Non-Linear Relation

Oct 2019
7
1
France
Could someone tell me about the relation in the attached image.
One in the top follows the linear relation and extended length can be obtained through multiplying the height*angle.

What would be relation for bottom one?
 

Attachments

skipjack

Forum Staff
Dec 2006
21,478
2,470
Can you define the curve in some way?
 
  • Like
Reactions: 2 people
Oct 2019
7
1
France
Concave

It can be of conic, but it has to be a concave shape.

Thank you for your response..
 
Jun 2019
493
262
USA
Is this a design problem? As in, you are supposed to choose a concave shape, and then calculate the length?

A parabola, a hyperbola, a circular arc, and an elliptical arc are all conics that can be concave and can fit there. Once you have the shape (whether it be given, or whether you design it yourself), just write the equation as y = f(x). x is the vertical distance, and y is the horizontal distance. Then the ? is just f(h).

For example, the surface of the top shape is described by the equation y = ax. Substitute x = h, and y(h) = ah.
 
  • Like
Reactions: 1 person
Oct 2019
7
1
France
Question Explained

Is this a design problem? As in, you are supposed to choose a concave shape, and then calculate the length?

A parabola, a hyperbola, a circular arc, and an elliptical arc are all conics that can be concave and can fit there. Once you have the shape (whether it be given, or whether you design it yourself), just write the equation as y = f(x). x is the vertical distance, and y is the horizontal distance. Then the ? is just f(h).

For example, the surface of the top shape is described by the equation y = ax. Substitute x = h, and y(h) = ah.
Yes This is a design problem. And I have to create a concave curve. Could you plaese specify how to obtain the concave curve points. When I tried with the parabolic equation it doesn't work.
 

Attachments

Jun 2019
493
262
USA
What do you mean by, "it didn't work?"
Your simple choices for conics are:
$y = \sqrt{r^2-x^2}$, where $r\geq h_5$
$y = \sqrt{a^2-\frac{a^2}{b^2}x^2}$, where $b \geq h_5$
$y = ax^2$, where $a > 0$
$y = \sqrt{\frac{a^2}{b^2}x^2+a^2}$, for any $a, b$
 
Last edited: