# Inkscape - Aligning imported image to axonometric grid

I am building a map of our outdoor retail store and am using a screenshot from Google Maps satellite view as a guide. I want this map to be isometric/axonometric and have set my grid up with a 26.5 degree angle on both the X and Y.

Since it is difficult to align a view in Google Maps to exactly a 26.5 degree angle, I would rather align this guide image in Inkscape; however, I can't figure out how to get it lined up at a reasonably precise position.

How can I align/transform the image on the right so that it matches the green object on the left?

Begin with a regular rectangular or square image. The original in the example below is an exact square, so you can see the distortion clearly.

1. Scale vertically 86.602%
2. Skew horizontal -30°
3. Rotate 30°

After this, you can scale it to any size by simply holding down Ctrl as you click and drag one of the scale handles to constrain the aspect ratio.

Example showing positioning and scaling to fit the grid.

• How do you get the scale amount? What is the significance of 86.062%? I was going for a 26.5 degree angle, but I suppose 30 degrees works as well. Jul 8, 2020 at 23:01
• @pspahn - sorry, I didn't calculate it. I just looked it up online. I'm no mathematician! Jul 8, 2020 at 23:13
• You can also install the isometric projection extension which does all the job. Jul 8, 2020 at 23:15
• @BillyKerr Ah yes, I see in the comments some discussion on the formula to find the scale factor (which they say should actually be 86.602). It seems to come from the difference in length of the sides after skewing - ie: 100px square skewed 30deg leaves sides of 100px and 115.47 - 100/115.47=86.6025808 Jul 8, 2020 at 23:21
• Aha. Functionally equivalent, instead of scaling -> skewing -> rotating, you can rotate -> scale. For a 30deg angle: rotate 45, scale 57.7350269 (which is simply `(tan30)`). For 26.5deg angle: rotate 45, scale 49.8581608 (which is `(tan26.5)`) I guess I learned math today. Jul 8, 2020 at 23:52

@BillyKerr gave me the answer I needed and through that I found out a slightly different method which pretty much works the same. I'm including this answer simply because it allows an easy way to figure the correct scale for any arbitrary angle.

Scale horizontally by finding the tangent of the angle you want to use - in my case, `(tan26.5)`

• You know you can actually choose your own answer as best. My answer is missing the actual maths you needed for your specific 26.5 degree grid, which you worked out yourself. I've upvoted it anyway. Jul 9, 2020 at 0:24
• There is a better way yet. Instead of doing all those individual operations you can just set the individual vectors in the matrix, which is faster an easier to do. No need to copy illustrator specific defiencies. See graphicdesign.stackexchange.com/questions/107509/… Jul 9, 2020 at 6:07

But there's still ways. This one doesn't need multiplications with numbers.

1. The wanted final fit of the photo. It's drawn by clicking 5 times with the Pen; snap to grid must be =ON.

2. Applied to 1. rotate 30 degrees CCW and skew +30 degrees horizontally. You must scale your photo to this size. It's easy if you have snap to points all = ON, but snap to grid and bounding boxes = OFF.

3-4. The transformations needed to the photo after scaling. 3=horizontal skew -30 degrees, 4=rotate 30 degrees CW.

In image 1. the proportions are Width:Height = 5:4. If you want to keep the proportions of the photo as undistorted as possible in Isometric projection, but it must still fit into the grid select the proportions of image 1 as closely as possible the same as the original proportions of your photo.

In Inkscape you can scale images and other shapes proportionally if you hold the Ctrl-key.