Dtjytyk

Accountant/ lawyer will not return my call

Is it necessary to separate DC power cables and data cables?

In the late 1940’s to early 1950’s what technology was available that could melt a LOT of ice?

Why would one plane in this picture not have gear down yet?

How to write ı (i without dot) character in pgf-pie

Should I take out a loan for a friend to invest on my behalf?

PTIJ: Should I kill my computer after installing software?

What's the "normal" opposite of flautando?

Can I pump my MTB tire to max (55 psi / 380 kPa) without the tube inside bursting?

Why was Goose renamed from Chewie for the Captain Marvel film?

Single word request: Harming the benefactor

Vocabulary for giving just numbers, not a full answer

List elements digit difference sort

How to detect if C code (which needs 'extern C') is compiled in C++

Conservation of Mass and Energy

How to secure an aircraft at a transient parking space?

How to draw cubes in a 3 dimensional plane

Grey hair or white hair

Are tamper resistant receptacles really safer?

Motivation for Zeta Function of an Algebraic Variety

Are there historical instances of the capital of a colonising country being temporarily or permanently shifted to one of its colonies?

How can I ensure my trip to the UK will not have to be cancelled because of Brexit?

What are the threaded holes in Manfrotto camera brackets?

Why does the negative sign arise in this thermodynamic relation?

Area of shape with a fixed offset around the perimeter

Find area bounded by distance from a curveMaximum area of rectangle with fixed perimeter.the maxium area of a fixed perimeter closed graph.Maximal area in fixed perimeterExpected area of a random triangle with fixed perimeterMaximum area / perimeter ratioIs it possible to find the area of a shape from its perimeter?The area to perimeter ratio and generalizationsArea covered by fixed perimeter around polygon.maximizing area when perimeter is fixed to determine what shape resultsWhat is the perimeter and area for this simple shape?

0

$begingroup$

I would like to find the calculated area of a shape, I will have the actual area and perimeter. There is a fixed offset around the shape, for example 3'. What would be the calculated area. The only known variables are actual area and perimeter.

geometry

share|cite|improve this question

asked Nov 7 '16 at 21:08

adr0327adr0327

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is your shape a polygon? Or is it bounded by a smooth curve?
  $endgroup$
  – Aretino
  Nov 7 '16 at 21:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is a polygon. Sometimes can be a smooth curve
  $endgroup$
  – adr0327
  Nov 7 '16 at 21:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To really solve this, there must be more information about the shape and the offset. For example, one way to put an "offset" of $3$ units around a square makes a larger square with sides that are $6$ units longer than the original square. But I think you're probably supposed to include only points that are within $3$ units of the original square, in which case you get a "square" with rounded corners. I also think you are probably supposed to require the shape's interior to be convex, or else you are only asked for an upper bound of the area including the offset.
  $endgroup$
  – David K
  Nov 7 '16 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this question might be a variation of math.stackexchange.com/questions/1751332 where the "curve" in that question is the boundary of the shape in your question, and you are interested in just the area on one side of the curve (the outside), not both sides (since you already know the total area inside the curve).
  $endgroup$
  – David K
  Nov 7 '16 at 21:27
  

  
  
  
  

add a comment | 

0

$begingroup$

I would like to find the calculated area of a shape, I will have the actual area and perimeter. There is a fixed offset around the shape, for example 3'. What would be the calculated area. The only known variables are actual area and perimeter.

geometry

share|cite|improve this question

asked Nov 7 '16 at 21:08

adr0327adr0327

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is your shape a polygon? Or is it bounded by a smooth curve?
  $endgroup$
  – Aretino
  Nov 7 '16 at 21:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is a polygon. Sometimes can be a smooth curve
  $endgroup$
  – adr0327
  Nov 7 '16 at 21:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To really solve this, there must be more information about the shape and the offset. For example, one way to put an "offset" of $3$ units around a square makes a larger square with sides that are $6$ units longer than the original square. But I think you're probably supposed to include only points that are within $3$ units of the original square, in which case you get a "square" with rounded corners. I also think you are probably supposed to require the shape's interior to be convex, or else you are only asked for an upper bound of the area including the offset.
  $endgroup$
  – David K
  Nov 7 '16 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this question might be a variation of math.stackexchange.com/questions/1751332 where the "curve" in that question is the boundary of the shape in your question, and you are interested in just the area on one side of the curve (the outside), not both sides (since you already know the total area inside the curve).
  $endgroup$
  – David K
  Nov 7 '16 at 21:27
  

  
  
  
  

add a comment | 

$begingroup$

I would like to find the calculated area of a shape, I will have the actual area and perimeter. There is a fixed offset around the shape, for example 3'. What would be the calculated area. The only known variables are actual area and perimeter.

geometry

share|cite|improve this question

asked Nov 7 '16 at 21:08

adr0327adr0327

1

$endgroup$

share|cite|improve this question

asked Nov 7 '16 at 21:08

adr0327adr0327

1

share|cite|improve this question

asked Nov 7 '16 at 21:08

adr0327adr0327

1

asked Nov 7 '16 at 21:08

adr0327adr0327

1

asked Nov 7 '16 at 21:08

adr0327adr0327

1

2 Answers
2

active

oldest

votes

0

$begingroup$

To give you another term to search for: you appear to be looking for the Minkowski sum of your given polygon and a disk of radius $3$ or whatever.

