Mixing
Rate of change when moving on a sphere is the same in all directions. So why different leg lengths of...
$begingroup$
Why can't points be plotted on a sphere equally such that a sphere could be divided where all the legs are equal say many triangles? The rate of change is the same in all directions so why are some legs longer in a geodesic dome? This is counter-intuitive for me.
geodesic
asked Jan 3 at 3:40
RICHRICH
1
$endgroup$
add a comment |
$begingroup$
Why can't points be plotted on a sphere equally such that a sphere could be divided where all the legs are equal say many triangles? The rate of change is the same in all directions so why are some legs longer in a geodesic dome? This is counter-intuitive for me.
geodesic
asked Jan 3 at 3:40
RICHRICH
1
$endgroup$
1
$begingroup$
Look up Platonic solids. The only polyhedra that can be constructed with equilateral triangles are the tetrahedron, the octahedron and the icosahedron.
$endgroup$
– John Douma
Jan 3 at 4:11
$begingroup$
The only regular polyhedra, that is.
$endgroup$
– amd
Jan 3 at 5:52
$begingroup$
Thank you John! I am familiar with polyhedras. If I place a equilateral triangal on a sphere, the points touching the surface, then hinge the next triangle so three point touch then the next will not work? Four points define a sphere but 3 define a plane? Or three planes define a sphere? I guess I just have to accept a residual shape.
$endgroup$
– RICH
Jan 3 at 14:49
$begingroup$
@RICH Try the experiment I suggest below.
$endgroup$
– John Douma
Jan 3 at 18:15
add a comment |
$begingroup$
Why can't points be plotted on a sphere equally such that a sphere could be divided where all the legs are equal say many triangles? The rate of change is the same in all directions so why are some legs longer in a geodesic dome? This is counter-intuitive for me.
geodesic
asked Jan 3 at 3:40
RICHRICH
1
$endgroup$
Why can't points be plotted on a sphere equally such that a sphere could be divided where all the legs are equal say many triangles? The rate of change is the same in all directions so why are some legs longer in a geodesic dome? This is counter-intuitive for me.
geodesic
geodesic
asked Jan 3 at 3:40
RICHRICH
1
asked Jan 3 at 3:40
RICHRICH
1
asked Jan 3 at 3:40
RICHRICH
1
asked Jan 3 at 3:40
RICHRICH
1
asked Jan 3 at 3:40
RICHRICH
1
1
1
$begingroup$
Look up Platonic solids. The only polyhedra that can be constructed with equilateral triangles are the tetrahedron, the octahedron and the icosahedron.
$endgroup$
– John Douma
Jan 3 at 4:11
$begingroup$
The only regular polyhedra, that is.
$endgroup$
– amd
Jan 3 at 5:52
$begingroup$
Thank you John! I am familiar with polyhedras. If I place a equilateral triangal on a sphere, the points touching the surface, then hinge the next triangle so three point touch then the next will not work? Four points define a sphere but 3 define a plane? Or three planes define a sphere? I guess I just have to accept a residual shape.
$endgroup$
– RICH
Jan 3 at 14:49
$begingroup$
@RICH Try the experiment I suggest below.
$endgroup$
– John Douma
Jan 3 at 18:15
add a comment |
1
$begingroup$
Look up Platonic solids. The only polyhedra that can be constructed with equilateral triangles are the tetrahedron, the octahedron and the icosahedron.
$endgroup$
– John Douma
Jan 3 at 4:11
$begingroup$
The only regular polyhedra, that is.
$endgroup$
– amd
Jan 3 at 5:52
$begingroup$
Thank you John! I am familiar with polyhedras. If I place a equilateral triangal on a sphere, the points touching the surface, then hinge the next triangle so three point touch then the next will not work? Four points define a sphere but 3 define a plane? Or three planes define a sphere? I guess I just have to accept a residual shape.
