Mixing
Calculate after how many bounces screen saver logo will hit a corners?
$begingroup$
I've always wondered if it is possible to calculate the number of bounces until the screen saver logo hit both sides of a tv at the same time? Or if it will hit at all?
Assuming that we know: The logos dimensions, it's starting position, angle of movement and the tv dimensions. Also assuming that after a bounce the ball leaves the wall with the same angle (just like light and mirror).
This is just a curiosity so I can know the limits of mathematics.
geometry
asked Feb 2 '17 at 21:53
João PedroJoão Pedro
342116
$endgroup$
add a comment |
$begingroup$
I've always wondered if it is possible to calculate the number of bounces until the screen saver logo hit both sides of a tv at the same time? Or if it will hit at all?
Assuming that we know: The logos dimensions, it's starting position, angle of movement and the tv dimensions. Also assuming that after a bounce the ball leaves the wall with the same angle (just like light and mirror).
This is just a curiosity so I can know the limits of mathematics.
geometry
asked Feb 2 '17 at 21:53
João PedroJoão Pedro
342116
$endgroup$
add a comment |
$begingroup$
I've always wondered if it is possible to calculate the number of bounces until the screen saver logo hit both sides of a tv at the same time? Or if it will hit at all?
Assuming that we know: The logos dimensions, it's starting position, angle of movement and the tv dimensions. Also assuming that after a bounce the ball leaves the wall with the same angle (just like light and mirror).
This is just a curiosity so I can know the limits of mathematics.
geometry
asked Feb 2 '17 at 21:53
João PedroJoão Pedro
342116
$endgroup$
I've always wondered if it is possible to calculate the number of bounces until the screen saver logo hit both sides of a tv at the same time? Or if it will hit at all?
Assuming that we know: The logos dimensions, it's starting position, angle of movement and the tv dimensions. Also assuming that after a bounce the ball leaves the wall with the same angle (just like light and mirror).
This is just a curiosity so I can know the limits of mathematics.
geometry
geometry
asked Feb 2 '17 at 21:53
João PedroJoão Pedro
342116
asked Feb 2 '17 at 21:53
João PedroJoão Pedro
342116
edited Feb 2 '17 at 22:03
João Pedro
asked Feb 2 '17 at 21:53
João PedroJoão Pedro
342116
asked Feb 2 '17 at 21:53
João PedroJoão Pedro
342116
asked Feb 2 '17 at 21:53
João PedroJoão Pedro
342116
342116
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Basic approach. Imagine an infinite grid. An imaginary ball, corresponding to the real ball on your screen, starts in the unit square. It starts moving in some direction and continues moving forever in that direction.
When it passes a line of the form $x = j$, where $j$ is an integer, that corresponds to the real ball bouncing off either the left or the right side of the screen. When it passes a line of the form $y = k$, where $k$ is an integer, that corresponds to the real ball bouncing off either the upper or the lower side of the screen.
The imaginary ball always moves in the same direction, but the real ball, of course, changes direction each time it passes one of these integer lines.
However, the moments at which the imaginary ball encounters a corner of the form $(j, k)$, where $j$ and $k$ are both integers, correspond to those moments when the real ball hits a corner as well. So the question is, given an initial starting point $(x_0, y_0)$ and a velocity vector $(v_x, v_y)$, if the line
$$
frac{y-y_0}{v_y} = frac{x-x_0}{v_x}
$$
has an integer solution. If so, the displacement between the initial point $(x_0, y_0)$ and the solution point $(j, k)$, along with the component velocities $v_x$ and $v_y$ will tell you how long it takes to get there.
I'll try to add more about this problem when I get more time.
answered Feb 2 '17 at 22:04
Brian TungBrian Tung
25.7k32554
$endgroup$
$begingroup$
Note that the imaginary ball has to be a single point. Otherwise, crossing a line takes time, while for the real ball, reflection is instantaneous. You have to make the distance between the lines narrower correspondingly.
$endgroup$
– Arthur
Feb 3 '17 at 18:20
1
$begingroup$
@Arthur: You're right, although I think that if you shrink the dimensions of the screen by the diameter of the ball (and adjust the initial location of the ball accordingly) it will all work out.
$endgroup$
– Brian Tung
Feb 3 '17 at 18:21
$begingroup$
I agree. It's easier to see how to do it when you adjust the original screen to accomodate a point-ball before expanding to the imaginary grid.
$endgroup$
– Arthur
Feb 3 '17 at 18:23
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%2f2126445%2fcalculate-after-how-many-bounces-screen-saver-logo-will-hit-a-corners%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Basic approach. Imagine an infinite grid. An imaginary ball, corresponding to the real ball on your screen, starts in the unit square. It starts moving in some direction and continues moving forever in that direction.
