NationStates Jolt Archive

Yeah, expected. I'm no math whiz, so don't confuse me with fancy jargon.

First off, what the hell is Concentric? I only found internet providers when I looked that up.

A concentric thing, like, say, a circle, is a circle that shares the same center as other circles, sort of like a "saturn's rings" pattern.

Yeah, expected. I'm no math whiz, so don't confuse me with fancy jargon.

First off, what the hell is Concentric? I only found internet providers when I looked that up.

http://www.merriam-webster.com/cgi-bin/dictionary?sourceid=Mozilla-search&va=concentric

Concentric means two (or more) shapes that have the same centre...unless it's an advanced maths term that I've not heard of.

Yeah, expected. I'm no math whiz, so don't confuse me with fancy jargon.

First off, what the hell is Concentric? I only found internet providers when I looked that up.

Approximately four seconds in Wikipedia gives:

Concentric objects share the same center or origin. Circles, disks, and spheres may be concentric; such concentric objects need not have the same radius.

Approximately four seconds in Wikipedia gives:

I have bad internet. I mean BAD.

you could always try google with
define:concentric
It'll give a host of definitions

I have bad internet. I mean BAD.
Yet you can post on a Jolt forum......hmmmmm

you could always try google with
define:concentric
It'll give a host of definitions

Ah. Always helpful. Learn something new everyday.

Yet you can post on a Jolt forum......hmmmmm

I'm on Sympatico right now, but some crazy shit is installed on my computer. Makes pictures out of sync, can't see anything in pictures whatsoever. So I go to something predominantly text.

I have bad internet. I mean BAD.

Fair enough. I assumed it would be something like that, rather than general assholishness, so I hedged my bets with a quote and a barbed comment

Fair enough. I assumed it would be something like that, rather than general assholishness, so I hedged my bets with a quote and a barbed comment

Yes, sarcasm is a very powerful weapon. This isn't the first time I made a 'I need help on math' thread. Some others can vouch for it.

I don't want people to give me the answers, since I want to figure this out myself, but I just need some hints, so something. I'm doing Conics for mathematics, and I'm stuck on this question:

The door of a house is in the shape of a semiellipse. The span is 100cm wide, and the maximum height of the door is 250 cm. A sofa has to fit through the door. The sofa's dimensions are 120 cm by 80 cm 250 cm. Will the sofa fit through the door? Justify your answer.

Okay, assuming somebody would have a door like that, I'm sure this door is half of an ellipse, but the points on it would never have to be under 120 by 80. I'm assuming the base of the sofa is 80cm, and that 250cm is irrelavent to the question (since it asks for whether the door could fit, not how much room is in the house). That 120cm by the 80cm would certainly form a square, but at the height of 120cm at the door, is the width between two points 80cm?

I'm not sure on how to figure out the width between two points on a certain part of the y-axis.

I don't want people to give me the answers, since I want to figure this out myself, but I just need some hints, so something. I'm doing Conics for mathematics, and I'm stuck on this question:

The door of a house is in the shape of a semiellipse. The span is 100cm wide, and the maximum height of the door is 250 cm. A sofa has to fit through the door. The sofa's dimensions are 120 cm by 80 cm 250 cm. Will the sofa fit through the door? Justify your answer.

Okay, assuming somebody would have a door like that, I'm sure this door is half of an ellipse, but the points on it would never have to be under 120 by 80. I'm assuming the base of the sofa is 80cm, and that 250cm is irrelavent to the question (since it asks for whether the door could fit, not how much room is in the house). That 120cm by the 80cm would certainly form a square, but at the height of 120cm at the door, is the width between two points 80cm?

I'm not sure on how to figure out the width between two points on a certain part of the y-axis.
First does span mean diameter or radius? I'll assume radius as it makes the question more difficult (if span=diameter the couch is wider that the door and won't fit). The easiest way to push the couch throught the door would to have its center directly verticle of the arch's center. The couch can be oriented one way: so it is 80 tall and 120 wide (120 tall would make it taller than the door, 250 would be wider any way it is oreinted). The top left and right corners are the two points of the couch most likely to touch the arch and not make it through, and if the couch is center either both will touch or neither (so you only need to find out if one touches). You need to find out how far the corners are from the center of the arch's base (all points of the arch are the same distance away from this point). You can use the pythagorian thyrum(a^2+b^2=c^2) to find the distance to the corners of the couch. A and B would be the height of the couch and half it's base, and c would be the distance to the corner. If c is greater than the radius (100cm) then the couch will not fit, it is less than the radius it will fit.