$endgroup$
– RICH
Jan 3 at 14:49
$begingroup$
@RICH Try the experiment I suggest below.
$endgroup$
– John Douma
Jan 3 at 18:15
1
1
$begingroup$
Look up Platonic solids. The only polyhedra that can be constructed with equilateral triangles are the tetrahedron, the octahedron and the icosahedron.
$endgroup$
– John Douma
Jan 3 at 4:11
$begingroup$
Look up Platonic solids. The only polyhedra that can be constructed with equilateral triangles are the tetrahedron, the octahedron and the icosahedron.
$endgroup$
– John Douma
Jan 3 at 4:11
$begingroup$
The only regular polyhedra, that is.
$endgroup$
– amd
Jan 3 at 5:52
$begingroup$
The only regular polyhedra, that is.
$endgroup$
– amd
Jan 3 at 5:52
$begingroup$
Thank you John! I am familiar with polyhedras. If I place a equilateral triangal on a sphere, the points touching the surface, then hinge the next triangle so three point touch then the next will not work? Four points define a sphere but 3 define a plane? Or three planes define a sphere? I guess I just have to accept a residual shape.
$endgroup$
– RICH
Jan 3 at 14:49
$begingroup$
Thank you John! I am familiar with polyhedras. If I place a equilateral triangal on a sphere, the points touching the surface, then hinge the next triangle so three point touch then the next will not work? Four points define a sphere but 3 define a plane? Or three planes define a sphere? I guess I just have to accept a residual shape.
$endgroup$
– RICH
Jan 3 at 14:49
$begingroup$
@RICH Try the experiment I suggest below.
$endgroup$
– John Douma
Jan 3 at 18:15
$begingroup$
@RICH Try the experiment I suggest below.
$endgroup$
– John Douma
Jan 3 at 18:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a fun experiment to do to understand how equilateral triangles can go together to make regular polyhedra. You will need construction paper and tape.
Cut equally sized equilateral triangles. You will need $32$ of them. To avoid writing "equilateral" every time, when I say triangle I mean equilateral triangle.
We are going to construct polyhedra by taping our triangles together with a fixed number of them meeting at a vertex.
If we insist that only one triangle can be at a vertex then the only shape we can construct is the triangle itself. Try $2$ triangles at a vertex and it is just as uninteresting.
Tape $3$ triangles together so they meet at a vertex. The three $60^{circ}$ angles that meet at the vertex will add up to less than $360^{circ}$ so when you tape the next triangle you will get a convex shape. In fact, you will only be able to tape one more triangle to get a tetrahedron.
Now do the same thing with $4$ triangles at a vertex. You will get an octahedron.
Five triangles at a vertex will give you an icosahedron.
What happens when we try $6$ triangles at a vertex? The $6$ interior angles add up to $360^{circ}$ so adding more triangles tiles the plane; it does not give you a convex shape.
Therefore, the tetrahedron, the octahedron and the icosahedron are the only regular polyhedra that can be formed with equilateral triangles and none of these approximate the sphere that well.
As a bonus, repeat the same experiment with squares and pentagons to get the cube and the dodecahedron.
Since regular hexagons already tile the plane, no Platonic solid can be formed with a regular $n$-gon where $nge 5$.
answered Jan 3 at 18:15
John DoumaJohn Douma
5,49711319
$endgroup$
$begingroup$
Thank you so much John I will look in to this.
$endgroup$
– RICH
Jan 4 at 5:12
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%2f3060227%2frate-of-change-when-moving-on-a-sphere-is-the-same-in-all-directions-so-why-dif%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
$begingroup$
Here is a fun experiment to do to understand how equilateral triangles can go together to make regular polyhedra. You will need construction paper and tape.
Cut equally sized equilateral triangles. You will need $32$ of them. To avoid writing "equilateral" every time, when I say triangle I mean equilateral triangle.
We are going to construct polyhedra by taping our triangles together with a fixed number of them meeting at a vertex.
