chart, find the angle of 40° and slowly slide down that line until you get to 20, which in this case is the wind speed. From that point draw a horizontal line and a vertical line that will intersect that point until you can read both the head and cross wind components on both axes. Click the chart to see result.

The answer: 15kt and 13kt

Simple, right?

Here is another rule of thumb, that is good to remember related to the wind component. It works well with wind angles of 15° increments (15, 30, 45, 60 and 75).

What is important to notice is that the head and crosswind component ratio is exactly the same at 45° at 2/3 and then is perfectly symmetrical. It should therefore a bit easier to remember. Just remember 1/4, 1/2, 2/3, 9/10 and 19/20.

Wind Angle	Crosswind	Headwind
15	1/4	19/20
30	1/2	9/10
45	2/3	2/3
60	9/10	1/2
75	19/20	1/4

Here is another question to test your skills:

For a take-off on runway 31 with the wind from 265° at 24 kt, the aircraft would be
subject to head wind and cross-wind components respectively of
(1) 20 and 15 kt.
(2) 16 and 16 kt.
(3) 16 and 20 kt.
(4) 13 and 15 kt.

We know the angles is now 45° (310 minus 265), we also know that the head and cross wind ratio is 2/3 of that, therefore the expected head and crosswind component is 16kt.
The answer: 2.
Here is another question I got wrong the first time I looked at it:

An aircraft flying an approach into a strong head wind encounters a sudden tailwind near
the ground. The wind shear hazard to be expected is a sudden
(1) increase in groundspeed and increase in lift.
(2) decrease in groundspeed and loss of lift.
(3) increase in airspeed and increase in lift.
(4) decrease in airspeed and loss of lift.

My initial reaction was that you can expect an increase of air and groundspeed and a loss of lift as you may think something is pushing you hence you will go faster. That's not exactly how it works.

For a second, think you are flying with calm wind or no wind at IAS of 100kts. As there is no wind in front of you or behind you the groundspeed is also 100kt.

When you encounter a sudden head wind sheer (wind picks up suddenly) directly ahead, your IAS will rise and so will lift. When airspeed increases, the ram air pressure is increased which gets through the pitot static system, which translates into an increase of airspeed on the indicator.

With a 20kt headwind shear, the IAS will rise instantly to 120, increasing the lift over the wing. The groundspeed will still be 100. However, after a moment due to the extra drag created by the extra lift, eventually this will slow the aircraft down to 100kts IAS again, at which speed the GS will have fallen to 80kts.

In the example above we expect the opposite to happen, where you encounter a tailwind (wind coming from behind you). In this case, when the tailwind hits you, the IAS will drop instantly to 80kts, resulting in a loss of lift as well. The ground speed will still be 100. After a little time due to drag, the plane's groundspeed will increase to 120 and the IAS will come back to 100kts. The key to the question however is what is to be expected immediately after it hits, not what happens later.

The answer therefore is 4.

This situation is particularly dangerous during landings when the IAS drops significantly getting the plane into airspeed range very close to a STALL.

And that is why folks, these air molecules in motion have to be well respected!

Good luck and safe flying.

Turns - useful rule of thumb

If you are a student pilot, you most likely have heard the term "Standard-Rate Turn" .

The definition is :

"A standard rate turn for (light) airplanes is defined as a 3° per second turn, which completes a 360° turn in 2 minutes. This is known as a 2-minute turn, or rate one (= 180°/minute)."

Thanks Wikipedia!

In layman's terms, turn at a rate at which you can complete a full circle in 2 minutes...

The bigger question why do I care...

Here are a few examples why it may be useful

• Air traffic Control - ATC uses standard turn rates in providing traffic separation.
• Instrument flying and IFR conditions - when you can't see where you are, using the standard rate is critical in telling you where you will be when making turns just using the correct turn bank angle and time piece.
• Intercepting, tracking, departures, approaches and other maneuvers are all accomplished using standard rate turns.
• If you are going for your IFR rating, you need to know all about the SRT and you need to perform it too.
  
• ....and my fave: It's always good to know you can be on your way back to the direction you came from in one minute:)
  

In not going to go into explaining how to use the turn coordinator or a turn and bank indicator but rather help you figure out the bank angle needed to get you into the standard rate turn....

Here is the easy to remember rule of thumb:

Divide your airspeed in knots by 10 and add 7.

For example, if your airspeed is 100 kt, your bank angle will be 17° (10 + 7).

Voila!

123s of Radio range

First of all, I want to wish you all a Happy New 2009! Do you have a New Year's resolution? I know I do.

All the best in your aviation studies and training. Remember to have fun and to stay safe!

Here is a short post I think will prove useful.

Here is a sample exam question:

You are flying at 8,500 feet and you are wondering for how many more miles you can expect to receive a readable VOR signal from the station.

A. 100 nm
B. 92 nm
C. 184 nm
D. 113 nm

What is the answer?

