1
$\begingroup$

a long time back at school, I had a difficult excercise which was named X, Y and Z bodies. They were drawings where both the front view and the top view were fully given. We had to find the view from the left plus the body in a sketch drawing. Unfortunately, I don't remember the solutions anymore and I don't find out them anymore. Could anyone help? As an example, I added the Y body with below link.

Y body

Ah, yes: the forms are squares and the middle of the Y is the middle of the squares - in both views.

Thanks for your help and hints in advance!

Best Tom

$\endgroup$
  • $\begingroup$ You may find this helps, if it does give it a vote... engineering.stackexchange.com/a/28582/10902 $\endgroup$ – Solar Mike Aug 12 '19 at 7:20
  • $\begingroup$ The problem in answerin this question is tgat i dont know if you are using a first or third angle projection. $\endgroup$ – joojaa Nov 4 '19 at 6:43
  • $\begingroup$ Many years ago this was a "descriptive geometry" problem solved on the drawing board. $\endgroup$ – blacksmith37 Nov 4 '19 at 21:03
6
$\begingroup$

Here is an attempt at going for the least amount of faces.

enter image description here

enter image description here

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Now as I see it... I think, this was the correct solution! $\endgroup$ – TomS Nov 5 '19 at 15:53
  • 1
    $\begingroup$ I think if you enforce the rules “must be planar faces and straight edges” and “hidden lines are included” then this is the only correct answer. The hidden lines are perfectly occluded by the visible edges $\endgroup$ – Jonathan R Swift Nov 5 '19 at 16:53
  • $\begingroup$ The more I look at it, the more I am sure that this was the body we had in high school! $\endgroup$ – TomS Nov 6 '19 at 18:52
2
$\begingroup$

The bottom right line is meant for transferring lines between the top view and the Left view. Same with the Front and Left vertical line. Ideally, the picture would also include dotted lines to indicate any hidden geometry inside or under the view, but let's assume that there aren't any hidden geometries - since there aren't any hidden lines.

It would help indexing the corners in each view, to transfer the geometry properly, and knowing where each line crosses one from another view. Indicate also the "bottom" corners on the front view as "under" the upper corners in the top view.

Draw straight lines out from the points from the top view, until you reach the bottom right line. Switch directions upwards, to project it to the left side view. Project the same point from the front view as well, and where the two lines intersect from the same points in the left view plane, is where you have the location of that point in the left view.

I can work through this exercise precisely, with steps when I get home from work in a few hours :)

|improve this answer|||||
$\endgroup$
  • $\begingroup$ It will basically be a "valley" like shape $\endgroup$ – bullfighter Nov 13 '19 at 15:51
1
$\begingroup$

This is not a trivial question. More trick, than straight-forward.

The only way to set two sides to look alike is if one face is non-linear. Intersection appears to be at approximately 40% to 45% from front and top. If it was at 50%, it would be a triangle from the sides.

I have distorted the curve to illustrate it. Top to intersection is straight. Intersection to bottom is curved.

Orthogonal Views.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ I remember that there wew no curved lines. But the longer I think about it, the more I assume that our teacher was wrong and we didn't see that back at that time. But I remember it seemed so straight-forward. On the other side I can't imagine that there could be a solution some of us found back in high school and now, where I am much more experienced, I am not able to find that again. $\endgroup$ – TomS Nov 4 '19 at 4:40
  • $\begingroup$ If junction was in middle, it would be a perfect triangle. Key is front bottom edge is T'ed, which makes bottom corners, top corner and center on a plane. Not possible without curved surface. $\endgroup$ – StainlessSteelRat Nov 4 '19 at 4:54
  • $\begingroup$ My last try was a variant, where the junction of the Y was 8n the middle of the body and the left part of the front view was completely vertical at the front of the body. Also, there was an edge going from the vertical wall to the middle of the body. That edge collapses on the front view, but is visible on the top view (stem of the Y). But this also does not really work with only straight edges... $\endgroup$ – TomS Nov 4 '19 at 5:16
  • $\begingroup$ I tried a couple approaches, but always ended up with a triangle not a T on bottom. I sort of knew front planes could not be flat, because top and front are similar (if not identical). I placed center point and used surface to connect up straight edge to curved edge. It is the only way. $\endgroup$ – StainlessSteelRat Nov 4 '19 at 5:27
1
$\begingroup$

One possible solution sketched below (the OnShape model is here):

Orthographic projections

3D

|improve this answer|||||
$\endgroup$
  • 1
    $\begingroup$ This highlights the real issue here, which is that there is not enough information in the question to present an unambiguous solution... $\endgroup$ – Jonathan R Swift Nov 5 '19 at 8:04
  • $\begingroup$ Thanks for this idea, but the solution qq jkztd showed was the one we saw back at school. But interesting to see another possible solution. $\endgroup$ – TomS Nov 6 '19 at 19:00
  • $\begingroup$ qq jkztd 's solution is far more elegant (I was just editing mine to clean it up) $\endgroup$ – tonys Nov 6 '19 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.