If we insist that only one triangle can be at a vertex then the only shape we can construct is the triangle itself. Try $2$ triangles at a vertex and it is just as uninteresting.
Tape $3$ triangles together so they meet at a vertex. The three $60^{circ}$ angles that meet at the vertex will add up to less than $360^{circ}$ so when you tape the next triangle you will get a convex shape. In fact, you will only be able to tape one more triangle to get a tetrahedron.
Now do the same thing with $4$ triangles at a vertex. You will get an octahedron.
Five triangles at a vertex will give you an icosahedron.
What happens when we try $6$ triangles at a vertex? The $6$ interior angles add up to $360^{circ}$ so adding more triangles tiles the plane; it does not give you a convex shape.
Therefore, the tetrahedron, the octahedron and the icosahedron are the only regular polyhedra that can be formed with equilateral triangles and none of these approximate the sphere that well.
As a bonus, repeat the same experiment with squares and pentagons to get the cube and the dodecahedron.
Since regular hexagons already tile the plane, no Platonic solid can be formed with a regular $n$-gon where $nge 5$.
answered Jan 3 at 18:15
John DoumaJohn Douma
5,49711319
$endgroup$
$begingroup$
Thank you so much John I will look in to this.
$endgroup$
– RICH
Jan 4 at 5:12
add a comment |
$begingroup$
Here is a fun experiment to do to understand how equilateral triangles can go together to make regular polyhedra. You will need construction paper and tape.
Cut equally sized equilateral triangles. You will need $32$ of them. To avoid writing "equilateral" every time, when I say triangle I mean equilateral triangle.
We are going to construct polyhedra by taping our triangles together with a fixed number of them meeting at a vertex.
If we insist that only one triangle can be at a vertex then the only shape we can construct is the triangle itself. Try $2$ triangles at a vertex and it is just as uninteresting.
Tape $3$ triangles together so they meet at a vertex. The three $60^{circ}$ angles that meet at the vertex will add up to less than $360^{circ}$ so when you tape the next triangle you will get a convex shape. In fact, you will only be able to tape one more triangle to get a tetrahedron.
Now do the same thing with $4$ triangles at a vertex. You will get an octahedron.
Five triangles at a vertex will give you an icosahedron.
What happens when we try $6$ triangles at a vertex? The $6$ interior angles add up to $360^{circ}$ so adding more triangles tiles the plane; it does not give you a convex shape.
Therefore, the tetrahedron, the octahedron and the icosahedron are the only regular polyhedra that can be formed with equilateral triangles and none of these approximate the sphere that well.
As a bonus, repeat the same experiment with squares and pentagons to get the cube and the dodecahedron.
Since regular hexagons already tile the plane, no Platonic solid can be formed with a regular $n$-gon where $nge 5$.
answered Jan 3 at 18:15
John DoumaJohn Douma
5,49711319
$endgroup$
$begingroup$
Thank you so much John I will look in to this.
$endgroup$
– RICH
Jan 4 at 5:12
add a comment |
$begingroup$
Here is a fun experiment to do to understand how equilateral triangles can go together to make regular polyhedra. You will need construction paper and tape.
Cut equally sized equilateral triangles. You will need $32$ of them. To avoid writing "equilateral" every time, when I say triangle I mean equilateral triangle.
We are going to construct polyhedra by taping our triangles together with a fixed number of them meeting at a vertex.
If we insist that only one triangle can be at a vertex then the only shape we can construct is the triangle itself. Try $2$ triangles at a vertex and it is just as uninteresting.
Tape $3$ triangles together so they meet at a vertex. The three $60^{circ}$ angles that meet at the vertex will add up to less than $360^{circ}$ so when you tape the next triangle you will get a convex shape. In fact, you will only be able to tape one more triangle to get a tetrahedron.
Now do the same thing with $4$ triangles at a vertex. You will get an octahedron.
