please cast you eye over this
Guest
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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.
"..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.
Guest
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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).]
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).]
Guest
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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
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
Guest
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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!
Yawn..
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Milk-Maid A-Hoy!
Guest
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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).]
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).]
Guest
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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).]
[This message has been edited by Davaar (edited 01 June 2001).]
Guest
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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).]
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).]
Guest
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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).]
[This message has been edited by Davaar (edited 02 June 2001).]
[This message has been edited by Davaar (edited 02 June 2001).]
Guest
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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?
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 $/shaekel/Franc/donkey/pina-colada
fx100 = nasty phone call from accounts
Clear as Bombay tapwater, eh?
Guest
Posts: n/a
'f' is what we used to use for acceleration; I understand there was a later change to 'a'.
v**2 is an old way of writing 'vee squared' as per BBC basic. It's the same as v^2, I guess.
We had to remember all 5 equations; there are 5 terms:
u (initial velocity)
v (final velocity)
f or a (acceleration)
s (distance)
t (time)
Each equation has 4 of the 5 terms - you need 3 terms to calculate the 4th in each. Hence you select the appropriate equation for the terms you know and for the term you wish to calculate.
Perhaps more than 3 equations to remember is too much for modern 'sdoinme edin' yoof who can't be ar$ed?
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?