First does span mean diameter or radius? I'll assume radius as it makes the question more difficult (if span=diameter the couch is wider that the door and won't fit). The easiest way to push the couch throught the door would to have its center directly verticle of the arch's center. The couch can be oriented one way: so it is 80 tall and 120 wide (120 tall would make it taller than the door, 250 would be wider any way it is oreinted). The top left and right corners are the two points of the couch most likely to touch the arch and not make it through, and if the couch is center either both will touch or neither (so you only need to find out if one touches). You need to find out how far the corners are from the center of the arch's base (all points of the arch are the same distance away from this point). You can use the pythagorian thyrum(a^2+b^2=c^2) to find the distance to the corners of the couch. A and B would be the height of the couch and half it's base, and c would be the distance to the corner. If c is greater than the radius (100cm) then the couch will not fit, it is less than the radius it will fit.

...I don't get it.

Assuming you're pushing the sofa straight in...

Ignore the length of the sofa (250cm)

The door, being in the shape of a semielipse, is going to be defined by the formula for an inverse parabola:

y=-ax^2, setting the origin at the top of the upside-down parabola.

At y=-250, x=50=half the span of the door

so, -250=-a50^2

solve for a, a=1/10, leaving y=-0.01x^2

Now, find the width of the door at the top of the sofa.

Y=-250+120=-130

-130=-.01x^2

solve for x

1300=x^2
x=sqrt1300 which is about 36

36x2=76<80

so it won't fit if you put it through on its side if we assume it's a square.

However, sofas aren't squares. We can easily gain those 4 extra centimeters by pushing the sofa a little in the direction that the part you sit on is pointing, or by tilting the sofa a bit.

Assuming you're pushing the sofa straight in...

Ignore the length of the sofa (250cm)

The door, being in the shape of a semielipse, is going to be defined by the formula for an inverse parabola:

y=-ax^2, setting the origin at the top of the upside-down parabola.

At y=-250, x=50=half the span of the door

so, -250=-a50^2

solve for a, a=1/10, leaving y=-0.01x^2

Now, find the width of the door at the top of the sofa.

Y=-250+120=-130

-130=-.01x^2

solve for x

1300=x^2
x=sqrt1300 which is about 36

36x2=76<80

so it won't fit if you put it through on its side if we assume it's a square.

However, sofas aren't squares. We can easily gain those 4 extra centimeters by pushing the sofa a little in the direction that the part you sit on is pointing, or by tilting the sofa a bit.

The hell.

Assuming you're pushing the sofa straight in...

Ignore the length of the sofa (250cm)

The door, being in the shape of a semielipse, is going to be defined by the formula for an inverse parabola:

y=-ax^2, setting the origin at the top of the upside-down parabola.

At y=-250, x=50=half the span of the door

so, -250=-a50^2

solve for a, a=1/10, leaving y=-0.01x^2

Now, find the width of the door at the top of the sofa.

Y=-250+120=-130

-130=-.01x^2

solve for x

1300=x^2
x=sqrt1300 which is about 36

36x2=76<80

so it won't fit if you put it through on its side if we assume it's a square.

However, sofas aren't squares. We can easily gain those 4 extra centimeters by pushing the sofa a little in the direction that the part you sit on is pointing, or by tilting the sofa a bit.
Parabolas can't be used to represent a semicircle.
My work would have been:
If C<r, the couch will fit.
c=sqrt(h^2+(b/2)^2)
c=sqrt(80^2+(120/2)^2)
c=100
c=r
It would have just barely fit.
Here (http://i14.photobucket.com/albums/a334/Yatt1/math.jpg) is what it would look like. The diagram is small, but to scale.