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I still dotn get itQuote:
Originally posted by alkatran
The trick is to get the lines to and from bottom to be symetrical, and reach the furthest point possible on circle like that.
how do you know that your path wont cross the circle, like the first picture in my last post?
and how do you get that angle :D:D:D
It does, well, it would.
As soon as you reach the circle you start to follow it.
hmm, alright, makes sense :)Quote:
Originally posted by alkatran
It does, well, it would.
As soon as you reach the circle you start to follow it.
but can you show me how you got that angle?
Hi MrPolite,
didin't you read my repy's?
I know, without a drawing you have to READ it!
See the attached circle.bmp.
I computed the solution of alkatran, with a change for the leg that goes inside rthe circle. I divided it in two parts, one is the tangent to the circle and the second is the walk along the perimeter. That way no possible tangent is missed and the total distance is less than 6400.
The angle length, would be around 1100.Quote:
Originally posted by alkatran
I really really did!
First you head off at 247.5 degrees (south south west), reach the bottom of the 1000m*1000m square. Now go north north west, until you reach the middle left of the square. Go around the circle until you are at the far east, then go straight down. Here's the formula
sqr(1000^2+500^2)*2+(1000*pi)+1000
this adds up to UNDER 6400! (about 33 under it)
http://www.vbforums.com/attachment.p...postid=1303106
Note: The line would pass through circle if you went directly to west side, so I used the 33 extra miles to go on the circle a bit longer.
Then to 45 degrees (1000 ft) from circle is 367.7.
8th of the circumference of the circle is 785.4.
half the circle is 3141.6.
going down after that is 1000.
6394.7
Yea. Close....
Here is the COMPUTED solution again:
Look at the circle2.gif to see what I'm talking about.
The start is at the centre A (X;Y are 1000;1000) and goes to point C (500,0). The lenght of this leg can be computed by:
SQR(1000^2 +500^2)=1118,04
To reach B you have to go at an angle (direction) of 206,57 degrees, that makes the angle between AB and AC 26,57.
The second Leg is from C to D. Since the triangle ACB is identicla to ACD (it's just mirrored at AC) the lenght of CD is 500
to compute the part on the circle from D to E we need the direction of the line AD. Since the triangles are identical, the angles in A for both triangles are the same. So covered already 2*26,57 degrees of the circle, to get to E we have to cover additional 36,86 degrees. That is a way of (36,87/360)*2*PI*1000=643,33
Now we have to cover the remaining half circle 3141,59 and the final 1000.
All adds up to 6392,96
Hopefully, everybody reads this one now! The credit goes to alkatran!
And MrPolite, please put a (SOLVED) in the original post!
Forgot the gif. Here it is
Opus, when I sum those numbers, I get a value of 6402.96. So the poor guy still has to walk nearly 3 miles.
I've put the calculations into an Excel sheet, and this returns 6403.128 for a horizontal distance of 500.
However, there are possible solutions for every horizontal distance between roughly 525 and 630.
The minimum is at about 577 miles. (Angle is 209.98 deg.) First part: diagonal line = 1154.525, second part: 577, third part (short parth along circle): 524.1243, fourth part (semicircle): 3141.59, fifth part: 1000. This sums to 6397.242.
Kind regards,
Riis
Blame on me, but my kids are up to % and trig, maybe I need anotherone doing the first class!
BTW: Riis could I get your Excel-sheet?
Here is the excel sheet, zipped since it is about 160k large.
First column, x, is horizontal distance (BC and CD)
Second column, a, is diagonal distance (AC)
Third column, b, is 'short' path over circle's edge (DE)
Fourth column, total, is sum of AC, CD, DE, semi-circle and 1000
I'm wondering if the graph is exactly symmetrical. I didn't do any research to it. Might be a nice question for this forum :)
the optimal solution would be, when the first checkpoint is at -1000,1000/sqrt(3)
if x=distance to tangent from first checkpoint then you have the first checkpoint distance sqrt(r^2+x^2), the second x, and the third r by the angle between the normal to the tangent trough origo and the other tangent, which is derived trough the angle to the first checkpoint, pi/2-2*arcsin(x/sqrt(r^2+x^2)), the minimum of this function can be found by testing when its derivative is 0.
