Position of points in an isometric image after rotationGetting point-coordinates after a rotationcalculate out-of-plane and in-plane rotation from virtual camera position.Find center of rotation after object rotated by known angle (2D)New angle formed after rotating pipeSquare's angle of rotation using 4 cornersImage after Rotation in Geometry???Geometric : Distance after Rotation 2DCalculate image rotation to correct for camera angleCalculate the position of two points based on direction and midpoint.How to rotate relative points in degrees?

How to deal with fear of taking dependencies

Finding files for which a command fails

What do the Banks children have against barley water?

Doomsday-clock for my fantasy planet

What's the difference between repeating elections every few years and repeating a referendum after a few years?

Is it legal to have the "// (c) 2019 John Smith" header in all files when there are hundreds of contributors?

Is it wise to focus on putting odd beats on left when playing double bass drums?

Crop image to path created in TikZ?

How can I fix this gap between bookcases I made?

Manga about a female worker who got dragged into another world together with this high school girl and she was just told she's not needed anymore

Hosting Wordpress in a EC2 Load Balanced Instance

What to wear for invited talk in Canada

What does 'script /dev/null' do?

What does it exactly mean if a random variable follows a distribution

"My colleague's body is amazing"

What are the advantages and disadvantages of running one shots compared to campaigns?

Why was the "bread communication" in the arena of Catching Fire left out in the movie?

Symmetry in quantum mechanics

How to make payment on the internet without leaving a money trail?

Eliminate empty elements from a list with a specific pattern

Does bootstrapped regression allow for inference?

If a centaur druid Wild Shapes into a Giant Elk, do their Charge features stack?

COUNT(*) or MAX(id) - which is faster?

Check if two datetimes are between two others



Position of points in an isometric image after rotation


Getting point-coordinates after a rotationcalculate out-of-plane and in-plane rotation from virtual camera position.Find center of rotation after object rotated by known angle (2D)New angle formed after rotating pipeSquare's angle of rotation using 4 cornersImage after Rotation in Geometry???Geometric : Distance after Rotation 2DCalculate image rotation to correct for camera angleCalculate the position of two points based on direction and midpoint.How to rotate relative points in degrees?













0












$begingroup$


I'm developing a game using isometric images for the units.
Each units has weapons represented in red on the first image below.



To attach animations to the weapons, I have to be able to calculate the position of each of those points after a rotation of X degrees (units move in 8 directions).



The isometric view is set with an angle of 45 degrees top down :



Points on the image for the main angle



My goal is to find a way to manually spot the red dots on one angle, and then apply a mathematical transformation to get the position of those points for all others angles.



Image after a rotation
Image after an other rotation



I have no clue on how to approach the problematic and I'd be happy to get some help here !



The 3 images are properly scaled if you want to give it a try.



Here is the first point coordinates in pixels on each image, from the top left corner (0,0) :



Point A - 0 degree rotation : (89,522)
Point A - 45 degree rotation : (573,550)
Point A - 90 degree rotation : (609,207)


So the goal is to have (89,522) and deduce (573,550) and (609,207) from that.



Thanks a lot !



G.



Edit :



It seems that a rotation on one frame around a given center point would not work as shown in that picture, probably because of the 45 degree top down angle.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I'm developing a game using isometric images for the units.
    Each units has weapons represented in red on the first image below.



    To attach animations to the weapons, I have to be able to calculate the position of each of those points after a rotation of X degrees (units move in 8 directions).



    The isometric view is set with an angle of 45 degrees top down :



    Points on the image for the main angle



    My goal is to find a way to manually spot the red dots on one angle, and then apply a mathematical transformation to get the position of those points for all others angles.



    Image after a rotation
    Image after an other rotation



    I have no clue on how to approach the problematic and I'd be happy to get some help here !



    The 3 images are properly scaled if you want to give it a try.



    Here is the first point coordinates in pixels on each image, from the top left corner (0,0) :



    Point A - 0 degree rotation : (89,522)
    Point A - 45 degree rotation : (573,550)
    Point A - 90 degree rotation : (609,207)


    So the goal is to have (89,522) and deduce (573,550) and (609,207) from that.



    Thanks a lot !



    G.



    Edit :



    It seems that a rotation on one frame around a given center point would not work as shown in that picture, probably because of the 45 degree top down angle.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I'm developing a game using isometric images for the units.
      Each units has weapons represented in red on the first image below.