When it passes a line of the form $x = j$, where $j$ is an integer, that corresponds to the real ball bouncing off either the left or the right side of the screen. When it passes a line of the form $y = k$, where $k$ is an integer, that corresponds to the real ball bouncing off either the upper or the lower side of the screen.
The imaginary ball always moves in the same direction, but the real ball, of course, changes direction each time it passes one of these integer lines.
However, the moments at which the imaginary ball encounters a corner of the form $(j, k)$, where $j$ and $k$ are both integers, correspond to those moments when the real ball hits a corner as well. So the question is, given an initial starting point $(x_0, y_0)$ and a velocity vector $(v_x, v_y)$, if the line
$$
frac{y-y_0}{v_y} = frac{x-x_0}{v_x}
$$
has an integer solution. If so, the displacement between the initial point $(x_0, y_0)$ and the solution point $(j, k)$, along with the component velocities $v_x$ and $v_y$ will tell you how long it takes to get there.
I'll try to add more about this problem when I get more time.
answered Feb 2 '17 at 22:04
Brian TungBrian Tung
25.7k32554
$endgroup$
$begingroup$
Note that the imaginary ball has to be a single point. Otherwise, crossing a line takes time, while for the real ball, reflection is instantaneous. You have to make the distance between the lines narrower correspondingly.
$endgroup$
– Arthur
Feb 3 '17 at 18:20
1
$begingroup$
@Arthur: You're right, although I think that if you shrink the dimensions of the screen by the diameter of the ball (and adjust the initial location of the ball accordingly) it will all work out.
$endgroup$
– Brian Tung
Feb 3 '17 at 18:21
$begingroup$
I agree. It's easier to see how to do it when you adjust the original screen to accomodate a point-ball before expanding to the imaginary grid.
$endgroup$
– Arthur
Feb 3 '17 at 18:23
add a comment |
$begingroup$
Basic approach. Imagine an infinite grid. An imaginary ball, corresponding to the real ball on your screen, starts in the unit square. It starts moving in some direction and continues moving forever in that direction.
When it passes a line of the form $x = j$, where $j$ is an integer, that corresponds to the real ball bouncing off either the left or the right side of the screen. When it passes a line of the form $y = k$, where $k$ is an integer, that corresponds to the real ball bouncing off either the upper or the lower side of the screen.
The imaginary ball always moves in the same direction, but the real ball, of course, changes direction each time it passes one of these integer lines.
However, the moments at which the imaginary ball encounters a corner of the form $(j, k)$, where $j$ and $k$ are both integers, correspond to those moments when the real ball hits a corner as well. So the question is, given an initial starting point $(x_0, y_0)$ and a velocity vector $(v_x, v_y)$, if the line
$$
frac{y-y_0}{v_y} = frac{x-x_0}{v_x}
$$
has an integer solution. If so, the displacement between the initial point $(x_0, y_0)$ and the solution point $(j, k)$, along with the component velocities $v_x$ and $v_y$ will tell you how long it takes to get there.
I'll try to add more about this problem when I get more time.
answered Feb 2 '17 at 22:04
Brian TungBrian Tung
25.7k32554
$endgroup$
$begingroup$
Note that the imaginary ball has to be a single point. Otherwise, crossing a line takes time, while for the real ball, reflection is instantaneous. You have to make the distance between the lines narrower correspondingly.
$endgroup$
– Arthur
Feb 3 '17 at 18:20
1
$begingroup$
@Arthur: You're right, although I think that if you shrink the dimensions of the screen by the diameter of the ball (and adjust the initial location of the ball accordingly) it will all work out.
$endgroup$
– Brian Tung
Feb 3 '17 at 18:21
$begingroup$
I agree. It's easier to see how to do it when you adjust the original screen to accomodate a point-ball before expanding to the imaginary grid.
$endgroup$
– Arthur
Feb 3 '17 at 18:23
add a comment |
$begingroup$
Basic approach. Imagine an infinite grid. An imaginary ball, corresponding to the real ball on your screen, starts in the unit square. It starts moving in some direction and continues moving forever in that direction.
When it passes a line of the form $x = j$, where $j$ is an integer, that corresponds to the real ball bouncing off either the left or the right side of the screen. When it passes a line of the form $y = k$, where $k$ is an integer, that corresponds to the real ball bouncing off either the upper or the lower side of the screen.
The imaginary ball always moves in the same direction, but the real ball, of course, changes direction each time it passes one of these integer lines.
However, the moments at which the imaginary ball encounters a corner of the form $(j, k)$, where $j$ and $k$ are both integers, correspond to those moments when the real ball hits a corner as well. So the question is, given an initial starting point $(x_0, y_0)$ and a velocity vector $(v_x, v_y)$, if the line
$$
frac{y-y_0}{v_y} = frac{x-x_0}{v_x}
$$
has an integer solution. If so, the displacement between the initial point $(x_0, y_0)$ and the solution point $(j, k)$, along with the component velocities $v_x$ and $v_y$ will tell you how long it takes to get there.