First thing to understand is that VOR uses VHF radio signal which stands for Very High Frequency. VHF is mainly used for FM radio, TV and two way radio communications. Unlike low, medium frequency radio signal, very or ultra high frequencies do not 'bend' much around the surface of the earth, Very High Frequencies can only be received by the receiver that is in line-of-sight to the sending station.


In order to calculate the max line-of-sight distance in nautical miles from the transmitter (ATC, VOR, DME) and the airplane, you use the following simple calculation:

1.23 * square root of airplane's altitude

Let's take the exam question and calculate the distance.

1.23 * square root of 8,500=113.4

You can also do an easy mental check. You know that if the altitude is 10,000, the answer will be 123 nm. Generally, you can assume an altitude lower than 10,000 will generate a shorter distance. On a practical level, remember that if you are out of comm range, one way to get into range is to climb. However it is also reasonable to expect that at higher altitudes you can encounter more interference from other transmitters on similar frequencies hence weaker signal etc.

Now that you know all that...

The answer to the exam question above is D.

It's all about the balance

For this post I'm going to assume you understand the concept of weight and balance in aviation. You can learn the basics by reading and watching the following: read here and watch this video.

Instead, I'm going to focus on what you need to know for the exam.

You can expect the following types of weight and balance questions in the exam:

A. Concerning aeroplane "x", one could _____, to bring the aeroplane back into the centre of gravity moment envelope.

1. remove weight from the aeroplane
2. move weight forward in the aeroplane
3. move weight towards the back of the aeroplane
4. do none of the above

B. What is the location of the CG, in relation to the datum, for aeroplane "x"?

1. 35 inches aft the datum.
2. 44 inches aft the datum.
3. 101 inches aft the datum.
   
4. 70 inches aft the datum.
   

C. The load details for aeroplane "x", indicates that this aeroplane

1. is within the weight limits for the utility category only.
2. is within the weight limits but is not within the C of G limits.
3. exceeds both the weight limits and the C of G limits.
4. is within both the weight and the C of G limits.

in most likelihood, you will be provided with three things to work with to solve the following problems:

1. Sample loading scenario
   
2. Loading Graph
3. Centre of Gravity Moment Envelope Chart

The starting point is your sample loading scenario calculation that will have incomplete portions that you will need to look up in the loading graph. You may be given the weight of the passenger and pilot but then you need to use the loading graph to find the corresponding load moment/1000 (pounds-inches). You will repeat this to find the total weight, CG and arm of the loaded plane.

Sample of the materials you will be provided with:




Below is a sample Loading Scenario.

The formulas you need to know are:

Moment 1000/lb-in = Weight * Moment Arm /1000

Hence: 51 = 1365 * 37/1000

Load Details	Weight	Moment Arm (inches)	Moment 1000/lb-in
Basic Weight (includes full oil/unusable fuel)	1365	37	51
Usable fuel at take-off	230	52	12
Pilot and Front Passenger	360	38	13.5
Two rear seat Passengers	282	71	20
Baggage	50	80	4
Total	2287	44	101
Moment arm inches from datum is (CG)	44

Let's solve the problems then:

Solution A.
Using the total of 2287 pounds and 101 as the Moment 1000/lb-in, you ensure both points are within the envelope. Using the chart above you will see both the weight and the C of G (in this question referring to the moment) are within limits so 4 is the answer. If they are not however, use the rule below to put it into balance:

The CG location is above the envelope - weight must be removed from the aircraft.
The CG location is ahead of the envelope - weight must be moved back in the aircraft.
The CG location is behind the envelope - weight must be moved forward in the aircraft.

So, if the moment was 80 instead of 100 at the same weight, the weight would have to be moved to the back in the aircraft.

Similarly, if the moment was 115 instead of 100 at the same weight, the weight would have to be moved forward in the aircraft.


Solution B.
This one can be tricky as one of the answers is 101, which you may think is the answer...Nope, 101 is the moment 1000/lb-in not the arm which is the distance in inches aft the datum.

The answer here is 44 as it is calculated by the sum of the moment arm divided by the total weight.

hence: 101,000/2287 = 44


Solution C.
Ensure that the total weight of the plane and the total moment where these two points cross each other are charted within the 'envelope' . In the case above with weight of 2287 pounds and moment of 101, the plane is within the envelope hence the answer is 4.

One tricky part about this question can be when the airplane is within the weight limits but is not within the C of G limits. An example of this would be a plane with weight of 2400 pounds and CG (moment 1000/lb-in) of 90. If you think about it, the max allowable weight for his plane is around 2550 lbs, hence 2400 is still below that. The CG however is too far left. The weight would need to move to the back of the plane to just over 100 for it to balance.

If you know how to answer the above questions, you will not have problems during the exam.

In addition have a look at this cool weight and balance calculator here.