      To attach animations to the weapons, I have to be able to calculate the position of each of those points after a rotation of X degrees (units move in 8 directions).



      The isometric view is set with an angle of 45 degrees top down :



      Points on the image for the main angle



      My goal is to find a way to manually spot the red dots on one angle, and then apply a mathematical transformation to get the position of those points for all others angles.



      Image after a rotation
      Image after an other rotation



      I have no clue on how to approach the problematic and I'd be happy to get some help here !



      The 3 images are properly scaled if you want to give it a try.



      Here is the first point coordinates in pixels on each image, from the top left corner (0,0) :



      Point A - 0 degree rotation : (89,522)
      Point A - 45 degree rotation : (573,550)
      Point A - 90 degree rotation : (609,207)


      So the goal is to have (89,522) and deduce (573,550) and (609,207) from that.



      Thanks a lot !



      G.



      Edit :



      It seems that a rotation on one frame around a given center point would not work as shown in that picture, probably because of the 45 degree top down angle.










      share|cite|improve this question











      $endgroup$




      I'm developing a game using isometric images for the units.
      Each units has weapons represented in red on the first image below.



      To attach animations to the weapons, I have to be able to calculate the position of each of those points after a rotation of X degrees (units move in 8 directions).



      The isometric view is set with an angle of 45 degrees top down :



      Points on the image for the main angle



      My goal is to find a way to manually spot the red dots on one angle, and then apply a mathematical transformation to get the position of those points for all others angles.



      Image after a rotation
      Image after an other rotation



      I have no clue on how to approach the problematic and I'd be happy to get some help here !



      The 3 images are properly scaled if you want to give it a try.



      Here is the first point coordinates in pixels on each image, from the top left corner (0,0) :



      Point A - 0 degree rotation : (89,522)
      Point A - 45 degree rotation : (573,550)
      Point A - 90 degree rotation : (609,207)


      So the goal is to have (89,522) and deduce (573,550) and (609,207) from that.



      Thanks a lot !



      G.



      Edit :



      It seems that a rotation on one frame around a given center point would not work as shown in that picture, probably because of the 45 degree top down angle.







      rotations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 23 at 5:47







      giviz

















      asked Mar 22 at 14:26









      givizgiviz

      33




      33




















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          You need to know not just the angle of the rotation, but also what axis the rotation is about. A 30 degree rotation about the x-axis is going to give a much different location than a 30 degree rotation about the y-axis. And a rotation about the y-axis passing through the ankles is going to give a different result than the same rotation about the y-axis passing through the shoulders.



          Game developers tend to prefer quaternion to matrix notation for these rotations, and I find it hard to believe you don't already such rotations built into your application. If you really haven't worked out how to move the camera around, then I urge you to put aside your immediate problem and start working your way through the wealth of material out there that explains how it's done, because it is far more than I (or anyone else) can adequately explain in a forum post.



          I don't have much need for 3D rotations, so I never remember how quaternion rotation works, but if you are interested, this is how it works by vectors:



          Suppose you want to rotate the point $mathbf p$ by $theta$ degrees about the axis passing through the point $mathbf q$ and extending in the direction of the vector $hat n$, where $|hat n| = 1$, and $theta$ is positive when rotating in the same direction as the fingers of your right-hand curl when your extended thumb points in the direction of $hat n$.



          Let $mathbf o = mathbf q + ((mathbf p - mathbf q)cdot hat n)hat n$ and let $vec p = mathbf p - mathbf o$ be the vector pointing to $mathbf p$ from $mathbf o$. Then $mathbf o$ is the point on the axis of rotation that is closest to $mathbf p$, and $vec p$ is perpendicular to $hat n$. Let $vec r = hat n times vec p$. The rotated point that you want is



          $$mathbf p' = mathbf o + vec pcos theta + vec rsintheta $$




          FYI - the reason for the somewhat unusual mixture of bold and $to$ notation is that the bold letters represent points in space, while $vec v$ represents a vector (direction and magnitude only). And the hat notation $hat n$ is a common way of denoting unit vectors.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you for your response, I should have mentioned that it's a 2D game so I haven't had to work with the camera. The sprites are made from a 3D model where a script rotate it and take an image on each angle. The rotation point is on the center of the model, so we know it's position. The 45 degree top down angle make things even more complicated. My initial though was that as the rotation point is fixed, all points might be located on a given circle around that center on the 1st image, and if we have one point we can determine the circle and so all others positions. Does that make any sense ?
            $endgroup$
            – giviz
            Mar 23 at 5:39










