• Create Account

## Land on a specific spot

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

6 replies to this topic

### #1kahl  Members

102
Like
0Likes
Like

Posted 29 September 2012 - 02:54 AM

Hi guys,

After a couple of test I've edited the question to make it simple and easy to understand.I have a fixed initial velocity V(Vx,Vy,VZ) and a spot on higher ground E(Ex,Ey,Ez), my character must jump up there and he stands still at S(Sx,Sy,Sz). How Can I find the closest point to the character that can make him achieve the jump?And If the spot Is not available how can check other solutions?

Edited by kahl, 29 September 2012 - 09:42 AM.

### #2luca-deltodesco  Members

637
Like
0Likes
Like

Posted 29 September 2012 - 05:07 AM

With no limit on angle and force, the quickest way to get somewhere is to go with the as much force as possible in a straight line

### #3kahl  Members

102
Like
0Likes
Like

Posted 29 September 2012 - 05:22 AM

Then you will miss the target, I need a parabola ;)

### #4luca-deltodesco  Members

637
Like
0Likes
Like

Posted 29 September 2012 - 05:44 AM

Why would you miss the target? You put no limit on the amount of force you can use, so the quickest way to get there is to go in a straight line with infinite force

### #5kahl  Members

102
Like
0Likes
Like

Posted 29 September 2012 - 05:55 AM

Because you will hit the border of the platform, I forgot to tell that the end location is the center of the platform. I need a parabola and a straight line is not realistic :never saw a platform game where you jump in straight line. I could use a simple vector force V(0,1,1) that makes the character jump forward but It's not my goal. I need to trace a parabola from startpoint to end point taking in account gravity, then plug the vector in my velocity vector so the character can move. . .

### #6kahl  Members

102
Like
0Likes
Like

Posted 29 September 2012 - 07:14 AM

Changed some things to make understandable and easier to answer. Sorry for the inconvenient. . .

### #7Freya  Members

831
Like
0Likes
Like

Posted 04 October 2012 - 02:10 AM

Find a direction vector to the platform:
vec direction = (E-S).normalized();


In order to jump up add some value to the upVector:
direction.y +=5; //y is upVector and 5 can whatever even (E.y-S.y)


Then multiply a direction vector with velocity.

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.