The rail guns fire at a speed of 0.99c, meaning to cross 2.5 light seconds it will take them longer. 2.5/0.99 is 2.525.



Yeah, I screwed up the wording. Though in my calculations I used light seconds. I can see no reason why a rail gun that fires a projectile in a straight line would form a cone. Idk about the nukes. The nukes aren't much larger than the blue sphere. I can imagine using all 6 of them and still have safe spots easily.

Because you haven't solved my problems yet.

Oh. I had just assumed the .99c was to facilitate the fact that a mass travelling at exactly 1c would be impossible. I just assumed the projectile would be there in 2.5 seconds lol. Since the target is so large, that extra .025 seconds doesn't make much of a difference anyway.

Err, maybe I confused you. In one of my prior attempts at a solution, I spaced the 19 railguns along a circle at the target around one of the nukes, and the shape they would make is a cone, since they're fired from a single point, but its approximately a cylinder since the target is so far away. The actual path of a single railgun would kill the target in a cylinder of 150m radius around it though, you're correct about that. Sorry to confuse you.

No, you can't. At least not in the configuration I used. I proved it by showing the closest spot to the target origin covered by the explosions is still too far out for the target to escape the blasts. Any spot further than the closest spot covered by the blasts will destroy it too, so we don't even have to worry about the rest, once we know that the closest spot will destroy the target. The closest spot will be at the intersection of 3 of the nukes, in the configuration I used, and there will be 8 such spots of equal distance. The blast radius at the closest spot to the origin is 866m, and since 937.5m(max radius traveled) - 150m(radius of the ship) = 787.5m, the blasts will hit the ship no matter where it flies, since 787.5m would then be the minimum distance needed to be 100% sure of hitting the target. Take a look at my picture above, you'll see what I mean. Also, keep in mind that the blast radius of each nuke is 750m, and since the target has a radius of 150m and can only travel 937.5m at max acceleration within the time frame, a single nuke detonated at the origin would cover 51% of the possible space the ship could reach.

I think we did. Maybe we haven't given you exactly what you wanted on #2, so if not give us a hint.

Alright, Here is the proof. This was what I originally had in mind. Course this is not the only way.

The sum of cubed from 1 to n can be expressed in a sum of permutations.

There are 4 elements x,y, z, u of set {1, 2, 3, ..., n+1}.
Define u as larger than the greatest of x, y, z.
( u > max{x, y, z} )
These are the steps to find the number of these sets.
(i) if u = k+1, (k = 1, 2, 3 ... n), then the number of permutations of (x, y, z, u) is k^3.
Therefore, the total number of these sets equate to the sum of cubed from 1 to n.
(ii) There are three possible cases.
a. When x, y, z are all same, the permutation of (x, y, z, u) becomes (n+1)!/2!(n-1)!
b. When only two of x,y,z are the same, the permutation of (x, y, z, u) becomes 6(n+1)!/3!(n-2)!
c. When all of x,y,z are different, the permutation of (x, y, z, u) becomes 6(n+1)!/4!(n-3)!

Thus, using (i) and (ii), we can see that the sum of cubed from 1 to n is (n+1)!/2!(n-1)! + 6(n+1)!/3!(n-2)! + 6(n+1)!/4!(n-3)!
That equates to n^2(n+1)^2/4. And the unit matches up.

Post has been edited 1 time(s), last time on Jun 19 2010, 2:15 pm by BeDazed.

You mean the number of permutations? For example, 5 is not a permutation of (x,y,z,u), but (u,x,z,y) and (y,u,z,x) are. So when you say "the permutation of (x,y,z,u) is k^3" I am confused. Do you mean "there are k^3 permutations of (x,y,z,u)"?

So we have n+1 consecutive integers, and we are choosing 4 values between 1 and n+1, such that each set always has one element greater than all the others. So if we call our max u, it can not be 1, and x y and z cannot be n+1, else we would not have u > max{x,y,z}. How many such sets can we find if we know n?

Is this the problem you're referring to?

Well yes.