Five triangles at a vertex will give you an icosahedron.
What happens when we try $6$ triangles at a vertex? The $6$ interior angles add up to $360^{circ}$ so adding more triangles tiles the plane; it does not give you a convex shape.
Therefore, the tetrahedron, the octahedron and the icosahedron are the only regular polyhedra that can be formed with equilateral triangles and none of these approximate the sphere that well.
As a bonus, repeat the same experiment with squares and pentagons to get the cube and the dodecahedron.
Since regular hexagons already tile the plane, no Platonic solid can be formed with a regular $n$-gon where $nge 5$.
answered Jan 3 at 18:15
John DoumaJohn Douma
5,49711319
$endgroup$
Here is a fun experiment to do to understand how equilateral triangles can go together to make regular polyhedra. You will need construction paper and tape.
Cut equally sized equilateral triangles. You will need $32$ of them. To avoid writing "equilateral" every time, when I say triangle I mean equilateral triangle.
We are going to construct polyhedra by taping our triangles together with a fixed number of them meeting at a vertex.
If we insist that only one triangle can be at a vertex then the only shape we can construct is the triangle itself. Try $2$ triangles at a vertex and it is just as uninteresting.
Tape $3$ triangles together so they meet at a vertex. The three $60^{circ}$ angles that meet at the vertex will add up to less than $360^{circ}$ so when you tape the next triangle you will get a convex shape. In fact, you will only be able to tape one more triangle to get a tetrahedron.
Now do the same thing with $4$ triangles at a vertex. You will get an octahedron.
Five triangles at a vertex will give you an icosahedron.
What happens when we try $6$ triangles at a vertex? The $6$ interior angles add up to $360^{circ}$ so adding more triangles tiles the plane; it does not give you a convex shape.
Therefore, the tetrahedron, the octahedron and the icosahedron are the only regular polyhedra that can be formed with equilateral triangles and none of these approximate the sphere that well.
As a bonus, repeat the same experiment with squares and pentagons to get the cube and the dodecahedron.
Since regular hexagons already tile the plane, no Platonic solid can be formed with a regular $n$-gon where $nge 5$.
answered Jan 3 at 18:15
John DoumaJohn Douma
5,49711319
answered Jan 3 at 18:15
John DoumaJohn Douma
5,49711319
answered Jan 3 at 18:15
John DoumaJohn Douma
5,49711319
answered Jan 3 at 18:15
John DoumaJohn Douma
5,49711319
5,49711319
$begingroup$
Thank you so much John I will look in to this.
$endgroup$
– RICH
Jan 4 at 5:12
add a comment |
$begingroup$
Thank you so much John I will look in to this.
$endgroup$
– RICH
Jan 4 at 5:12
$begingroup$
Thank you so much John I will look in to this.
$endgroup$
– RICH
Jan 4 at 5:12
$begingroup$
Thank you so much John I will look in to this.
$endgroup$
– RICH
Jan 4 at 5:12
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%2f3060227%2frate-of-change-when-moving-on-a-sphere-is-the-same-in-all-directions-so-why-dif%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
1
$begingroup$
Look up Platonic solids. The only polyhedra that can be constructed with equilateral triangles are the tetrahedron, the octahedron and the icosahedron.
$endgroup$
– John Douma
Jan 3 at 4:11
$begingroup$
The only regular polyhedra, that is.
$endgroup$
– amd
Jan 3 at 5:52
$begingroup$
Thank you John! I am familiar with polyhedras. If I place a equilateral triangal on a sphere, the points touching the surface, then hinge the next triangle so three point touch then the next will not work? Four points define a sphere but 3 define a plane? Or three planes define a sphere? I guess I just have to accept a residual shape.
$endgroup$
– RICH
Jan 3 at 14:49
$begingroup$
@RICH Try the experiment I suggest below.
$endgroup$
– John Douma
Jan 3 at 18:15