I doubt this can be answered if all you have is area, perimeter and offset. Take for example a square of edge length $10$ (so area $10^2=100$ and perimeter $4times 10=40$). An offset of $3$ around this will lead to an area of

$$10^2+4times10times3+pitimes3^2approx248$$

Now take a circle of the same area, i.e. of radius $r=sqrt{frac{10^2}pi}approx5.64$. It has perimeter $2pi rapprox35.45$, so that is smaller than the radius of the sphere. Now make a very narrow but deep cut into that circle. That can easily add to the parameter, but if it's narrow enough won't affect the area much. What's more important is that offsetting this resulting shape will look pretty much identical to offsetting the uncut circle, since the boundaries of the cut are so close together that the whole cut vanishes with almost no extra area. So the area of the offset shape would be essentially a circle of radius $r+3$, leading to an area of approximately $235$.

So you'd have two different shapes, with equal area and equal perimeter, but different area for the offset shape. These are not polygons, but they could be approximated by polygons arbitrarily well.

It should be possible to make this work if you only consider convex polygons. There you can consider the outline of the offset shape to be composed of straight runs which form rectangles with the original perimeter, and curved corners which must sum up to one whole disk. So for original perimeter $p$ and area $A$, adding an offset of $w$ will lead to this new area and perimeter:

$$A_w = A + pw + pi w^2qquad p_w = p + 2pi w$$

And since you can approximate other convex shapes by polygons, the above should hold for other convex shapes as well. There should be a proof to that effect somewhere out there, but I don't have a reference at hand.

share|cite|improve this answer

answered Nov 9 '16 at 10:00

MvGMvG

31k450105

$endgroup$

add a comment | 

0

$begingroup$

Area = A1; New Area = A2; perimeter = p; offset= w ;

Approximately, i.e., if $w^2<< A1,$ and the offset is outside the convex figure

$$ A2= A1+ pw $$

is if inside the convex figure $$ A2= A1- p w $$ where p*w is the growth or reduction in area.

The co-ordinates or profile of A1 should be known for a more accurate calculation.

share|cite|improve this answer

edited Dec 27 '17 at 5:15

answered Dec 27 '17 at 4:44

NarasimhamNarasimham

20.9k62158

$endgroup$

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2004162%2farea-of-shape-with-a-fixed-offset-around-the-perimeter%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

To give you another term to search for: you appear to be looking for the Minkowski sum of your given polygon and a disk of radius $3$ or whatever.

I doubt this can be answered if all you have is area, perimeter and offset. Take for example a square of edge length $10$ (so area $10^2=100$ and perimeter $4times 10=40$). An offset of $3$ around this will lead to an area of

$$10^2+4times10times3+pitimes3^2approx248$$

Now take a circle of the same area, i.e. of radius $r=sqrt{frac{10^2}pi}approx5.64$. It has perimeter $2pi rapprox35.45$, so that is smaller than the radius of the sphere. Now make a very narrow but deep cut into that circle. That can easily add to the parameter, but if it's narrow enough won't affect the area much. What's more important is that offsetting this resulting shape will look pretty much identical to offsetting the uncut circle, since the boundaries of the cut are so close together that the whole cut vanishes with almost no extra area. So the area of the offset shape would be essentially a circle of radius $r+3$, leading to an area of approximately $235$.

So you'd have two different shapes, with equal area and equal perimeter, but different area for the offset shape. These are not polygons, but they could be approximated by polygons arbitrarily well.

It should be possible to make this work if you only consider convex polygons. There you can consider the outline of the offset shape to be composed of straight runs which form rectangles with the original perimeter, and curved corners which must sum up to one whole disk. So for original perimeter $p$ and area $A$, adding an offset of $w$ will lead to this new area and perimeter:

$$A_w = A + pw + pi w^2qquad p_w = p + 2pi w$$

And since you can approximate other convex shapes by polygons, the above should hold for other convex shapes as well. There should be a proof to that effect somewhere out there, but I don't have a reference at hand.

share|cite|improve this answer

answered Nov 9 '16 at 10:00

MvGMvG

31k450105

$endgroup$

add a comment | 

0

$begingroup$

To give you another term to search for: you appear to be looking for the Minkowski sum of your given polygon and a disk of radius $3$ or whatever.

I doubt this can be answered if all you have is area, perimeter and offset. Take for example a square of edge length $10$ (so area $10^2=100$ and perimeter $4times 10=40$). An offset of $3$ around this will lead to an area of

$$10^2+4times10times3+pitimes3^2approx248$$

Now take a circle of the same area, i.e. of radius $r=sqrt{frac{10^2}pi}approx5.64$. It has perimeter $2pi rapprox35.45$, so that is smaller than the radius of the sphere. Now make a very narrow but deep cut into that circle. That can easily add to the parameter, but if it's narrow enough won't affect the area much. What's more important is that offsetting this resulting shape will look pretty much identical to offsetting the uncut circle, since the boundaries of the cut are so close together that the whole cut vanishes with almost no extra area. So the area of the offset shape would be essentially a circle of radius $r+3$, leading to an area of approximately $235$.

