Equation for the amplitude of a line sweeping across a triangle

[nordmoon]

Homework Statement

I have a triangle with top sides of 5 mm (base is then 5*sqrt(2)). The top triangle has a line cut through it, at a height of 5/sqrt(2)/2. I want to compute the amplitude as function of position as a line sweeps into the triangle that is cut off.

Requirement is A(0) = 0 and A(5/sqrt(2)/2)) = 5/2/L. After this point the amplitude is constant at this value. I need to find a formula that is in accordance to the requirement, but I can't think...

Homework Equations

The formula should increase linearly between the positions x = 0 and x = 5/sqrt(2)/2. f(0) = 0 and f(5/sqrt(2)/2) = 5/2/L. The line has a length of L, so the max amplitude becomes 5/2*(1/L)

The Attempt at a Solution

if h = 5/sqrt(2)/2
f(x) = h*sqrt(2)/L1*g(x)
f(x) = 5/(2*L)*g(x)
when
f(h) = 5/2/L, g(h) = 1 and
f(0)=0, g(0) =0

.. ok I am stuck

g(x) = kx+m
k*h+m = 1
k*0+m=0

m = 0
k*h = 1
k = 1/h

f(x)=5/2/L*(1/h*x)

-- but this is not correct.

1/h = 2*sqrt(2)/5

 

[FactChecker]

It's hard to picture this. What is L? Can you draw a picture and post it?

 

[Nidum]

Your problem descriptions are chaotic . I unravelled the last one and showed you one way of getting a solution but you didn't even look at it as far as I can tell .

If you want help on these problems please describe them clearly and provide a diagram in your opening post and then reply to answers given .

In the present problem finding the required answer is not difficult once you've drawn the diagram . You will find it easier to get a general answer in algebraic form before substituting actual numbers in .

Please make another attempt at getting a solution yourself and post your workings for us to see .

 

[Mark44]

I have a triangle with top sides of 5 mm (base is then 5*sqrt(2)).

It doesn't necessarily follow that the base is $5\sqrt 2$. From your description, you have an isosceles triangle with two sides of length 5 mm. Here's a drawing of two such triangles, neither of which has a base of $5 \sqrt 2$..

The only way the base could be $5\sqrt 2$ is if the angle at the top was 90°. That would have to be the situation in the triangle you describe, but you didn't mention that fact.

The top triangle has a line cut through it, at a height of 5/sqrt(2)/2.

What top triangle? You have described only a single triangle.

As others have mentioned, your description is incomplete, and doesn't give us a good idea of the problem you are attempting to solve. While we generally discourage posting an image of a problem, in this case a drawing would be very helpful and would be encouraged. I made the drawing above in Paint in less than a minute.

BTW, the vertical height of a triangle is called the altitude, not the amplitude.

 

[nordmoon]

Yes, sorry,.. I can overcomplicate things sometimes, but I think I solve the problem eventually.

A square is plot in two, two triangles. Top of the lines is placed at centre of the square. The square is rotated 45 degrees.
The triangle has a top angle of 90 degrees, in this case, the base is 5*√2 (with sides of lengths 5) .
It all came down to finding the correct equations that describes the amplitude as function of the position as three different lines positioned differently enters the triangle.
if I start with scanning the line from the centre line (height = 0), from right to left... the lower triangle is being scanned (top triangle is empty)
In the first case I want the amplitude to be following the requirement, f(0) = max amplitude = 5/L1 and f(5*√2) = 0. L1,L2,L3 is just a line (with different lengths) that is only partially inside the rectangle. It is just an normalisation factor.
The second case is the same but opposite, f(delay) =0 and f(delay+ 5*√2) = 5/L3.
The third case is just a vertical line with f(0) = 0 and f(5/√2) = amplitude max = 5/(√2*L2) and f(5√2) = 0.
I just needed to find the equations for this.

The next step was cutting the top triangle at a height h, and recalculate the new amplitude (with the top and lower added together).