D[x+sqrt(r^2+x^2)+r*(pi/2-2*arcsin(x/sqrt(r^2+x^2)))]=
1+x/sqrt(r^2+x^2)-(2r(-(x^2/(r^2+x^2)^(3/2))+1/sqrt(r^2+x^2)))/sqrt(1-x^2/(r^2+x^2))=0
of which 2 are imaginary and one is negative and the positive is:
x=r/sqrt(3)
Thanks Kedaman for the derivation ( maybe I had derivation of trig'S in school, but only maybe)
BTW Your Avatar looks nice, should I try to hunt those sub'S done their in the med? I'll sure try using my application
sorry I'm asking this over and over :(Quote:
Originally posted by kedaman
the optimal solution would be, when the first checkpoint is at -1000,1000/sqrt(3)
if x=distance to tangent from first checkpoint then you have the first checkpoint distance sqrt(r^2+x^2), the second x, and the third r by the angle between the normal to the tangent trough origo and the other tangent, which is derived trough the angle to the first checkpoint, pi/2-2*arcsin(x/sqrt(r^2+x^2)), the minimum of this function can be found by testing when its derivative is 0.
D[x+sqrt(r^2+x^2)+r*(pi/2-2*arcsin(x/sqrt(r^2+x^2)))]=
1+x/sqrt(r^2+x^2)-(2r(-(x^2/(r^2+x^2)^(3/2))+1/sqrt(r^2+x^2)))/sqrt(1-x^2/(r^2+x^2))=0
of which 2 are imaginary and one is negative and the positive is:
x=r/sqrt(3)
I understood what you said about the first checkpoint, but nothing after it. Here;s what I do understand:
http://www.vbforums.com/attachment.php?postid=1304869
but I dont know how you get pi/2 - 2teta..... and I dont understand what angle that is.
And in the derivitive, why do you have an X in there? -> D[ x + ...]
I will really appriciate if you show me how you got those...
I tried the other solution that opus was saying before, and I was getting 6403 miles. So I guess your solution is better
Thanks for the help everyone... this is an interesting problem, but I still dont understand kedaman's solution:D(maybe I'm dumb:D:D)
AND: is my diagram right? is x where it's supposed to be :confused:
(I've been sleeping at 3 am all days this week, and umm...ahem... my brain isnt working very well at the moment:p sorry if I make a dumb mistake)
aah alright, did I do this right?:D
http://www.vbforums.com/attachment.php?postid=1305061
Yup :)
alright I think I'm being annoying... one more thing:D:D:D:D:D:DQuote:
Originally posted by riis
Yup :)
why are both sides X? why are they equal?
Short answer: Yup
Long answer:
Theyare equal, since they are both start at the end a line coming from the centre of the circle and both are tangents to that circle. So they have to be identical in length.
aah got it! :) thanks :)Quote:
Originally posted by opus
Short answer: Yup
Long answer:
Theyare equal, since they are both start at the end a line coming from the centre of the circle and both are tangents to that circle. So they have to be identical in length.
A good ending to your story
http://www.math.yorku.ca/Who/Faculty.../Stewart1.html
one thing still confuses me, if he didn't for once look at the compass during his long trip, how did he manage to go in a straight line?
thanks opus, my av inspired by the diplomacy game i'm currently playing
http://www.math.yorku.ca/Who/Facult...h/Stewart1.html
COOL! I found a beter solution then Sherlock Holmes ;)
Oh darn, he didn't have to start at center :mad:
This is true! I don't think it's a math problem at all.Quote:
Posted by: kedaman
one thing still confuses me, if he didn't for once look at the compass during his long trip, how did he manage to go in a straight line?
If he had an accurate compass AND
he is a space travler AND
he knew he was traveling perpindicular to a straight river for 1000 miles,
then he obviously knew which direction he was going.
So his solution is to get out his compass and travel the opposite direction for 1000 miles - and there's the river...