Users browsing this thread: 2 Guest(s)
Thread Rating:
  • 1 Vote(s) - 5 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Hyperjumping calculation (server part)
02-13-2010, 05:57 PM, (This post was last modified: 02-13-2010, 05:58 PM by MajTom.)
#11
RE: Hyperjumping calculation (server part)
I almost never targeted when I used hyperjump, only targeted when it would say I could not jump
[Image: l_d36f07faf65349c5acd4bf7b7b373b54.png]
Reply
02-13-2010, 09:30 PM, (This post was last modified: 02-13-2010, 09:40 PM by Neo VII.)
#12
RE: Hyperjumping calculation (server part)
(02-13-2010, 04:29 PM)rajkosto Wrote: i rebound mine to T
so i can just HJ everywhere, very useful, much faster than running if theres a lot of obstacles in between (like DT)

(02-13-2010, 05:57 PM)MajTom Wrote: I almost never targeted when I used hyperjump, only targeted when it would say I could not jump

This is what I was wondering about as well. I always bound mine to V and rarely target HJ'd anywhere.

Edit: Another note about that, when you HJ with a keybind 'char' rather than targeting, you have to hold down that 'char'. So often I would let go of the keybound 'char' to drop out of HJ shorter than where I would end up if I'd held it in the whole time.
[Image: SigVII-1.gif]
Reply
02-13-2010, 09:57 PM,
#13
RE: Hyperjumping calculation (server part)
Non targeted has the same pattern if i recall correctly.

As long as client sends the same data, protocol will be the same.

The exception to this calculation is Hyperjump in place, as going up and down doesnt make a parabola.

But anyway, detecting that initial point and end point are the same, we can do a list of packets going up, and the reverse, like a elevator, just changing Y axis.

Got it grabbed by the balls Tongue.
Reply
02-13-2010, 10:19 PM,
#14
RE: Hyperjumping calculation (server part)
its not the same for the key shortcut
since there is no target, you keep going untill you release the key (or you have traveled the maximum distance), then it calculates the falling parabola, and reverses you a bit so it can execute it and you land where you have been before the reversing
Reply
02-13-2010, 10:40 PM, (This post was last modified: 02-13-2010, 10:50 PM by Neo VII.)
#15
RE: Hyperjumping calculation (server part)
(02-13-2010, 09:57 PM)Morpheus Wrote: The exception to this calculation is Hyperjump in place, as going up and down doesnt make a parabola.

Ahh, okay... in place, like the vertical hyperjump thingy. Took me a second to figure out what that meant.

Could it be written similarly to the way the landing packet will be sent with the right end coordinates on the second ball of the parabola? Or would it be a different packet and animation.
(02-13-2010, 10:19 PM)rajkosto Wrote: its not the same for the key shortcut
since there is no target, you keep going untill you release the key (or you have traveled the maximum distance), then it calculates the falling parabola, and reverses you a bit so it can execute it and you land where you have been before the reversing

Wouldn't the maximum distance of the keybind be the same as the target at the end of the parabola?

Though I get your explanation for the calculation of what happens when you release the key bind.
[Image: SigVII-1.gif]
Reply
02-13-2010, 11:03 PM,
#16
RE: Hyperjumping calculation (server part)
if you hold and dont release the key, you are on a parabola with length of the maximum HJ distance with your HJ level
if you stop, thats what happens
also, the in place jump, is the parabola with length 0 and height max height Big Grin
Reply
02-13-2010, 11:07 PM,
#17
RE: Hyperjumping calculation (server part)
Nice.
[Image: SigVII-1.gif]
Reply
02-13-2010, 11:13 PM,
#18
RE: Hyperjumping calculation (server part)
So, what about small jumps over fences, walls, garbage bags etc, using just the space bar? Is it the same calculation? Like if you dont hit the ctrl it just hyperjumps you the minimum height for your level? Very interesting stuff....
Reply
02-14-2010, 12:03 AM, (This post was last modified: 02-14-2010, 12:07 AM by Morpheus.)
#19
RE: Hyperjumping calculation (server part)
(02-13-2010, 10:40 PM)Neo VII Wrote: Could it be written similarly to the way the landing packet will be sent with the right end coordinates on the second ball of the parabola? Or would it be a different packet and animation.
Cant use same thing for calculation. The ecuation and Gauss-Jordan needs 3 diff points. Otherwise Gaussian elimination will remove one of the ecuations as repeated and you cant calculate the 3rd value. Math things xD.
More like just going up in a line and then, when at the max height, go down.


possibly the same calculation for jumps, but with less altitude and different animation, i suppose.

anyway, having done the thing, is just changing the maths parameters. ^^
Reply
02-14-2010, 12:47 AM, (This post was last modified: 02-14-2010, 03:51 AM by Neo VII.)
#20
RE: Hyperjumping calculation (server part)
(02-13-2010, 04:56 AM)Morpheus Wrote: [Image: testes.png]

So the three points are the beginning, end, and the highest point in the arch? Y is cut off in this picture at the highest point.
[Image: SigVII-1.gif]
Reply


Forum Jump: