How to calculate the dimensions of an ellipse in a multi-view projection

The following image is an example that will help to demonstrate what i am saying and this below is a blender image version of an object with a slant surface at the top.

to draw the multi view projection of this view all the necessary dimensions are given, so the top and the front will be as follows

top: front:

i very well understand how the front and the top projections worked and here is where my question arises, as seen both from the pictorial view and the two side projection there are two thoroughly subtracted cylinders in the slant surface which means when i try to do the right side projection i will be seeing two ellipses rather than the original circles i have seen by the top view as shown below, so how could i figure out the dimensions of the major and minor axis of the ellipses.

right view:

• Dont have a compuer in front of me so i can not show you how to do this but ill give you a hint. Draw the hole hidden lines in the front projection. You can now project the holes intersection points to your desired view and they mark the extremes of the ellipse. Mar 14 '19 at 21:22

This is a simple trigonometric exercise. You really only need the "front" view to solve this.

Here's the cleaned up version of that view (I have omitted the holes for now, and I assumed the right-hand "vertical" is 1.50 m thick, which isn't explicitly stated in your diagrams):

So, really, the only things we need to observe here is the slope of the incline, which can be trivially found to be

$$\dfrac{\text{d}y}{\text{d}} = \dfrac{0.99}{9.00} = 0.11$$

So, for every meter traveled horizontally, the incline has traveled 11 centimeters vertically.

Now, one of the cylindrical holes is shown with a diameter of 1 m[1]. I'm assuming this is valid for both.

Knowing the cylinder's diameter and the slope of the incline, it's easy to see that the vertical drop will be of 0.11 meters. After all, if you're at the top-most lip of the hole and move horizontally to the other end, you'll need to drop 0.11 meters to get to the incline once again.

So when you're drawing the holes in the "right" view, your ellipses need to have a major diameter of 1 meter (just like the actual circle, since there's no "warping" at this angle) and a minor diameter of 0.11 meters, since that's the vertical drop between the top and bottom lips of the circle.

[1] There's a point of uncertainty here as to whether the cylinder's cross-section is circular or whether the hole on the inclined surface is circular. Only one of these can be true (if the cylinder is circular, then the inclined hole will be an ellipse; if the inclined hole is circular, then the cylinder is elliptical). I have further assumed here that the cylinder is circular since, well, that seems much more likely.

For completeness, in the unusual case where the inclined hole is circular (and the cylinder and hole at the bottom face are therefore elliptical), the solution would require a tiny bit more math: we'd need to use the Pythagorean Theorem to find the hypotenuse associated with the given slope, and then just use proportions to find the vertical step equivalent to a 1 m hyotenuse with the same slope.

\small\begin{align} h &= \sqrt{0.11^2 + 1^2} = 1.00603\text{ m} \\ \dfrac{x}{1} &= \dfrac{0.11}{1.00603} \\ \therefore x &= 0.10934\text{ m} \end{align}