Looking for correlation between length and 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
asked Nov 22 '18 at 16:25
SectorSector
254
add a comment |
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
asked Nov 22 '18 at 16:25
SectorSector
254
add a comment |
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
asked Nov 22 '18 at 16:25
SectorSector
254
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 '18 at 16:25
SectorSector
254
asked Nov 22 '18 at 16:25
SectorSector
254
asked Nov 22 '18 at 16:25
SectorSector
254
asked Nov 22 '18 at 16:25
SectorSector
254
asked Nov 22 '18 at 16:25
SectorSector
254
254
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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 '18 at 16:48
Robert HowardRobert Howard
1,9161822
add a comment |
Use Cosine(x) = adjacent / hypotenuse
answered Nov 22 '18 at 16:28
John McGeeJohn McGee
1361
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009322%2flooking-for-correlation-between-length-and-angle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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}$$
answered Nov 22 '18 at 16:48
Robert HowardRobert Howard
1,9161822
add a comment |
The portion of the blue line segment between the vertex and the intersection with the dashed line must have length $x+h$, and that allows you to find $x$ like so:
$$begin{align}
cos(alpha)&=frac{h}{x+h} \[0.2ex]
x+h&=hsec(alpha) \[0.7ex]
x&=h(sec(alpha)-1)
end{align}$$
answered Nov 22 '18 at 16:48
Robert HowardRobert Howard
1,9161822
add a comment |
The portion of the blue line segment between the vertex and the intersection with the dashed line must have length $x+h$, and that allows you to find $x$ like so:
$$begin{align}
cos(alpha)&=frac{h}{x+h} \[0.2ex]
x+h&=hsec(alpha) \[0.7ex]
x&=h(sec(alpha)-1)
end{align}$$
answered Nov 22 '18 at 16:48
Robert HowardRobert Howard
1,9161822
The portion of the blue line segment between the vertex and the intersection with the dashed line must have length $x+h$, and that allows you to find $x$ like so:
$$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 '18 at 16:48
Robert HowardRobert Howard
1,9161822
answered Nov 22 '18 at 16:48
Robert HowardRobert Howard
1,9161822
answered Nov 22 '18 at 16:48
Robert HowardRobert Howard
1,9161822
answered Nov 22 '18 at 16:48
Robert HowardRobert Howard
1,9161822
1,9161822
add a comment |
add a comment |
Use Cosine(x) = adjacent / hypotenuse
answered Nov 22 '18 at 16:28
John McGeeJohn McGee
1361
add a comment |
Use Cosine(x) = adjacent / hypotenuse
answered Nov 22 '18 at 16:28
John McGeeJohn McGee
1361
add a comment |
Use Cosine(x) = adjacent / hypotenuse
answered Nov 22 '18 at 16:28
John McGeeJohn McGee
1361
Use Cosine(x) = adjacent / hypotenuse
answered Nov 22 '18 at 16:28
John McGeeJohn McGee
1361
answered Nov 22 '18 at 16:28
John McGeeJohn McGee
1361
answered Nov 22 '18 at 16:28
John McGeeJohn McGee
1361
answered Nov 22 '18 at 16:28
John McGeeJohn McGee
1361
1361
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009322%2flooking-for-correlation-between-length-and-angle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