So you'd have two different shapes, with equal area and equal perimeter, but different area for the offset shape. These are not polygons, but they could be approximated by polygons arbitrarily well.

It should be possible to make this work if you only consider convex polygons. There you can consider the outline of the offset shape to be composed of straight runs which form rectangles with the original perimeter, and curved corners which must sum up to one whole disk. So for original perimeter $p$ and area $A$, adding an offset of $w$ will lead to this new area and perimeter:

$$A_w = A + pw + pi w^2qquad p_w = p + 2pi w$$

And since you can approximate other convex shapes by polygons, the above should hold for other convex shapes as well. There should be a proof to that effect somewhere out there, but I don't have a reference at hand.

share|cite|improve this answer

answered Nov 9 '16 at 10:00

MvGMvG

31k450105

$endgroup$

add a comment | 

$begingroup$

To give you another term to search for: you appear to be looking for the Minkowski sum of your given polygon and a disk of radius $3$ or whatever.

I doubt this can be answered if all you have is area, perimeter and offset. Take for example a square of edge length $10$ (so area $10^2=100$ and perimeter $4times 10=40$). An offset of $3$ around this will lead to an area of

$$10^2+4times10times3+pitimes3^2approx248$$

Now take a circle of the same area, i.e. of radius $r=sqrt{frac{10^2}pi}approx5.64$. It has perimeter $2pi rapprox35.45$, so that is smaller than the radius of the sphere. Now make a very narrow but deep cut into that circle. That can easily add to the parameter, but if it's narrow enough won't affect the area much. What's more important is that offsetting this resulting shape will look pretty much identical to offsetting the uncut circle, since the boundaries of the cut are so close together that the whole cut vanishes with almost no extra area. So the area of the offset shape would be essentially a circle of radius $r+3$, leading to an area of approximately $235$.

So you'd have two different shapes, with equal area and equal perimeter, but different area for the offset shape. These are not polygons, but they could be approximated by polygons arbitrarily well.

It should be possible to make this work if you only consider convex polygons. There you can consider the outline of the offset shape to be composed of straight runs which form rectangles with the original perimeter, and curved corners which must sum up to one whole disk. So for original perimeter $p$ and area $A$, adding an offset of $w$ will lead to this new area and perimeter:

$$A_w = A + pw + pi w^2qquad p_w = p + 2pi w$$

And since you can approximate other convex shapes by polygons, the above should hold for other convex shapes as well. There should be a proof to that effect somewhere out there, but I don't have a reference at hand.

share|cite|improve this answer

answered Nov 9 '16 at 10:00

MvGMvG

31k450105

$endgroup$

share|cite|improve this answer

answered Nov 9 '16 at 10:00

MvGMvG

31k450105

answered Nov 9 '16 at 10:00

MvGMvG

31k450105

answered Nov 9 '16 at 10:00

MvGMvG

31k450105

0

$begingroup$

Area = A1; New Area = A2; perimeter = p; offset= w ;

Approximately, i.e., if $w^2<< A1,$ and the offset is outside the convex figure

$$ A2= A1+ pw $$

is if inside the convex figure $$ A2= A1- p w $$ where p*w is the growth or reduction in area.

The co-ordinates or profile of A1 should be known for a more accurate calculation.

share|cite|improve this answer

edited Dec 27 '17 at 5:15

answered Dec 27 '17 at 4:44

NarasimhamNarasimham

20.9k62158

$endgroup$

add a comment | 

0

$begingroup$

Area = A1; New Area = A2; perimeter = p; offset= w ;

Approximately, i.e., if $w^2<< A1,$ and the offset is outside the convex figure

$$ A2= A1+ pw $$

is if inside the convex figure $$ A2= A1- p w $$ where p*w is the growth or reduction in area.

The co-ordinates or profile of A1 should be known for a more accurate calculation.

share|cite|improve this answer

edited Dec 27 '17 at 5:15

answered Dec 27 '17 at 4:44

NarasimhamNarasimham

20.9k62158

$endgroup$

add a comment | 

$begingroup$

Area = A1; New Area = A2; perimeter = p; offset= w ;

Approximately, i.e., if $w^2<< A1,$ and the offset is outside the convex figure

$$ A2= A1+ pw $$

is if inside the convex figure $$ A2= A1- p w $$ where p*w is the growth or reduction in area.

The co-ordinates or profile of A1 should be known for a more accurate calculation.

share|cite|improve this answer

edited Dec 27 '17 at 5:15

answered Dec 27 '17 at 4:44

NarasimhamNarasimham

20.9k62158

$endgroup$

share|cite|improve this answer

edited Dec 27 '17 at 5:15

answered Dec 27 '17 at 4:44

NarasimhamNarasimham

20.9k62158

answered Dec 27 '17 at 4:44

NarasimhamNarasimham

20.9k62158

answered Dec 27 '17 at 4:44

NarasimhamNarasimham

20.9k62158