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Topic, The Laws of Paths

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Ahroo 14 years ago
  See? Even the person that's MADE of paths agrees. |:p
demonicyoshi 14 years ago
  I just adjust the path until it feals like a good speed. I never calculate anything with them other then basic addition.

I dont think I will ever need this...
But good work, this will confuse people and then I will have all my competition eliminated!
Ahroo 14 years ago
  The power of Garygoh making equations that make no sense for things that make complete sense. |:p
Hexicube 14 years ago
  oh wait, I just realised that the formula does account for average speed changing XD
also, in the formal proof, then less-than signs should really be less-than-or-equal-to signs
azz 14 years ago
  Wow i dont understand this
Hexicube 14 years ago
  I was complaining that the average speed of the whole line was used for calculating time to only travel part of it, when that causes the average speed to be different
Garygoh884 14 years ago
  Solving the Law 1 equation:
Hexicube 14 years ago
  alright...1 sec...
P(speed at L)=(y-x)/2+x
(ds/dt) = s|x-P|/L
not sure...:P
AK 14 years ago
  .....What?

This topic is confusing, and i'm in Trig...
allyally 14 years ago
  HA! Hexicube FAILS again!
Garygoh884 14 years ago
  Wrong. If s has reached the endpoint (L), then s = L. Now, Hexicube's adjustment is wrong because

(ds/dt) = (x + (y-x))/2 + x = y/2 + x.
Hexicube 14 years ago
  now, for law 1, a fix:
(ds/dt)=(x + s(y-x)/L)/2+x
jasperpostema 14 years ago
  Ok, what does this mean? I cannot scroll over the page!
Hexicube 14 years ago
  s(y-x)/L
incorrect - assumes same speed throughout(equal to avg speed)
allyally 14 years ago
  Gis, If 2 node speeds are different, you might want to know what time it will be a point between them, as it wont be linear to to the acceleration/ deceleration in the node speeds, So this formulae works out the time,
Elizea 14 years ago
  These are cool I guess, but I can't see anyone calculating path things like these while making a level.
gameinsky 14 years ago
  Quest- ion.
What is the law ?
allyally 14 years ago
  Ooh... This could be useful... Nice work Gary,

[and i guess you use √a^2 + b^2 to get L]
Garygoh884 14 years ago
  Law 1
SPOILER

In this diagram, the position of s against time t is
(ds/dt) = x + s(y-x)/L. First-order differential equation

If y = x, then
(ds/dt) = x
s = xt (C is ignored since as t = 0, s = 0)

Formal proof
SPOILER
At s = 0, v = x.
At s = L, v = y.
If s holds 0 < s < L, then v = (L-s)x/L + sy/L.
Simplifying gives v = x - s(y-x)/L.
The velocity, v, is given by v = ds/dt. Also, s = ∫v dt.
Substituting v into the previous equation gives
(ds/dt) = x - s(y-x)/L.












Click the spoiler to open!


Law 2
SPOILER

In this diagram, a/b = A/B.













Click the spoiler to open!


Although the law(s) are usually developed by me, so PM me if you have developed a true law of paths.

[1]

General

First post of the topic

Garygoh884 14 years ago
  Law 1
SPOILER

In this diagram, the position of s against time t is
(ds/dt) = x + s(y-x)/L. First-order differential equation

If y = x, then
(ds/dt) = x
s = xt (C is ignored since as t = 0, s = 0)

Formal proof
SPOILER
At s = 0, v = x.
At s = L, v = y.
If s holds 0 < s < L, then v = (L-s)x/L + sy/L.
Simplifying gives v = x - s(y-x)/L.
The velocity, v, is given by v = ds/dt. Also, s = ∫v dt.
Substituting v into the previous equation gives
(ds/dt) = x - s(y-x)/L.












Click the spoiler to open!


Law 2
SPOILER

In this diagram, a/b = A/B.













Click the spoiler to open!


Although the law(s) are usually developed by me, so PM me if you have developed a true law of paths.
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