Geometry -Trigonometry without CAD
Geometry -Trigonometry without CAD
(OP)
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
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





RE: Geometry -Trigonometry without CAD
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Geometry -Trigonometry without CAD
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)
RE: Geometry -Trigonometry without CAD
2*A*tangent(alpha/2)
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Geometry -Trigonometry without CAD
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)
RE: Geometry -Trigonometry without CAD
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: Eng-Tips.com Forum Policies
RE: Geometry -Trigonometry without CAD
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
RE: Geometry -Trigonometry without CAD
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: Eng-Tips.com Forum Policies
RE: Geometry -Trigonometry without CAD
RE: Geometry -Trigonometry without CAD
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
RE: Geometry -Trigonometry without CAD
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.
RE: Geometry -Trigonometry without CAD
-handleman, CSWP (The new, easy test)
RE: Geometry -Trigonometry without CAD
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Geometry -Trigonometry without CAD
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
RE: Geometry -Trigonometry without CAD
Ted
RE: Geometry -Trigonometry without CAD
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)xcosβ/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.
RE: Geometry -Trigonometry without CAD
In the earlier post, read cosβ/2 as cos(β/2)
RE: Geometry -Trigonometry without CAD
.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.
RE: Geometry -Trigonometry without CAD
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: Eng-Tips.com Forum Policies
RE: Geometry -Trigonometry without CAD