          • $begingroup$
            Well nan, I've added a picture 4 with such circle, and that's not a valid method, probably because of the top down angle.
            $endgroup$
            – giviz
            Mar 23 at 5:49






          • 1




            $begingroup$
            You have a 3D model. You HAVE to treat the points you are trying to track as 3D points. Once you've thrown out that 3rd dimension of information, you can't get it back, and it is necessary to get the locations right. Identify the points in your 3D model where your weapons are. Then for each orientation, rotate the 3D points first, then project it to your iso view to get the 2D coordinates.
            $endgroup$
            – Paul Sinclair
            Mar 23 at 14:06










          • $begingroup$
            Yeah that make sense. Thank you, I'd look into the 3D software to see if I can find a way to get those points coordinates there instead of using rotation calculations.
            $endgroup$
            – giviz
            Mar 23 at 14:46










          • $begingroup$
            @giviz The projected path of a point on the rotating model is going to be an ellipse in general.
            $endgroup$
            – amd
            Mar 24 at 19:50











          Your Answer





          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158204%2fposition-of-points-in-an-isometric-image-after-rotation%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          You need to know not just the angle of the rotation, but also what axis the rotation is about. A 30 degree rotation about the x-axis is going to give a much different location than a 30 degree rotation about the y-axis. And a rotation about the y-axis passing through the ankles is going to give a different result than the same rotation about the y-axis passing through the shoulders.



          Game developers tend to prefer quaternion to matrix notation for these rotations, and I find it hard to believe you don't already such rotations built into your application. If you really haven't worked out how to move the camera around, then I urge you to put aside your immediate problem and start working your way through the wealth of material out there that explains how it's done, because it is far more than I (or anyone else) can adequately explain in a forum post.



          I don't have much need for 3D rotations, so I never remember how quaternion rotation works, but if you are interested, this is how it works by vectors:



          Suppose you want to rotate the point $mathbf p$ by $theta$ degrees about the axis passing through the point $mathbf q$ and extending in the direction of the vector $hat n$, where $|hat n| = 1$, and $theta$ is positive when rotating in the same direction as the fingers of your right-hand curl when your extended thumb points in the direction of $hat n$.



          Let $mathbf o = mathbf q + ((mathbf p - mathbf q)cdot hat n)hat n$ and let $vec p = mathbf p - mathbf o$ be the vector pointing to $mathbf p$ from $mathbf o$. Then $mathbf o$ is the point on the axis of rotation that is closest to $mathbf p$, and $vec p$ is perpendicular to $hat n$. Let $vec r = hat n times vec p$. The rotated point that you want is



          $$mathbf p' = mathbf o + vec pcos theta + vec rsintheta $$




          FYI - the reason for the somewhat unusual mixture of bold and $to$ notation is that the bold letters represent points in space, while $vec v$ represents a vector (direction and magnitude only). And the hat notation $hat n$ is a common way of denoting unit vectors.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you for your response, I should have mentioned that it's a 2D game so I haven't had to work with the camera. The sprites are made from a 3D model where a script rotate it and take an image on each angle. The rotation point is on the center of the model, so we know it's position. The 45 degree top down angle make things even more complicated. My initial though was that as the rotation point is fixed, all points might be located on a given circle around that center on the 1st image, and if we have one point we can determine the circle and so all others positions. Does that make any sense ?
            $endgroup$
            – giviz
            Mar 23 at 5:39










          • $begingroup$
            Well nan, I've added a picture 4 with such circle, and that's not a valid method, probably because of the top down angle.
            $endgroup$
            – giviz
            Mar 23 at 5:49






          • 1




            $begingroup$
            You have a 3D model. You HAVE to treat the points you are trying to track as 3D points. Once you've thrown out that 3rd dimension of information, you can't get it back, and it is necessary to get the locations right. Identify the points in your 3D model where your weapons are. Then for each orientation, rotate the 3D points first, then project it to your iso view to get the 2D coordinates.
            $endgroup$
            – Paul Sinclair
            Mar 23 at 14:06










          • $begingroup$
            Yeah that make sense. Thank you, I'd look into the 3D software to see if I can find a way to get those points coordinates there instead of using rotation calculations.
            $endgroup$
            – giviz
            Mar 23 at 14:46










          • $begingroup$
            @giviz The projected path of a point on the rotating model is going to be an ellipse in general.
            $endgroup$
            – amd
            Mar 24 at 19:50















          0












          $begingroup$

          You need to know not just the angle of the rotation, but also what axis the rotation is about. A 30 degree rotation about the x-axis is going to give a much different location than a 30 degree rotation about the y-axis. And a rotation about the y-axis passing through the ankles is going to give a different result than the same rotation about the y-axis passing through the shoulders.



