Calculating Drift On Approach

OneNineHundy

OneNineHundy

Well-Known Member
Hi there, I have a question regarding drift on approach. Below is an example, how you would go about answering this question.

Wind is 150/20kt what is the crosswind on RWY10? = 14kts

For the above runway what is the drift required at 180kt G/S?

On approach you note you are holding 10' drift and that your G/S 180kts, what is the X/W?

Im sure this has an incredibly simple solution, but I have never seen a question like this.

One last thing, when an aircraft is assymetric the co-ordination ball is not an indication of whether the aircraft is in balance. When using the rudder and bank method to control the aircraft what would a piece of string indicate to show that the aircraft is in balance?

Thanks in advance.
 
I know the answers to your math problem but I must say, why bother? You are not going to calculate all this while you are flying. If you have a crosswind component you simply sideslip, or crab, until the airplane has zero drift.

Wind is 150/20kt what is the crosswind on RWY10? = 14kts
For the above runway what is the drift required at 180kt G/S?
5 degrees, with a 15kt tailwind I might add.

On approach you note you are holding 10' drift and that your G/S 180kts, what is the X/W?
30kts
 
I spent 10 years in the submarine service figuring out angles and speed components. Of course, I used the knowledge to potentially shoot torpedos at other ships and subs but still, the math is same.

This is hard to explain without a whiteboard but basically it is just algebra. You solve for x. There are 3 values and you need two of them to solve for the third.

a= the angle between the runway and the wind compontent, or the angle between the runway heading and the aircraft heading (drift correction),
x= the crosswind component of the wind, or the aircraft component of drift correction.
y= the overall speed component of the wind, or the aircraft.

Formula :

x = sine(a) * y

example= to figure out the crosswind component of wind:
150/20kt, what is the crosswind for RWY10?
a=wind is at a 40 degree angle to the runway
x= unknown variable
y=wind is at 20kts

x = sine(40) * 20
sine 40 is .64, so
x = .64 * 20
x= 13kts

Problem 2:
For the above runway, what is the drift required at 180kt G/S?

We already know the crosswind is 14kts(I used his 14kts when I worked it, vice the 13kts I came up with above), so what we need to do is make the airplanes drift match 14kts to zeroize any drift.
a= unknown variable (the amount of drift correction)
x= 14Kts
y=180kts

14kts = sine(a) * 180kts
divide both sides bt 180
.078 = sine(a)
.078 = 4.5 degrees (rounded up to 5 degrees)

Ok, I spent way to much time on this and I don't know if many will follow but I do stand behind my answers. I also stand behind my original comment that you don't need to know this stuff. Just sideslip or crab until drift stops and hold it all the way to the runway.
 
Ok, you officially outdorked me, tgrayson. :)

I guess the answer is there is no answer as there is no way to calculate the winds on the approach to an exact constant.
 
I find that staying on the localizer or exended runway centerline with a crab, followed by a transition to a slight wing low in the flare seems to work for the 757 & 767. I guess that those GA airplanes require something a little more precise?
 
subpilot said:
I guess the answer is there is no answer as there is no way to calculate the winds on the approach to an exact constant.
Click to expand...

Based on the information he provided, I didn't see any way to calculate what he was looking for, but I was prepared to be amazed. ;) I've seen many analysts pull an extra equation out of the air and solve an apparently unsolvable problem.

If he supplies his airspeed, which he ought to have, we should be able to solve for the wind velocity, using the Law of Sines I posted a few threads ago.
 
Cptnchia said:
I find that staying on the localizer or exended runway centerline with a crab, followed by a transition to a slight wing low in the flare seems to work for the 757 & 767. I guess that those GA airplanes require something a little more precise?
Click to expand...
Exactly my original point. You just fly the plane, you don't compute this type of stuff while flying. This is for engineers to figure out, not pilots.
 
tgrayson said:
Based on the information he provided, I didn't see any way to calculate what he was looking for, but I was prepared to be amazed. ;) I've seen many analysts pull an extra equation out of the air and solve an apparently unsolvable problem.
Click to expand...
It would be a pretty cool equation to account for the winds almost constantly changing as you moved along the FAC both horizontally and vertically. Of course, I probably wouldn't understand it.
 
MidlifeFlyer said:
It would be a pretty cool equation to account for the winds almost constantly changing as you moved along the FAC both horizontally and vertically. Of course, I probably wouldn't understand it.
Click to expand...


Here's one for you. If you have a 30 degree angle between where the wind is coming from and you're course, half of the wind velocity is your cross wind component.

