On board ship would be a difficult environment for making the observations, but is there something more fundamental than that?

]]>1. Maximum number of triangles and any 9 lines can create cannot be greater than 9C3 = 84. Therefore, if counting beyond 84 you are definitely duplicating.

2. If three or more lines go through the same point those lines cannot contribute to a triangle. (which is my second step above)

3. If there are two lines not intersecting with each other, they arent going to form a triangle (atleast within the picture), which is my third step.

Max triangles – triangles that cannot be created by more than 2 intersecting lines – triangles that cannot be created by non-intersecting lines ]]>

Identify

Total number of unique straight lines in the triangle

All intersection points with more than two lines

Total number of lines parallel (or not touching within the triangle)

Steps:

1. Calculate max number of triangles all ‘N’ lines can produce = Nmax = NC3

2. Calculate sum total of max triangles using number of lines (n) passing through each intersection with more than two lines = ∑nC3 (for intersections with more than 2 lines), where n = number of lines passing through that intersection > 2

For example, if an intersection has 5 lines, the total no. triangles from this intersection point is 5C3 = 10.

3. Calculate total number of non intersecting lines sets – if two lines dont meet then treat it as one set = S

The formula is

Total triangles = NC3 – ∑nC3 – (S*(N-2))

N – total number of lines

n – number of lines at any intersection with more than two lines

S – total number of a set of non intersecting line

Using the formula in the above example

Step 1

N = 9 lines

Nmax = NC3 = 9C3 = 84

Step 2

Total intersections with more than two lines 2

Intersection 1 (bottom left corner) – n1 = 5 lines

Intersection 2 (bottom right corner) = n2 = 5 lines

∑nC3 = n1C3 + n2C3

= 5C3 + 5C3

= 10 + 10

Step 3:

No. of non intersecting line sets = 0

S* (N-2) = 0 * (9-2) = 0

Total number of triangels in the above picture = NC3 – ∑nC3 – (S*(N-2))

= 84 – (10+10) – 0

= 64

What is more significant is the speed of light! The difference in distance to Jupiter between opposition (4 AU) and conjunction (12 AU) is 1.2e+9 km. That corresponds to a light travel-time of about 4000 seconds, or more than an hour! If you used ephmerides derived at opposition, you could (near conjunction) be off by an hour in your calculation of time, or as much as 1000 km in distance!

In fact, it was this effect which led to the first observation of the finite speed of light in 1676, by Romer. Huygens combined the time delay with an estimate of Earth’s orbit to derive a speed of 220,000 km/s (somewhat less than the currently accepted value).

]]>http://www.screwattack.com/shows/originals/death-battle/death-battle-goku-vs-superman

and for more writing about the philosophy of DBZ, check out http://www.thedaoofdragonball.com

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