So a new FTL drive is invented but it has a hitch: It can jump you exactly 10 light years in any direction. So you want to get to Alpha Centauri, 4.3 light years away.
How many jumps and in what direction does it take to get to Alpha Centauri, within say 1/000th of a light year?
If I understand you right, we're looking for a triangle with side lengths 10, 10 and 4.3 light years, respectively? If so, you could do it with two jumps - the first one 76.703 degrees to any direction from a straight course to the target, which takes you to exactly 10 lightyears from where you want to be?
Originally posted by talzamir If I understand you right, we're looking for a triangle with side lengths 10, 10 and 4.3 light years, respectively? If so, you could do it with two jumps - the first one 76.703 degrees to any direction from a straight course to the target, which takes you to exactly 10 lightyears from where you want to be?
Yeah, except I made the first angle 77.58 degrees. I did two right triangles with 10, 2.15 as two sides then the arc sin of .215 gave me 12.41 degrees so 90 degrees minus 12.41 is 77.58, hang a right 77.58 degrees do the 10 ly jump then hang a left 155.16 degrees do the 10 ly jump. So the three sides would be 2.15, 10 and 9.766 with 10 as hypotenuse. Sound right?
Makes sense. What happens if there is an extra condition that the entire course most be plotted at one go, and has to consist of exactly 10-lightyear distances and 90-degree turns?
Originally posted by talzamir Makes sense. What happens if there is an extra condition that the entire course most be plotted at one go, and has to consist of exactly 10-lightyear distances and 90-degree turns?
I don't see how that could possibly work, the whole idea of angles less or more than 90 degrees allows oddball distances to be available. I did a few routes with just 90 degree bends and all you get is a matrix of dots 10 ly apart, not very good for getting to a destination unless you are happy getting within 10 ly of something. If you had to go say 200 LY, then you could get within 10 ly of that distance but still left with a significant travel time even doing the 10 ly at say 0.9999999c.
And 10 years would go by on Earth for that leg and another 10 for the return to the same point where the 20 hops could get you back home.
It would still put you a lot further ahead timewise over just going 200 LY at that close to c. 20 years or so away from your time instead of 400 years in your personal future when you get back home.
Originally posted by sonhouse I don't see how that could possibly work, the whole idea of angles less or more than 90 degrees allows oddball distances to be available. I did a few routes with just 90 degree bends and all you get is a matrix of dots 10 ly apart, not very good for getting to a destination unless you are happy getting within 10 ly of something. If you had to go say 200 LY, ...[text shortened]... s or so away from your time instead of 400 years in your personal future when you get back home.
To go to Alpha Centauri with 10 LY steps and then turn 90 degress in any direction, it suffices with 4 steps and 3 turns.
Originally posted by FabianFnas To go to Alpha Centauri with 10 LY steps and then turn 90 degress in any direction, it suffices with 4 steps and 3 turns.
How do you turn 90 degrees in any direction? Ah, I think you mean think 3 dimensionally. I was plotting 2D courses. Maybe if you can do a ninety up and then a ninety sideways you can end up where you want to go, but harder to visualize!
So could you also plot a course if you were limited to doing 10 ly in the exact direction of the target? Like AC, 4.3 LY away, you aim in that direction but over shoot by 5.7 ly then do your twisty 3D 90 degree thing several times and get back to the target?
The way I intended was this.. to put it in easier to picture terms:
you have a set of sticks, all of them of equal 10" length; and then angle pieces that consist of a short length of tube with a 90-degree angle in the middle so you can push one stick into each end of the tube. To link points A and B, 4.3" apart, with a construct that consists of sticks linked together with those tube pieces, how many sticks at least do you need?
Originally posted by talzamir uh, could you explain that again?
The way I intended was this.. to put it in easier to picture terms:
you have a set of sticks, all of them of equal 10" length; and then angle pieces that consist of a short length of tube with a 90-degree angle in the middle so you can push one stick into each end of the tube. To link points A and B, 4.3" apart, with ...[text shortened]... s linked together with those tube pieces, how many sticks at least do you need?
with three steps, we have an arch with equal sides of length x. If each join can swivel, keeping all angles 90 then it is easy to see that the two ends can never come closer together than x (especially if you make the arch out of your fingers).
With 4 steps we have a square, now both ends meet, by hinging at the point of the square opposite to the ends (again we can do this with our fingers) it is easy to see that the distance between two ends two ends can be any number between 0 and sqrt(8).x
with three steps, we have an arch with equal sides of length x. If each join can swivel, keeping all angles 90 then it is easy to see that the two ends can never come closer together than x (especially if you make the arch out of your fingers).
With 4 steps we have a square, now both ends meet, by hinging at the point of the square opposite to the ...[text shortened]... asy to see that the distance between two ends two ends can be any number between 0 and sqrt(8).x
Is that supposed to be square root of 8 times x? You have a period between the (8) and the x. (8).x Is it supposed to be square root of (8x)?
No, the longest distance is a sort of W with all angles right angles
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So by pythagoras each hypotenuse is sqrt(2x^2)
there are two "hypoteni"? so the total distance from end to end is 2sqrt(2).x
which can be written as sqrt(8).x
if we rotate the bottom horizontal line out of the page and back to the left, keeping it attached at 90 degrees to the joint, we can turn that W to:
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And then we can rotate the top vertical line out of the page and down, again keeping it attached at 90 degrees to the joint we can turn the shape above to:
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this has distance 0 between the ends, and since all moves were continuous the ends have gone through all separations between 0 and sqrt(8).x
20 Mar '14 12:42
Problem dealing with interstellar travel:
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