My answer to this is,First case:
f(x) = 5/L1*(1-x/5√2)
I am setting zero as zero, .. if I have a height, the this line is delayed with 2* height, so at position -height, the line enters the triangle already.
For the top triangle, is a special case. The amplitude increases to position: + height and then is constant.
f(x) = √2/L1*x, at position height f(x) = height*√2/L1

Third case is the same as first case but opposite in amplitude. Maximal amplitude at position 2*delay+height*2. and at position 5√2+2*delay+height its is contant and then decreases with f(x) = √2/L3*(5√2-x+2*delay+2*height) to position 5√2+2*delay+height*2. The delay is the time between the lines as they enter the area investigated.

Second case:
f(x) = 1/L2*(x-delay-height) for x < 5/√2 + delay + height
f(x) = 1/L2*(delay+height+5√2-x) for x > 5/√2 + delay + height
if a height is added in top triangle, one of the triangle amplitudes is simply cut at height h.

I hope I got that correct, its all in MATLAB code. =)

 

[Mark44]

Yes, sorry,.. I can overcomplicate things sometimes, but I think I solve the problem eventually.

A square is plot in two, two triangles. Top of the lines is placed at centre of the square. The square is rotated 45 degrees.
The triangle has a top angle of 90 degrees, in this case, the base is 5*√2 (with sides of lengths 5) .
It all came down to finding the correct equations that describes the amplitude as function of the position as three different lines positioned differently enters the triangle.
if I start with scanning the line from the centre line (height = 0), from right to left... the lower triangle is being scanned (top triangle is empty)
In the first case I want the amplitude to be following the requirement, f(0) = max amplitude = 5/L1 and f(5*√2) = 0. L1,L2,L3 is just a line (with different lengths) that is only partially inside the rectangle. It is just an normalisation factor.
The second case is the same but opposite, f(delay) =0 and f(delay+ 5*√2) = 5/L3.
The third case is just a vertical line with f(0) = 0 and f(5/√2) = amplitude max = 5/(√2*L2) and f(5√2) = 0.
I just needed to find the equations for this.

The next step was cutting the top triangle at a height h, and recalculate the new amplitude (with the top and lower added together).

My answer to this is,First case:
f(x) = 5/L1*(1-x/5√2)
I am setting zero as zero, .. if I have a height, the this line is delayed with 2* height, so at position -height, the line enters the triangle already.
For the top triangle, is a special case. The amplitude increases to position: + height and then is constant.
f(x) = √2/L1*x, at position height f(x) = height*√2/L1

Third case is the same as first case but opposite in amplitude. Maximal amplitude at position 2*delay+height*2. and at position 5√2+2*delay+height its is contant and then decreases with f(x) = √2/L3*(5√2-x+2*delay+2*height) to position 5√2+2*delay+height*2. The delay is the time between the lines as they enter the area investigated.

Second case:
f(x) = 1/L2*(x-delay-height) for x < 5/√2 + delay + height
f(x) = 1/L2*(delay+height+5√2-x) for x > 5/√2 + delay + height
if a height is added in top triangle, one of the triangle amplitudes is simply cut at height h.

I hope I got that correct, its all in MATLAB code. =)

A drawing would be very helpful...
There's an old saying: "A picture is worth a thousand words."

 

[nordmoon]

Okay, I made a drawing with the amplitudes and uploaded. There are two cases I am looking at,. if the area is a rectangle or if the area is a square. Lines L1 and L2 is sweeping past the area. The question is how much of L1 or L2 respectively is part of the area. So I am looking at the amplitudes.

One case is a rectangle with sides a_x and a_y. The line has length L1 and is angled 45 degrees. How much of L1 is within the area?

One case is a square with sides a and the square is rotated 45 degrees to be aligned with L1.
The first line is L1 (as above) and the second has length L2. How much of L1 and L2 respectively is part of the area?

Thus, my question is...What will the amplitudes be for the last two cases and does the amplitudes agree with the ones I already calculated?

I am a bit confused with (1) the case when the square is larger than the length (a>L1 or L2) of the line. (2) the case when L2 in the largest part of the square, what is the amplitude? The length of the rotated square is then a*sqrt(2),.. does this mean the amplitude is larger than 1? Can the amplitude be larger than 1? Can a/L2 >1? or any ratio for example ay/sqrt(2)/L1.. We can't get larger line??