          Game developers tend to prefer quaternion to matrix notation for these rotations, and I find it hard to believe you don't already such rotations built into your application. If you really haven't worked out how to move the camera around, then I urge you to put aside your immediate problem and start working your way through the wealth of material out there that explains how it's done, because it is far more than I (or anyone else) can adequately explain in a forum post.



          I don't have much need for 3D rotations, so I never remember how quaternion rotation works, but if you are interested, this is how it works by vectors:



          Suppose you want to rotate the point $mathbf p$ by $theta$ degrees about the axis passing through the point $mathbf q$ and extending in the direction of the vector $hat n$, where $|hat n| = 1$, and $theta$ is positive when rotating in the same direction as the fingers of your right-hand curl when your extended thumb points in the direction of $hat n$.



          Let $mathbf o = mathbf q + ((mathbf p - mathbf q)cdot hat n)hat n$ and let $vec p = mathbf p - mathbf o$ be the vector pointing to $mathbf p$ from $mathbf o$. Then $mathbf o$ is the point on the axis of rotation that is closest to $mathbf p$, and $vec p$ is perpendicular to $hat n$. Let $vec r = hat n times vec p$. The rotated point that you want is



          $$mathbf p' = mathbf o + vec pcos theta + vec rsintheta $$




          FYI - the reason for the somewhat unusual mixture of bold and $to$ notation is that the bold letters represent points in space, while $vec v$ represents a vector (direction and magnitude only). And the hat notation $hat n$ is a common way of denoting unit vectors.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you for your response, I should have mentioned that it's a 2D game so I haven't had to work with the camera. The sprites are made from a 3D model where a script rotate it and take an image on each angle. The rotation point is on the center of the model, so we know it's position. The 45 degree top down angle make things even more complicated. My initial though was that as the rotation point is fixed, all points might be located on a given circle around that center on the 1st image, and if we have one point we can determine the circle and so all others positions. Does that make any sense ?
            $endgroup$
            – giviz
            Mar 23 at 5:39










          • $begingroup$
            Well nan, I've added a picture 4 with such circle, and that's not a valid method, probably because of the top down angle.
            $endgroup$
            – giviz
            Mar 23 at 5:49






          • 1




            $begingroup$
            You have a 3D model. You HAVE to treat the points you are trying to track as 3D points. Once you've thrown out that 3rd dimension of information, you can't get it back, and it is necessary to get the locations right. Identify the points in your 3D model where your weapons are. Then for each orientation, rotate the 3D points first, then project it to your iso view to get the 2D coordinates.
            $endgroup$
            – Paul Sinclair
            Mar 23 at 14:06










          • $begingroup$
            Yeah that make sense. Thank you, I'd look into the 3D software to see if I can find a way to get those points coordinates there instead of using rotation calculations.
            $endgroup$
            – giviz
            Mar 23 at 14:46










          • $begingroup$
            @giviz The projected path of a point on the rotating model is going to be an ellipse in general.
            $endgroup$
            – amd
            Mar 24 at 19:50













          0












          0








          0





          $begingroup$

          You need to know not just the angle of the rotation, but also what axis the rotation is about. A 30 degree rotation about the x-axis is going to give a much different location than a 30 degree rotation about the y-axis. And a rotation about the y-axis passing through the ankles is going to give a different result than the same rotation about the y-axis passing through the shoulders.



          Game developers tend to prefer quaternion to matrix notation for these rotations, and I find it hard to believe you don't already such rotations built into your application. If you really haven't worked out how to move the camera around, then I urge you to put aside your immediate problem and start working your way through the wealth of material out there that explains how it's done, because it is far more than I (or anyone else) can adequately explain in a forum post.



          I don't have much need for 3D rotations, so I never remember how quaternion rotation works, but if you are interested, this is how it works by vectors:



          Suppose you want to rotate the point $mathbf p$ by $theta$ degrees about the axis passing through the point $mathbf q$ and extending in the direction of the vector $hat n$, where $|hat n| = 1$, and $theta$ is positive when rotating in the same direction as the fingers of your right-hand curl when your extended thumb points in the direction of $hat n$.



          Let $mathbf o = mathbf q + ((mathbf p - mathbf q)cdot hat n)hat n$ and let $vec p = mathbf p - mathbf o$ be the vector pointing to $mathbf p$ from $mathbf o$. Then $mathbf o$ is the point on the axis of rotation that is closest to $mathbf p$, and $vec p$ is perpendicular to $hat n$. Let $vec r = hat n times vec p$. The rotated point that you want is



          $$mathbf p' = mathbf o + vec pcos theta + vec rsintheta $$




          FYI - the reason for the somewhat unusual mixture of bold and $to$ notation is that the bold letters represent points in space, while $vec v$ represents a vector (direction and magnitude only). And the hat notation $hat n$ is a common way of denoting unit vectors.






          share|cite|improve this answer









          $endgroup$



          You need to know not just the angle of the rotation, but also what axis the rotation is about. A 30 degree rotation about the x-axis is going to give a much different location than a 30 degree rotation about the y-axis. And a rotation about the y-axis passing through the ankles is going to give a different result than the same rotation about the y-axis passing through the shoulders.



          Game developers tend to prefer quaternion to matrix notation for these rotations, and I find it hard to believe you don't already such rotations built into your application. If you really haven't worked out how to move the camera around, then I urge you to put aside your immediate problem and start working your way through the wealth of material out there that explains how it's done, because it is far more than I (or anyone else) can adequately explain in a forum post.



          I don't have much need for 3D rotations, so I never remember how quaternion rotation works, but if you are interested, this is how it works by vectors:



          Suppose you want to rotate the point $mathbf p$ by $theta$ degrees about the axis passing through the point $mathbf q$ and extending in the direction of the vector $hat n$, where $|hat n| = 1$, and $theta$ is positive when rotating in the same direction as the fingers of your right-hand curl when your extended thumb points in the direction of $hat n$.



          Let $mathbf o = mathbf q + ((mathbf p - mathbf q)cdot hat n)hat n$ and let $vec p = mathbf p - mathbf o$ be the vector pointing to $mathbf p$ from $mathbf o$. Then $mathbf o$ is the point on the axis of rotation that is closest to $mathbf p$, and $vec p$ is perpendicular to $hat n$. Let $vec r = hat n times vec p$. The rotated point that you want is



          $$mathbf p' = mathbf o + vec pcos theta + vec rsintheta $$




          FYI - the reason for the somewhat unusual mixture of bold and $to$ notation is that the bold letters represent points in space, while $vec v$ represents a vector (direction and magnitude only). And the hat notation $hat n$ is a common way of denoting unit vectors.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 22 at 23:25









          Paul SinclairPaul Sinclair

          20.8k21543




          20.8k21543











          • $begingroup$
            Thank you for your response, I should have mentioned that it's a 2D game so I haven't had to work with the camera. The sprites are made from a 3D model where a script rotate it and take an image on each angle. The rotation point is on the center of the model, so we know it's position. The 45 degree top down angle make things even more complicated. My initial though was that as the rotation point is fixed, all points might be located on a given circle around that center on the 1st image, and if we have one point we can determine the circle and so all others positions. Does that make any sense ?
            $endgroup$
            – giviz
            Mar 23 at 5:39










          • $begingroup$
            Well nan, I've added a picture 4 with such circle, and that's not a valid method, probably because of the top down angle.
            $endgroup$
            – giviz
            Mar 23 at 5:49






          • 1




            $begingroup$
            You have a 3D model. You HAVE to treat the points you are trying to track as 3D points. Once you've thrown out that 3rd dimension of information, you can't get it back, and it is necessary to get the locations right. Identify the points in your 3D model where your weapons are. Then for each orientation, rotate the 3D points first, then project it to your iso view to get the 2D coordinates.
            $endgroup$
            – Paul Sinclair
            Mar 23 at 14:06










          • $begingroup$
            Yeah that make sense. Thank you, I'd look into the 3D software to see if I can find a way to get those points coordinates there instead of using rotation calculations.
            $endgroup$
            – giviz
            Mar 23 at 14:46










          • $begingroup$
            @giviz The projected path of a point on the rotating model is going to be an ellipse in general.
            $endgroup$
            – amd
            Mar 24 at 19:50
















          • $begingroup$
            Thank you for your response, I should have mentioned that it's a 2D game so I haven't had to work with the camera. The sprites are made from a 3D model where a script rotate it and take an image on each angle. The rotation point is on the center of the model, so we know it's position. The 45 degree top down angle make things even more complicated. My initial though was that as the rotation point is fixed, all points might be located on a given circle around that center on the 1st image, and if we have one point we can determine the circle and so all others positions. Does that make any sense ?
            $endgroup$
            – giviz
            Mar 23 at 5:39










          • $begingroup$
            Well nan, I've added a picture 4 with such circle, and that's not a valid method, probably because of the top down angle.
            $endgroup$
            – giviz
            Mar 23 at 5:49






          • 1




            $begingroup$
            You have a 3D model. You HAVE to treat the points you are trying to track as 3D points. Once you've thrown out that 3rd dimension of information, you can't get it back, and it is necessary to get the locations right. Identify the points in your 3D model where your weapons are. Then for each orientation, rotate the 3D points first, then project it to your iso view to get the 2D coordinates.
            $endgroup$
            – Paul Sinclair
            Mar 23 at 14:06










          • $begingroup$
            Yeah that make sense. Thank you, I'd look into the 3D software to see if I can find a way to get those points coordinates there instead of using rotation calculations.
            $endgroup$
            – giviz
            Mar 23 at 14:46










          • $begingroup$
            @giviz The projected path of a point on the rotating model is going to be an ellipse in general.
            $endgroup$
            – amd
            Mar 24 at 19:50















          $begingroup$
          Thank you for your response, I should have mentioned that it's a 2D game so I haven't had to work with the camera. The sprites are made from a 3D model where a script rotate it and take an image on each angle. The rotation point is on the center of the model, so we know it's position. The 45 degree top down angle make things even more complicated. My initial though was that as the rotation point is fixed, all points might be located on a given circle around that center on the 1st image, and if we have one point we can determine the circle and so all others positions. Does that make any sense ?
          $endgroup$
          – giviz
          Mar 23 at 5:39




          $begingroup$
          Thank you for your response, I should have mentioned that it's a 2D game so I haven't had to work with the camera. The sprites are made from a 3D model where a script rotate it and take an image on each angle. The rotation point is on the center of the model, so we know it's position. The 45 degree top down angle make things even more complicated. My initial though was that as the rotation point is fixed, all points might be located on a given circle around that center on the 1st image, and if we have one point we can determine the circle and so all others positions. Does that make any sense ?
          $endgroup$
          – giviz
          Mar 23 at 5:39












          $begingroup$
          Well nan, I've added a picture 4 with such circle, and that's not a valid method, probably because of the top down angle.
          $endgroup$
          – giviz
          Mar 23 at 5:49




          $begingroup$
          Well nan, I've added a picture 4 with such circle, and that's not a valid method, probably because of the top down angle.
          $endgroup$
          – giviz
          Mar 23 at 5:49




          1




          1




          $begingroup$
          You have a 3D model. You HAVE to treat the points you are trying to track as 3D points. Once you've thrown out that 3rd dimension of information, you can't get it back, and it is necessary to get the locations right. Identify the points in your 3D model where your weapons are. Then for each orientation, rotate the 3D points first, then project it to your iso view to get the 2D coordinates.
          $endgroup$
          – Paul Sinclair
          Mar 23 at 14:06




          $begingroup$
          You have a 3D model. You HAVE to treat the points you are trying to track as 3D points. Once you've thrown out that 3rd dimension of information, you can't get it back, and it is necessary to get the locations right. Identify the points in your 3D model where your weapons are. Then for each orientation, rotate the 3D points first, then project it to your iso view to get the 2D coordinates.
          $endgroup$
          – Paul Sinclair
          Mar 23 at 14:06












          $begingroup$
          Yeah that make sense. Thank you, I'd look into the 3D software to see if I can find a way to get those points coordinates there instead of using rotation calculations.
          $endgroup$
          – giviz
          Mar 23 at 14:46




          $begingroup$
          Yeah that make sense. Thank you, I'd look into the 3D software to see if I can find a way to get those points coordinates there instead of using rotation calculations.
          $endgroup$
          – giviz
          Mar 23 at 14:46












          $begingroup$
          @giviz The projected path of a point on the rotating model is going to be an ellipse in general.
          $endgroup$
          – amd
          Mar 24 at 19:50




          $begingroup$
          @giviz The projected path of a point on the rotating model is going to be an ellipse in general.
          $endgroup$
          – amd
          Mar 24 at 19:50

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158204%2fposition-of-points-in-an-isometric-image-after-rotation%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

          random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

          Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye