#### 1. Angle based method

A arched lintel maintains it's angle if it has no parameters. It's just simple.

This can be efficient when proportion is a priority. And it's the easiest way.

But it makes difficult to predict total height of the window. It goes higher or lower depends on the width of the window.

#### 2. Radius based method

With this way, height of the lintel can be controlled separately by changing radius. But it's also hard to know peak of the lintel.

#### 3. Height based method with Pythagorean Theorem.

I'll explain it again for someone who can't remember what teacher said in boring math class.

At the above image, we can get the formula with Pythagorean Theorem as follow.

*radius^2 = (width/2)^2 + (radius-h)^2*

*so...*

~~radius^2~~ = (width/2)^2 ~~+ radius^2~~ -2*radius*h + h^2

..........

*so...*

**radius = ((width / 2) ^ 2 + h ^ 2) / (2 * h)**This formula can find radius with given width and height of arch.

Now we can control the height of arched lintels like this.

Enjoy. :-)

I like your method. It is intelligent, yet simple. That said, I think there is an even easier way to acomplish this for those who don't want to write equations.

ReplyDeleteIf you draw two reference lines along the "top of arch lintel" reference plane - each line starts at the center plane (left/right) and goes outward towards the two jamb planes that define the window's width. Then you can use these reference lines as starting poins for two "Tangent End Arc" reference lines coming out from the top of the arch and down to where the lintel meets the jambs.

One of the benefits of this approach is that it will allow you to model a full Round/Semicircular arch in addition to the segmental arches allowed by the Pythagorean theorem method. I'm not sure if you noticed, but when you set the lintel height to be equal to half the window's width (to create a semicircular arch) the constrains break.

Good idea!!, I think both have pros and cons. Your method can be easier if the exact number of radius is not necessary. (radius can be noticed in my method you know)

DeleteAnd I suppose curved reference lines are necessary at only one side(left or right) in your method. Because both left & right arch have same center point and radius.(I haven't tried, just guess)

And about the case of semicircular, I knew it, but I thought it's far different case when we 'design' windows. Anyway, it can be solved in my method with 'if' formula if necessary.

Thanks for your kind words and sharing your thought.