please cast you eye over this

[hmm nice flaps]

http://www.pprune.org/ubb/NonCGI/confused.gif
I have just been to OASC for FAA aptitude and really struggled on questions like this:

If a ship leaves its port and sail 130 miles north, then turns west and sails 35 miles, then turn south and sails 170, and then north 20 mile how far away is it from its original destination?

I believe i may encounter imilar questions at the AIB. I realise it's a tad sad asking such questions on this forum, and must by rather a bore , but can any kind soul over any advice?

[BEagle]

Are you sure you copied the question correctly? For in the true tradition of the Navy, this ship seems to have wandered around for several hundred miles without any idea of where it was supposed to be going in the first place! Keep a look out for obvious 3/4/5 or 5/12/13 triangles (if they still teach such things at school, you'll know what I mean!).

[This message has been edited by BEagle (edited 26 May 2001).]

[AllTrimDoubt]

No need to work it out...the ship won't be where it said it would and will definitely not be on the planned radio frequency! Just mutter something about typical fishead ineptitude and you'll be fine for FAA!

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Just Lookout and Fly Accurately Bloggs!

[Talking Radalt]

shouldn't it read, a BOAT is in dry dock at a rate of £x per day awaiting a contract to replace a part which cost $y. The part could have been manufactured on board but the mechanic's post was disestablished to save £z per year. Considering a working group spent £A per week to decide on the contract, how much did the MOD plc lose by laying off Bloggs?

[Davaar]

From the information given, God and the Lords Commissioners alone know, because the original destination is not stated, nor can it be deduced. Triangles will not help.

[tony draper]

surly its just simple trig,if you mean how far from the point of origin, ie the port
its just a right angle triangle with sides 35x20 work out the length of the hypotenus.
that should do it.
Or it would if the earth was flat.

[BEagle]

......which is why I asked whether you'd copied the question down correctly! If indeed the courses are as stated and you meant the port of departure, not destination, the answer is pretty obviously just over 40 miles. Divide the courses by 5 and you have the hypoteneuse as 5((7*7)+(4*4))**0.5 ; in other words 5*(65**0.5). 8 eights are 64, so the answer is a tad over 40. Simplify things like this, keep an eye out for pythagoras, know simple squares of 1 to 13 and it's much easier!!


[This message has been edited by BEagle (edited 27 May 2001).]

[mriya225]

Wouldn't this ship now be 75mi. SW of the port?



[This message has been edited by mriya225 (edited 27 May 2001).]

[Gentleman Aviator]

BA ask similar questions in their aptitude tests. If it makes you any happier they also give directions in terms of NE, SE, SW & NW as well as N, E, S & W!

Suggest you break it into bits:

1. 130nm North.
Sails 35nm West so ship is

2. 130nm North, 35nm West.
Sails 170nm South so ship is

3. 40nm South, 35nm West.
Sails 20nm North so ship is

4. 20nm South, 35nm West.

Hope this helps.

GA :)

[mriya225]

...and don't second guess yourself!!!
my original post was 55mi. SW of port.

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"You did WHAT"?!

[hmm nice flaps]

many thanks for the replies - useful and amusing.

The question wasn't copied down exactly, but at OASC there was a whole batch that were similar to this.

Cheers anyway!
http://www.pprune.org/ubb/NonCGI/cool.gif

[BigBulge]

I make it 40.3 (to nearest decimal place).

As Tony Draper pointed out, (35 x 35) + (20 x20) = (40.3 x 40.3).

BEagle, your approximation was very close.

[Rusty Cessna]

Its just all about being able to mentally picture the route in your head, how did you get on anway?

Rusty

[BEagle]

BigBulge - yes, I knew that. But unless you use a calculator or know how to do 'long square rooting' (which wasn't even taught when I went to school - I learned it from an ancient mathematician some time later), 'a tad over 40' is quicker to work out than 40.31128874149!!

[BigBulge]

A fair reply. However, I too didn't use a calculator. A piece of paper, long multiplication, and good old trial and error. I know it's sad, but I was bored......... mmmmmm, very bored.

[mriya225]

Okay look, can we get a ruling on this--one way or the other--it'll drive me nuts until we do.

[Davaar]

By "long square rooting" (which I never knew by that name, but I know what you mean)I got it to 40.31, to two places of decimals. Of course, that is to point of origin, not to unstated destination. No trial and error involved, or having to memorise lists of squares and square roots.

[Avtrician]

Dont forget, that the answer needs the direction to be correct.(Tho the way this guy is driving he is obviously lost,drunk,blonde or all of the above)

So the Marie Celeste is 40 and a tad miles South Westish of his starting point.

[RRAAMJET]

It's obvious:

the fishheads were moseying around looking for a bit of "long square rooting"...

ahem....now what was the question?

[aerostude]

This can't have been a question with those exact numbers. That took a calcualtor!! And furthermore, the bearing of the final position from the start is a bit obscure. I seem to recall all these problems breaking down into 3,4,5 triangles etc.

[Sonic Boozer]

Stuff the answer. Anybody called 'hmm nice flaps' should be allowed to join the military as a pilot anyway.