I'll try to add more about this problem when I get more time.
answered Feb 2 '17 at 22:04
Brian TungBrian Tung
25.7k32554
$endgroup$
Basic approach. Imagine an infinite grid. An imaginary ball, corresponding to the real ball on your screen, starts in the unit square. It starts moving in some direction and continues moving forever in that direction.
When it passes a line of the form $x = j$, where $j$ is an integer, that corresponds to the real ball bouncing off either the left or the right side of the screen. When it passes a line of the form $y = k$, where $k$ is an integer, that corresponds to the real ball bouncing off either the upper or the lower side of the screen.
The imaginary ball always moves in the same direction, but the real ball, of course, changes direction each time it passes one of these integer lines.
However, the moments at which the imaginary ball encounters a corner of the form $(j, k)$, where $j$ and $k$ are both integers, correspond to those moments when the real ball hits a corner as well. So the question is, given an initial starting point $(x_0, y_0)$ and a velocity vector $(v_x, v_y)$, if the line
$$
frac{y-y_0}{v_y} = frac{x-x_0}{v_x}
$$
has an integer solution. If so, the displacement between the initial point $(x_0, y_0)$ and the solution point $(j, k)$, along with the component velocities $v_x$ and $v_y$ will tell you how long it takes to get there.
I'll try to add more about this problem when I get more time.
answered Feb 2 '17 at 22:04
Brian TungBrian Tung
25.7k32554
answered Feb 2 '17 at 22:04
Brian TungBrian Tung
25.7k32554
answered Feb 2 '17 at 22:04
Brian TungBrian Tung
25.7k32554
answered Feb 2 '17 at 22:04
Brian TungBrian Tung
25.7k32554
25.7k32554
$begingroup$
Note that the imaginary ball has to be a single point. Otherwise, crossing a line takes time, while for the real ball, reflection is instantaneous. You have to make the distance between the lines narrower correspondingly.
$endgroup$
– Arthur
Feb 3 '17 at 18:20
1
$begingroup$
@Arthur: You're right, although I think that if you shrink the dimensions of the screen by the diameter of the ball (and adjust the initial location of the ball accordingly) it will all work out.
$endgroup$
– Brian Tung
Feb 3 '17 at 18:21
$begingroup$
I agree. It's easier to see how to do it when you adjust the original screen to accomodate a point-ball before expanding to the imaginary grid.
$endgroup$
– Arthur
Feb 3 '17 at 18:23
add a comment |
$begingroup$
Note that the imaginary ball has to be a single point. Otherwise, crossing a line takes time, while for the real ball, reflection is instantaneous. You have to make the distance between the lines narrower correspondingly.
$endgroup$
– Arthur
Feb 3 '17 at 18:20
1
$begingroup$
@Arthur: You're right, although I think that if you shrink the dimensions of the screen by the diameter of the ball (and adjust the initial location of the ball accordingly) it will all work out.
$endgroup$
– Brian Tung
Feb 3 '17 at 18:21
$begingroup$
I agree. It's easier to see how to do it when you adjust the original screen to accomodate a point-ball before expanding to the imaginary grid.
$endgroup$
– Arthur
Feb 3 '17 at 18:23
$begingroup$
Note that the imaginary ball has to be a single point. Otherwise, crossing a line takes time, while for the real ball, reflection is instantaneous. You have to make the distance between the lines narrower correspondingly.
$endgroup$
– Arthur
Feb 3 '17 at 18:20
$begingroup$
Note that the imaginary ball has to be a single point. Otherwise, crossing a line takes time, while for the real ball, reflection is instantaneous. You have to make the distance between the lines narrower correspondingly.
$endgroup$
– Arthur
Feb 3 '17 at 18:20
1
1
$begingroup$
@Arthur: You're right, although I think that if you shrink the dimensions of the screen by the diameter of the ball (and adjust the initial location of the ball accordingly) it will all work out.
$endgroup$
– Brian Tung
Feb 3 '17 at 18:21
$begingroup$
@Arthur: You're right, although I think that if you shrink the dimensions of the screen by the diameter of the ball (and adjust the initial location of the ball accordingly) it will all work out.
$endgroup$
– Brian Tung
Feb 3 '17 at 18:21
$begingroup$
I agree. It's easier to see how to do it when you adjust the original screen to accomodate a point-ball before expanding to the imaginary grid.
$endgroup$
– Arthur
Feb 3 '17 at 18:23
$begingroup$
I agree. It's easier to see how to do it when you adjust the original screen to accomodate a point-ball before expanding to the imaginary grid.
$endgroup$
– Arthur
Feb 3 '17 at 18:23
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.
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%2f2126445%2fcalculate-after-how-many-bounces-screen-saver-logo-will-hit-a-corners%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