E.g. Landing runway 25, wind 280@20, you have a 10 knot crosswind
 
tgray, sub, Midlife, and all other geeks,

You simply read the G1000's CW read out, hit APR on the autopilot, and let the dang thing fly itself (using it's own WCA calculations) until you see the runway, 'click' autopilot comes off and you land.

Next...

:)
 
meritflyer said:
tgray, sub, Midlife, and all other geeks,

You simply read the G1000's CW read out, hit APR on the autopilot, and let the dang thing fly itself (using it's own WCA calculations) until you see the runway, 'click' autopilot comes off and you land.

Next...

:)
Click to expand...

With that in mind, with any GPS if you have a trk in degrees that's being displayed, you can shoot an exceedingly precise approach if you match that number with the final approach course, even if you're hand flying.
 
Thanks for your help guys, the reason I ask the question is that it is a potential interview question I had from some gouge. I'm not sure if there is a simple solution because it is meant to be answered in your head, not sure at all how to go about it simply. Obviously this is not relevent when actually flying the aircraft.

Once again thanks for your help.
 
Here is a god thumbrule to figure out crosswind component.

If the angle between the runway and the wind is less than 40 degrees:
Angle / 60 = percentage of crosswind component

If the angle is greater than 40 degrees:
Angle + 25 = percentage of crosswind component

example:
runway 180, winds 210/20
30 degrees (less than 40), so 30 / 60 = .5 or (50%)

20 * .5 = 10kts of crosswind component.

Easy to do in your head with a little practice.
 
subpilot said:
I spent 10 years in the submarine service figuring out angles and speed components. Of course, I used the knowledge to potentially shoot torpedos at other ships and subs but still, the math is same.

This is hard to explain without a whiteboard but basically it is just algebra. You solve for x. There are 3 values and you need two of them to solve for the third.

a= the angle between the runway and the wind compontent, or the angle between the runway heading and the aircraft heading (drift correction),
x= the crosswind component of the wind, or the aircraft component of drift correction.
y= the overall speed component of the wind, or the aircraft.

Formula :

x = sine(a) * y

example= to figure out the crosswind component of wind:
150/20kt, what is the crosswind for RWY10?
a=wind is at a 40 degree angle to the runway
x= unknown variable
y=wind is at 20kts

x = sine(40) * 20
sine 40 is .64, so
x = .64 * 20
x= 13kts

Problem 2:
For the above runway, what is the drift required at 180kt G/S?

We already know the crosswind is 14kts(I used his 14kts when I worked it, vice the 13kts I came up with above), so what we need to do is make the airplanes drift match 14kts to zeroize any drift.
a= unknown variable (the amount of drift correction)
x= 14Kts
y=180kts

14kts = sine(a) * 180kts
divide both sides bt 180
.078 = sine(a)
.078 = 4.5 degrees (rounded up to 5 degrees)

Ok, I spent way to much time on this and I don't know if many will follow but I do stand behind my answers. I also stand behind my original comment that you don't need to know this stuff. Just sideslip or crab until drift stops and hold it all the way to the runway.
Click to expand...

Really appreciate the help Subpilot, and others. From the information you have provided I have come up with a possible way of calculating this answer. Not to say this is correct but seems to come close to the answer, possible by coincidence.

Wind is 150/20kt what is the crosswind on RWY10? = 14kts

For the above runway what is the drift required at 180kt G/S?
(A/C is doing 3nm/min, 14/3 = just under 5' drift required.

On approach you note you are holding 10' drift and that your G/S 180kts, what is the X/W?
(A/C still doing 3nm/min, 3 X 10 = 30, ANS X/W 30kts)

The way I use to calculate the X/W is work out the angle between wind and runway direction, then add 20, this is the percentage of the total wind which is X/W
e.g. Wind 250/30 kts, what is X/W on runway 28? 30' + 20 = 50, 30 x 50% = 15kts

Cheers again
 
tgrayson said:
That's why they invented Calculus. ;)
Click to expand...
Thanks for the confirmation. I always suspected that they invented calculus just to make sure I became a lawyer. Good to know I was right :D
ppragman said:
Here's one for you. If you have a 30 degree angle between where the wind is coming from and you're course, half of the wind velocity is your cross wind component.
Click to expand...
Yup know that RoT. And 15° is 1/4 and 45° is 3/4. Works well (but not as well as watching what the airplane does with respect to the runway centerline). But a RoT is not what it sounds like the OP was looking for.
 
subpilot said:
Wind is 150/20kt what is the crosswind on RWY10? = 14kts
For the above runway what is the drift required at 180kt G/S?
5 degrees, with a 15kt tailwind I might add.

Click to expand...

I still don't see how you can have a tailwind with the wind 50 degrees off the nose.:)
 
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