Geometry -Trigonometry without CAD

[RogerBryenton]

I am trying to figure out the distance involved:
Take a square sheet of paper, fold upper right corner (A)to lower left corner - gives a triangle. The fold angle -alpha, is 45 degrees, sloping up to your left. Unfold the paper. Now lift the upper right corner (A), by say 45 degrees, lift angle - beta. How do I calculate the distance from the flat position of point(A) to the new, lifted point (A), using fold angle alpha and lift angle beta?
By observation, when fold angle alpha is 45 degrees, and lift angle beta is 90 degrees, the distance moved is half of the paper width. But for different fold and lift angles, there is a way using sine and cosine functions, that elude me. I do not have CAD to figure this one out...
Thanks, Roger

 

[IRstuff]

Is this for school?

TTFN

FAQ731-376

 

[btrueblood]

Find the distance from the fold line to point A, call this distance "R", as in radius.

As the fold angle (let's call it X) increases, the y and z coordinates (y = lateral distance from fold line, z = vertical distance from flat position) are found by

z = R*sin(X)
y = R*cos(X)

 

[IRstuff]

You really need to review your textbooks

2*A*tangent(alpha/2)

TTFN

FAQ731-376

 

[rb1957]

its just R*theta ...

if alpha = 45deg, then R = a/sqrt(2), a = side length of original square. note this is a special case; generally, R is the distance between A and the fold, normal to the fold.

if beta (+ theta) = 90deg, then A has moved a/sqrt(2)*(pi/2)

 

[IRstuff]

oops, should have been 2*S*cos(alpha)*tangent(beta/2)

S*cos(alpha) = distance from point to fold. Lifting the fold results in isoceles triangle, whose base is twice its side multiplied by the tangent of half the center angle.

TTFN

FAQ731-376

 

[RogerBryenton]

Oops - distance sought not clear... sorry ... Let's try this again.
I am trying to figure out the distance involved:
The distance is the horizontal distance that point (A) in the upper right moves toward me, (y direction) in the flat xy plane.

Take a square sheet of paper, fold upper right corner (A)to lower left corner - gives a triangle. The fold angle -alpha, is 45 degrees, sloping up to your left. Unfold the paper. Now lift the upper right corner (A), by say 45 degrees, lift angle - beta. How do I calculate the HORIONTAL DISTANCE ONLY, IN THE Y DIRECTION, from the flat position of point(A) to the new, lifted point (A), using fold angle alpha and lift angle beta?

By observation, when fold angle alpha is 45 degrees, and lift angle beta is 90 degrees, the distance moved is half of the paper width. But for different fold and lift angles, there is a way using sine and cosine functions, that eludes me. I do not have CAD to figure this one out... (but I do have you folks, thanks)
Thanks, Roger

IRStuff - I think you are close, but I asked the wrong distance - see above. Is "S" the paper side length?

(I am re-designing how rubber dams are installed. To date, the toes of the end walls (side slopes) have been parallel to the stream direction - at 90 degrees to the dam. This causes a big crease/fold when the dam is inflated, because the inflated part effectively moves upstream, but the upper end point does not because it is clamped along the side slope. If the side slopes are placed at 45 to 60 degrees (fold angle), then the top end point of the rubber (which is clamped to the side slope) effectively is upstream, now in-line with the downstream edge of the inflated rubber - no crease) thx Roger

 

[IRstuff]

S is the side of the square.

so S*cos(alpha)(1-cos(beta) is the distance from original A to the projection of current A in the plane. Multiply by cos(alpha) to get y-change.

TTFN

FAQ731-376

 

[RogerBryenton]

IRstuff - thanks for working on this. I think you are really close, but when the lift angle is 90 degrees, we know that the point moves closer by half. In your formula, it becomes 1.44 for fold angle alpha 45 degrees, and lift angle beta of 90 degrees. My mind is foggy ... back to you... thx

 

[GBor]

You need a carpenter's wheel, not an engineer I did something similar...it took about 2 days before I worked through the geometry. A carpenter did it in about 10 seconds if I'm understanding your question correctly.

You are looking for the compound (miter) angle formed by two planes at an angle to each other. The first plane is flat, the second is at a skewed angle. This is not unlike where the gable end of a roof joins the hip.

A good roof man can answer your question easily. As engineers, we overthink this one

 

[dvd]

Roger:

Define some geometry. If you always have a square, won't the fold angle always be 45 degrees? If the sheet isn't always square, then define the sides better so everyone uses the same nomenclature.

What does "HORIONTAL DISTANCE ONLY, IN THE Y DIRECTION" mean? If I were defining a system on a 2-d plane the horizontal direction would be x-direction.

By rotating the sheet to maintain the fold-line in the vertical direction, I believe that your calculations will be simpler and also yield the "horizontal" movement. But I am still struggling with your meaning above.

 

[handleman]

If the fold angle is 45 degrees, sloping up to the left, does that mean that as the fold angle changes, the fold line always intersects with the bottom right corner? Is the paper always square, or does it change shape such that the fold always goes corner-to-corner?

-handleman, CSWP (The new, easy test)

 

[IRstuff]

Note that it's cos(alpha)^2 in the final equation, which results in 1/sqrt(2)^2 = 1/2, while sin(90) = 1, resulting in S/2

TTFN

FAQ731-376

 

[GBor]

Just for fun, I dropped it in to AutoCAD. Here is a table of values that may be useful:

Angle "X" Distance to corner
0 10 (I started with a 10x10 square)
10 9.924
20 9.6985
30 9.3301
45 8.5355
60 7.5
75 6.2941
90 5.0000

 

[hydtools]

The x-distance to projected point A is 0.5*S + 0.5*S*cos(beta). Beta being the fold up angle. S = length of a side of the square. The distance projected point A moves from flat to folded is S - x.

Ted

 

[quark]

When the lift angle is 90 and fold angle is 45, the distance (linear) should be equal to the side of the square and not half of it.

If the side is x, then the diagonal of the square is [√]2x and half of the diagonal is x/[√]2

The 90 deg lift forms a right angle triangle with two half diagonals as sides and the linear distance as hypotenuse. So, the linear distance is [√][(x^/2)+(x^/2)] = x

If alpha is 45 then the linear distance can be measured as ([√]2)xcos[β]/2

When the lift angle is [β], the triangle formed by the linear distance and two half diagonals as sides, has an angle of 180-[β] and other two angles are [β]/2 (since this is an equilateral triangle). The median is altitude for this two right angle triangles are formed with hypotenuse as half diagonal of the original square and one angle as [β]/2.

 

[quark]

For other fold angles of alpha (provided the sheet is folded upto the top left corner), the linear distance is 2([√]2)xcot[α]cos([β]/2)

In the earlier post, read cos[β]/2 as cos([β]/2)

 

[zekeman]

I get for the unit square

.5*(1-cos(beta))

Testing this

for: beta ans
0 0
45 1/2-sqrt(2)/4
90 1/2
180 1
Looks OK. Just multiply thre answer by the square dimension.

 

[IRstuff]

S = side

S*cos([α]) = distance from A to fold

S*cos([α])*cos([β]) = projection of line from A to fold onto plane

S*cos([α])*[1-cos([β])] = distance from A to projection of A on plane

S*cos([α])^2*cos([β]) = vertical displace from A to projection of A on plane





TTFN

FAQ731-376

 

[quark]

I presumed linear length as the line (or its projection) perpendicular to the fold line.