Seems like a top bloke! (assuming it's a bloke?)

[Talking Radalt]

As Tony D says it assumes the world is flat. I've seen Monty Python and the Holy Grail...
"..and that, my liege, is how we know the world to be banana shaped" which considering the way things are going these days probably isn't too far off the spot.

[tony draper]

whole new ball game working problems like that on curved surfaces, don't suppose it would make much difference with the distances involved in that prob,
hell I can barely remember pythagorus. http://www.pprune.org/ubb/NonCGI/frown.gif

[FlapsOne]

Are the directions true or mag?

At what latitude does the ship start?

Who cares?

[Ganf]

Flaps One makes a good point - who cares?

However, hmm nice flaps, to answer the question: you cannot work this out unless you know the latitude of the start point, as Flaps One also mentioned. If the ship started north of the equator its westerly course would take it over some number of degrees of longitude. As it headed south, the new longitude would remain constant but the number of actual miles west of the start point would increase since the lines of longitude grow further apart as you travel further from the poles. Similarly, if the start point was south of the equator, the number of actual miles west of the origin would decrease as the ship travelled south again. Admittedly, over the distances mentioned, the error would be quite small unless the start point was, say, 150 miles south of the North Pole so the Westerly leg would take it over a large number of degrees of longitude before it headed south over the other side of the globe. But then it would actually not move anywhere anyway because it would be stuck in the middle of a f****** great expanse of solid ice. I suspect you have missed some of the information from the original question.

Which brings us back to Flaps One's point - who cares? Sorry to blah on - christ, I'm turning into an anorak!!

Good luck!

Edited for spelling

[This message has been edited by Ganf (edited 28 May 2001).]

[only1leftmate!]

Its a trick question!!!!

You ran aground on the third leg and are probably drinking it off in a pub!!

Since when has mental agility and accuracy been a pre-requisite for the Queen's Comission anyway? Total rigidity of thought and iron inflexibility should see you through OASC or AIB. If its wrong - justify it!!!! Baaaaaaaaaaaaaaaaaaaa

[Batwing]

Can only assume ship civilian charter as must have had fuel on board-ergo not a British warship-unless chartered German submarine of course.

[Per Ardua Ad Asda]

SNAFU. As GANF suggests, not enough info has been supplied to give an exact answer. If the info quoted is ALL that has been given then the examiner is only after an answer using basic trig on a flat projection.

Yawn..


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Milk-Maid A-Hoy!

[obnoxio f*ckwit]

Doing it in my head, i got the same as BEagle, (a tad over 40 miles), but what's 'long square rooting' and how do you do it?

(I know: 'get a life etc'...)

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Its not only the depth that varies...

[Davaar]

Obnoxio. Sides A (20 miles) and B (35 miles) subtend a right angle. Per Pythagoras, the sum of the squares of A and B is the square of the hypotenuse H.

A squared is 400 and B squared is 1225. The sum is 1625. H is the square root of 1625. Here is the math:

4 16 25 .00 00 00 40.310
4 16
80 25 00
0
803 24 09
3 91 00
8061 80 61
1 59 00

The answer is 40.31.

Step 1: write the number in blocks of two each way from the decimal
Step 2: by trial, find out which number, squared, is closest to but less than the first couple (here, to the 16). Write that number down on the left and on the right. The figure is 4.
Step 3: subtract the square of that number from the first couple (here leaving zero)furthest to the left of the decimal.
Step 4: add the number you have just squared to itself where you wrote it on the left (here, the 4 becomes 4+4=8).
Step 5: drop the second couple to follow on the remainder from the first couple (here 25 drops to follow a remainder of zero, i.e., 16-16=0.
Step 6: Add any number after the 8 that will enable you to divide the 2500. There is none. So put 0 after the 8, which now becomes 80. Also write 0 to the left and right.
Step 7: drop another couple. The 8 is now 80. What amount added to the 80 allows you to divide the 2500? Answer is 3 (803x3=2409). Write down the 3 after the 80, which becomes 803. Divide 2500 by 803. The remainder is 91. Write down the 3 to left and right. At the right the answer is 40.31, which is H. To the left you have 806, which allows you to continue as far as you care into the decimals.

If I could draw lines on this machine I would add those, which go pretty much where you would imagine them.

Courtesy of George Watson's Boys' College, Edinburgh, 1946.

I know. Get a life. Still, this lets you find square roots without calculator, log tables, or other artificial aids.

Edit: This does not come out in the final presentation much at all as I type it. The final version compresses the columns. There should be three columns (separated by lines when I have a pencil). The left column is the 4, 4, 8, 80, 803, 806 and so on. The middle is the 16 25 .00 00 and the various subtractions. The right is the 40.31.



[This message has been edited by Davaar (edited 01 June 2001).]

[BEagle]

Heck, Davaar, I didn't think that there were many of us still alive who know how to do 'long square roooting'! My method is exactly the same as yours!!

One wonders how many od the 'sdoinmeedin' yoof-cuwcha kids could do it!

[Davaar]

Yes, BEagle, we are a rare and endangered species. Years ago I worked on the 34th floor of a high-rise, and with nothing better to do one day I speculated on how fast a body would be going at impact, ignoring air resistance, if it jumped out of the window. One of my group of heavy thinkers at the time was a PhD (Mathematics)from Stanford, no slouch at all. He knew nothing of the three speed, acceleration, time, distance, equations, or how to find a square root without his toys. As a simple farm boy/arts man/lawyer I could but shake my head.

[This message has been edited by Davaar (edited 01 June 2001).]

[BEagle]

The 3 speed/acceleration/distance/time equations? Always thought that there were 5:

v=u+ft
s=ut+(ft**2)/2
s=vt-(ft**2)/2
s=(u+v)t/2
v**2=(u**2)+2fs

[tony draper]

I am speechless with admiration,
I need a calculator to add up a colomn of five figures now.
All that stuff has disapeared into the mists along with most of the other stuff I used to know. http://www.pprune.org/ubb/NonCGI/frown.gif

[Davaar]

In mitigation I can only plead that it HAS been a long time.

[Rusty Cessna]

Beagle,

The 3 equations of motion are:

V=U+at
V^2=U^2+2aS
S=Ut+1/2at^2

When V=Final Velocity
U=Initial Velocity
a=Acceleration
S=displacement from origin (distance)
t=Time

and what on earth is f?

Davaar,

Id go about that problem with V^2=U^2+2as

Hence step 1, find out how tall the building is, initial velocity is 0, which means V^2 will be 2 times 9.81(vertically downwards) times the height of the building.

So say there are 10 feet per floor, which is near on 3.5 metres, thats 113.5 metres,so V^2=2(9.81X113.5)
V^2=2226.81
V=root 2226.81
V=47.2 meters per second

This is roughly 95 knots!

Thanks for waking up the brain Davaar.



[This message has been edited by Rusty Cessna (edited 02 June 2001).]

[Davaar]

Those are the three I had in mind, Rusty, and the one we used was the one you selected. I was accustomed to a slightly different enunciation, as in v(2)-u(2)= 2as. No substantive difference, of course. Same ship, different tally-band. The banter became a tad brittle as we wrestled with the v(2), and how to derive v, I assuring the proletariat that I could do it with paper and pencil away back in Form 1 for Goodness' sake, Oh Yes I could, Sure I could (if only I could remember how!); and what kind of mathematicians were they anyway? and the PhD offering that I could not too, and why did I not just jump out of the damned window while I was at it. We knew this as the spirit of scientific inquiry and policy formulation.

[This message has been edited by Davaar (edited 02 June 2001).]

[This message has been edited by Davaar (edited 02 June 2001).]

[RRAAMJET]

Rusty:

the "f" that BEagle refers to, as a long time truckie, is of course the " f*ck-up, or f*dge " factor applied to correct and balance imprests when you were too ****-faced to remember what happened last night, and the reciept is in that t@rt's room down the corridor with your shreddies.

Everyone knows that f-squared = -1 $/shaekel/Franc/donkey/pina-colada

fx100 = nasty phone call from accounts

Clear as Bombay tapwater, eh?

[RRAAMJET]

i before e except after c when the sound is e....my spelling is decaying, lucky my long square rooting skills still pulls 'em in...

write out "receipt" one hundred times...

[BEagle]

'f' is what we used to use for acceleration; I understand there was a later change to 'a'.

v**2 is an old way of writing 'vee squared' as per BBC basic. It's the same as v^2, I guess.

We had to remember all 5 equations; there are 5 terms:

u (initial velocity)
v (final velocity)
f or a (acceleration)
s (distance)
t (time)

Each equation has 4 of the 5 terms - you need 3 terms to calculate the 4th in each. Hence you select the appropriate equation for the terms you know and for the term you wish to calculate.

Perhaps more than 3 equations to remember is too much for modern 'sdoinme edin' yoof who can't be ar$ed?

[Charlie Luncher]

Anorak cx

Charlie sends

[Rusty Cessna]

Ram!

Its all so clear now!

Beags,

I kinda figured out what you were doing, it's not that we can't be arsed, we just aren't taught the other two, but we are taught how to derive them and other such usefulls as: S=Vt-1/2at^2
and S=(U+V)t/2

which are probably the two you suggested, I cant remember, haha :)

Davaar,

You are indeed right there is no difference to the working, but the major difference is you can do it without a calculator, I can't! maybe if I could be arsed eh beags? :) :)

Thanks guys,
Rusty

[Perky Penguin]

There was a selection question for Navigators some years ago.

A sheet of A4 with two points, 'A' and 'B' on it. The question was "Draw the shortest line between points A and B. (Extra paper is available if required)!

[STANDTO]

As an ex navy police pilot pointed out to me some years ago:

You are working in radio silence, take of from your ship and go of pinging for subs. The deal from the captain of the ship is that he will be roughly at X at such and such a time.

Whilst you are off, a sub comes and scares off your ship, who sods off in the opposite direction.

You come back, low on juice. No ship..........

Jion the RAF, airfields are easier to find and there is more of them