Urgetnt - Revised ETA's

[aviator_hyd]

Can someone please help me with this problem.

If you are 300 Nm from your destination flying at 260kts. You are advised to slow down to 200kts so that you can delay your ETA by 8 minutes. How far from your destination will you reduce your speed to 200Kts to delay your ETA by 8 Mins.


Any kind of help is Appreciated.

[Rainboe]

Interesting problem I have met before. I think this is worth looking at. I shall try and get a more elegant solution.

Imagine 2 different aeroplanes flying in formation. One slows down so they diverge. The divergence must be 8 minutes, or 8/60 hours by the time the slower one gets to destination.

Speed=Distance/Time. Or Time= Distance/Speed
Let 'x' be faster aeroplane and 'y' slower aeroplane.

So as Time=Distance/Speed, but slower aeroplane time, Ty, is Tx+8(minutes, or 8/60 hours).
Therefore D/200= D/260 + 8/60

Multiply top line by 200 x 260 x60 to get rid of the nasty fractions,

(D x 260 x 60) = (D x 200 x 60) + (8 x 200 x 260)

3600 x D = 416000

D= 115.5

Testing, 260kts, 115.5 miles takes 26 2/3 minutes
At 200 kts, 115.5 miles takes 34 2/3 minutes

Bingo. Harder one please!

[aviator_hyd]

Thanks!

Thanks a lot rainboe.. your help is greatly appreciated and so is the way you took the problem. The thought of taking them as 2 aircrafts flying formation did the trick.. Once again thanks a lot.. well ill try to shoot harder ones at you the next time!!! hehehe

[Rainboe]

I have to admit I fudged the answer slightly because the numbers weren't nice. You actually have to slow down 82.5 metres earlier than the 115.5 nm point. Usually when you are set problems like this, a very clean number pops out as the solution.

I suggest you shoot the problem setter.

[C-N]

dry textbooks

and this could be the dry, direct answer: (again without calculus)

d=300nm (assuming thats ground distance you have given)
v=260nm/hr (knots)

with that v you'll arrive 1hr 9.2mins
but you want it delayed 8mins.

Hence, new ETA 1hr 17.2mins or 1.28667hrs

let x be the whole distance where v=200nm/hr in order to meet your revised ETA:

x/200 = 1.286667
x=257.3334nm (ground distance)

or 9.8461mins from now, so act quickly.

challenge: post a one liner differential calculus solution

[Rainboe]

It goes wrong where you apply Dist/Speed=Time.

You have used 200kts for the time from now......300 miles away for 1 hr 17.2. It gives the wrong answer that does not apply. However an interesting way of looking at the problem. You didn't test the answer. 257.3 miles at 260kts takes just about an hour. At 200kts it takes 1 hr 17.2 minutes.

[powerstall]

Must have been dozing off, when this was taught by our GI's!

[Rainboe]

I don't think they taught this stuff! I was brought up on 'you leave Hawaii at 0800LT on Tuesday, fly 14 hrs to London. How high will the moon be in the sky on arrival, and will the Bear be brown or white?'

[fireflybob]

More than one way of skinning a cat but this is how I did it:-

Let distance from "speed change point" to destination be x

time (in mins) from start to "speed change point" t1 =

(300 - x)/260 (times by 60 to get minutes)

time (in mins) from "speed change point" to destination t2 =

(x/200) (times by 60 to get minutes)

Total time = 300/260 (x 60) + 8 = (900/13) + 8

Total time = t1 + t2

Therefore solve for:-

(900/13) + 8 = t1 + t2

Substitute for t1 and t2 and then resolve to get

x= 115.55555555 recurring!

[Rainboe]

After the 115.5 nms, I refused to get sucked into the .05555555555 nm bit, which I now work out is actually 337.77777 feet, or 102.955 metres, not 82.5 metres, I was being sloppy above.
Therefore the correct speed reduction point is 155.55555 nms, or 102.9555metres before the 115.5 nm point.

This does assume the aeroplane slows instantly. It does not help mixing units, therefore the whole world should go Imperial. Nothing is as elegant as a nautical mile for distance purposes. We could do it if we nuke Paris.

[Otto Throttle]

Bah and humbug. Tell them you'll keep 260kts and take up the hold. Twice round should do the trick.

[C-N]

oh my.. gave flawed logic indeed. thanks for pointing out.
around 115.05333nm for 200kts and 184.94667nm for 260kts

but yes, this also assumes the speed will be reduced to 200kts in an instant, as the speed can only be reduced gradually.

[C-N]

'you leave Hawaii at 0800LT on Tuesday, fly 14 hrs to London. How high will the moon be in the sky on arrival, and will the Bear be brown or white?'

strange riddle, but looks like you'll also be touching down on a Tuesday at 0800LT in London, 3hrs past sunrise (assuming you mean this coming Tuesday 11 aug'09). this is new to me, I'll share this to those GI's.

I cant answer the collor of the Bear, please explain the color.

[Rainboe]

115.05333nm

Back to the calculator! The problem set was a joke to illustrate the point of the sort of problems that used to be set for us!

[powerstall]

Is this one of those trick questions?

[Jetdriver]

Subject detour. Please follow the diversion signs.