Looking for correlation between length and angle
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1
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The problem I'm facing might be rather easy to solve, but I can't think of a way how to do it atm. I want to clip straight 90-degree and some other degree lines. If I clip them at a fixed height (like h in the graphic) the 90-degree lines are too long.
So all I need to know is how to calculate the difference (x) which occurs if the line is not rotated by angle alpha.
Variables I know: alpha and h
geometry angle
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up vote
1
down vote
favorite
The problem I'm facing might be rather easy to solve, but I can't think of a way how to do it atm. I want to clip straight 90-degree and some other degree lines. If I clip them at a fixed height (like h in the graphic) the 90-degree lines are too long.
So all I need to know is how to calculate the difference (x) which occurs if the line is not rotated by angle alpha.
Variables I know: alpha and h
geometry angle
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The problem I'm facing might be rather easy to solve, but I can't think of a way how to do it atm. I want to clip straight 90-degree and some other degree lines. If I clip them at a fixed height (like h in the graphic) the 90-degree lines are too long.
So all I need to know is how to calculate the difference (x) which occurs if the line is not rotated by angle alpha.
Variables I know: alpha and h
geometry angle
The problem I'm facing might be rather easy to solve, but I can't think of a way how to do it atm. I want to clip straight 90-degree and some other degree lines. If I clip them at a fixed height (like h in the graphic) the 90-degree lines are too long.
So all I need to know is how to calculate the difference (x) which occurs if the line is not rotated by angle alpha.
Variables I know: alpha and h
geometry angle
geometry angle
asked Nov 22 at 16:25
Sector
254
254
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2 Answers
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The portion of the blue line segment between the vertex and the intersection with the dashed line must have length $x+h$, and that allows you to find $x$ like so:
$$begin{align}
cos(alpha)&=frac{h}{x+h} \[0.2ex]
x+h&=hsec(alpha) \[0.7ex]
x&=h(sec(alpha)-1)
end{align}$$
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1
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Use Cosine(x) = adjacent / hypotenuse
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The portion of the blue line segment between the vertex and the intersection with the dashed line must have length $x+h$, and that allows you to find $x$ like so:
$$begin{align}
cos(alpha)&=frac{h}{x+h} \[0.2ex]
x+h&=hsec(alpha) \[0.7ex]
x&=h(sec(alpha)-1)
end{align}$$
add a comment |
up vote
1
down vote
accepted
The portion of the blue line segment between the vertex and the intersection with the dashed line must have length $x+h$, and that allows you to find $x$ like so:
$$begin{align}
cos(alpha)&=frac{h}{x+h} \[0.2ex]
x+h&=hsec(alpha) \[0.7ex]
x&=h(sec(alpha)-1)
end{align}$$
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The portion of the blue line segment between the vertex and the intersection with the dashed line must have length $x+h$, and that allows you to find $x$ like so:
$$begin{align}
cos(alpha)&=frac{h}{x+h} \[0.2ex]
x+h&=hsec(alpha) \[0.7ex]
x&=h(sec(alpha)-1)
end{align}$$
The portion of the blue line segment between the vertex and the intersection with the dashed line must have length $x+h$, and that allows you to find $x$ like so:
$$begin{align}
cos(alpha)&=frac{h}{x+h} \[0.2ex]
x+h&=hsec(alpha) \[0.7ex]
x&=h(sec(alpha)-1)
end{align}$$
answered Nov 22 at 16:48
Robert Howard
1,9181822
1,9181822
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up vote
1
down vote
Use Cosine(x) = adjacent / hypotenuse
add a comment |
up vote
1
down vote
Use Cosine(x) = adjacent / hypotenuse
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up vote
1
down vote
up vote
1
down vote
Use Cosine(x) = adjacent / hypotenuse
Use Cosine(x) = adjacent / hypotenuse
answered Nov 22 at 16:28
John McGee
1361
1361
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