Taurian's Forum Posts

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 16 Mar, 2014
  • Quote

  > Hey Taurian,

  > thanks for the tut.

  >

  > Always wanted to try out Ejecta for doing the iOs Export.

  > Going to try it right now <img src="{SMILIES_PATH}/icon_e_smile.gif" alt=":)" title="Smile">

  >

  > Do you have any experiences regarding audio, ajax and ads?

  >

  Please post your experience Ejecta. I'm very curious about how that works out.

  I made a video about it, it's in the tutorial. But if I could describe it in 3 words:

  Pretty fvut@!$%g easy

  At the bottom: https://www.scirra.com/tutorials/907/de ... e-easy-way

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 15 Mar, 2014
  • Quote

  It is possible, it's just difficult there. It's under Target >> Info

  "Bundle Settings"

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 15 Mar, 2014
  • Quote

  Sure -- But I haven't tested Ajax with Ejecta + Xcode, but I don't see a reason why it wouldn't work. Especially when WebSockets do.

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 15 Mar, 2014
  • Quote

  I'm been pretty upset with CocoonJS. Sure you can deploy with it, but what about making money? I found a better solution: EJecta and wrote up a tutorial about it today.

  https://www.scirra.com/tutorials/907/de ... e-easy-way

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• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 15 Mar, 2014
  • Quote

  I'm been pretty upset with CocoonJS. Sure you can deploy with it, but what about making money? I found a better solution: EJecta and wrote up a tutorial about it today.

  https://www.scirra.com/tutorials/907/de ... e-easy-way

  Ejecta is extensible and you should be able to call various Objective-C methods from it, so it's very possible to do what you're trying to do with it in XCode + Construct 2.

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 10 Mar, 2014
  • Quote

  It's the angles that was jamming me up. 95 is southwest. Thanks. I'm still curious if this equation seems correct: Rojo mentioned I had too many parenthesis:

  2000 + cos(95 degrees) * (30*10)

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 10 Mar, 2014
  • Quote

  I have removed Physics from the equation, it is and always has been a constant speed of 10 pixels per second.

  Take a look here:

  https://www.google.com/search?q=2000+%2B+cos(95+degrees)+*+(30*10)&rlz=1C1CHFX_enUS576US576&oq=2000+%2B+cos(95+degrees)+*+(30*10)&aqs=chrome..69i57j0.4111j0j7&sourceid=chrome&espv=210&es_sm=93&ie=UTF-8

  I'm getting the same result, any idea what's wrong?

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 10 Mar, 2014
  • Quote

  Where?

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 10 Mar, 2014
  • Quote

  oldX + cos(yourangle) * (30*10)= new x

  oldY + sin(yourangle) * (30*10)= new y

  Should work for what ever angle you want to point at.

  Give or take some dt. Just so long as its a straight line.

  You'll have to call R0J0hound back for curves.

  It's not adding up see below:

  NewX = 2000 + cos(95 degrees) * (30*10) = 1973.85327718

  NewY = 528 + sin(95 degrees) * (30*10) = 826.858409428

  So after traveling for 30 seconds on a heading of 95 degrees, the dot will be at 1973, 826 ?

  The NewX seems to be incorrect as the dot is going South East, the X value should be increasing not decreasing.

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 9 Mar, 2014
  • Quote

  oldX + cos(yourangle) * (30*10)= new x

  oldY + sin(yourangle) * (30*10)= new y

  Cool, will this work if we change the heading from 315 degrees to 90?

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 9 Mar, 2014
  • Quote

  I've got:

  NewX = OldX + (30 * 10)

  NewY = OldY - (30 * 10)

  But this would only work if the dot is heading 315 degrees. I'll need to put more thought into what I'm trying to do.

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 9 Mar, 2014
  • Quote

  It's a start, but where does the 30 seconds fit in? We're using c2 for this example really.

  Where will the dot be 30 or so seconds later?

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 9 Mar, 2014
  • Quote

  It would need a mathematical solution. The server needs to calculate this in the end.

  We have the speed, it's starting point, we need a way to figure out it's ending point 30 or so seconds later.

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 9 Mar, 2014
  • Quote

  Hello,

  I have this diagram and I'm trying to work out the solution.

  

  I have a ship, traveling at 10 pixels per second with an angle of 315 degrees (North East)

  His position:

  X: 5000

  Y: 528

  How do I calculate where he'll be in 30 or so seconds?

  Keep in mind we're in C2 and he's about to hit 0y and go into the negative.

• • Taurian
  • 
  • • Joined 11 May, 2012
    • 
    • 43 topics • 169 posts
  • 1

  • 26 Feb, 2014
  • Quote

  I'm wondering the same. I can't find any CSS3 code. Or is this something that isn't possible?