How good is your math???

gowest · Post by **gowest** » Tue Aug 16, 2005 1:05 pm

Give me a hand finding the formula for that question:

flying direct to CYST on a Mag heading of 020, 15 nm away the captain asks me what heading I should take in order to intercept a 5 nm final on rwy 04 (QFU 040):

What is the formula to figure that out?
what mental calculation can you do to figure that out while you're flying ?
to make things even harder on the formula, the captains name is Chris and the plane is a Dornier...

Thanks for your help!

Gowest

FL410 · Post by **FL410** » Tue Aug 16, 2005 3:22 pm

Easiest way would be to sketch out your heading and distance on a piece of paper, and then at the end of the line you'd draw a line 1/3 of the length of your heading (since it's 5nm). Then connect the lines to get a rough heading and distance to the extended centerline.

It's just a basic vector diagram, and the airport would be at the top right.

Like so:

scotothedoublet · Post by **scotothedoublet** » Tue Aug 16, 2005 4:23 pm

Another way sans paper (same idea as above) is to use your VOR/HSI (if the points are from the same VOR). Obviously you are on the radial corresponding to the tail, picture the radial/DME on the HSI/VOR and WAG a hdg to get there and adjust it for drift. Ex, if you're on the 180R@20 and want to get to the 270R@10 you would want to fly a hdg of approx 335. Pretty crude but it works. I think a formula might be more work than it's worth.

Out of Control · Post by **Out of Control** » Tue Aug 16, 2005 6:42 pm

Just get it on the ground, don't make life so hard

hazatude · Post by **hazatude** » Tue Aug 16, 2005 6:47 pm

Your radio calls should ALWAYS end with, "Comin' in hot"!

bush pilot · Post by **bush pilot** » Tue Aug 16, 2005 7:00 pm

I believe you would call it the sixty in one rule.

At 60 miles out if you want to be on a 5 mile final turn 5 deg.
At 30 miles turn 10 deg for 5 mile final
At 15 miles turn 20 deg for 5 mile final etc.

I have used it a few times usually at 30 or 15 miles out.

gowest · Post by **gowest** » Tue Aug 16, 2005 7:23 pm

Thank you guys for your comments, drawing the "vector diagram I found a 10 degre angle in that example, I sure they is an accurate trigonometri formula
we know 2 sides of the triangle 15 and 5: what is the 3rd one knowing that one of the angles is 20 deg?

cheers

bush pilot · Post by **bush pilot** » Tue Aug 16, 2005 8:09 pm

Well since I do not know how to put my image like FL410 did we will just use his example. Lets start off with the distance, I don't know if this will only work for right angle triangle or not(so lets assume it is), but a2=b2+c2 or since #2 side is smaller a=1,2=b,c=3. b2=a2-c2 therefor b=14.1miles(I think). I thought I had the angles worked out but my math is rusty for that. I would have to get out the old text books out and find out if you have to do sin or cosin or tan or whatever you have to do! but trust me just use the one in sixty rule. Its quick and easy. By the time you tried using the other while doing 200kts you will be passing overhead.

Pachanga · Post by **Pachanga** » Tue Aug 16, 2005 8:46 pm

Mario · Post by **Mario** » Tue Aug 16, 2005 11:07 pm

I haven't been in grade 11 for a while but you can go nuts with the formula

(I was trying to put a picture of an obtuse triangle here but I'm not so smart. So just assume that A B and C relate to angles and a b and c to sides)

If you are given two sides and the included angle, say a, b, and C:

You can compute the third side c by using the Law of Cosines. Then the Law of Sines can be used to find the sines of the other two angles sin(A) = a sin(C)/c and sin(B) = b sin(C)/c. The angles opposite the two shortest sides are then acute, and uniquely determined from their sines, and the third, largest angle is found from A + B + C = pi

Alternatively, you can use the Law of Tangents. You know that (A+B)/2 = (-C)/2, which is easily computable. Then by the Law of Tangents, tan[(A-B)/2] = cot(C/2) (a-b)/(a+b), so you can find (A-B)/2 uniquely. Then A = (-C)/2 + (A-B)/2, and B = (-C)/2 - (A-B)/2. Then c = a sin(C)/sin(A

seriously though who cares...1 in 60's sounding pretty sweet eh?

Mostly Harmless · Post by **Mostly Harmless** » Wed Aug 17, 2005 8:27 am

http://www.sosmath.com/tables/tables.html

http://www.math.com/tables/tables.htm

Try these.

flyinhigh · Post by **flyinhigh** » Wed Aug 17, 2005 11:14 am

well you have got my brain working now, I haven't thought about this kinda stuff in awhile.

KAG · Post by **KAG** » Wed Aug 17, 2005 11:31 am

the 1 in 60 rule would work fine.
For every 1 degree off, 60 miles out you'll miss your destination by 1 mile.

So off the top of my head a heading of north would put you on a 5 mile final for 04.

To figure it out 15 miles is 4 times the turn what you would normally need, so 5 time 4 = 20 degrees to the left. North.

Cheers.

Nomad · Post by **Nomad** » Wed Aug 17, 2005 8:36 pm

I can't believe all the math geniuses here. We're just flying an airplane

Here's what I would do:

Turn left a little,
see the runway,
turn right a little more than the left,
Land

It's worked for me for years

Ralliart · Post by **Ralliart** » Wed Aug 17, 2005 10:10 pm

Ask Chris?

Wild Cat · Post by **Wild Cat** » Fri Aug 19, 2005 1:26 pm

I hate math!