Silly question?

[danielj]

(can you say why?)

From the free pdf, a Polygon is defined in Greek as ‘many-angle’, and with one straight line there are no angles or vertices, and with two straight lines there is one internal and one external. Also, one or two straight lines cannot define an area. Also, to try it out, I gave n a value of 1 and then 2 and in both cases it gives an answer of an internal angle of 0, which is nonsense I think?

I can understand your teaching, it just takes me awhile and several reads through to digest it, thank you. 
 

I understand reasonably well x,y and z axes, positive and negative as I have to occasionally use a laser cnc machine at work. I can visualise it.

[studiot]

Just now, danielj said:

I can understand your teaching, it just takes me awhile and several reads through to digest it, thank you.

I'm glad we are making progress.

Just now, danielj said:

From the free pdf, a Polygon is defined in Greek as ‘many-angle’, and with one straight line there are no angles or vertices, and with two straight lines there is one internal and one external. Also, one or two straight lines cannot define an area. Also, to try it out, I gave n a value of 1 and then 2 and in both cases it gives an answer of an internal angle of 0, which is nonsense I think?

A plane figure is the perimeter only of a shape that lies all in one plane (ie flat not like the surface of the Earth).

Such a shape may have any combination of curved or straight or straight sides, but it must be 'closed'. That is it must come back to its starting point and enclose some area, although the area does not form part of the figure.

The only closed shape with a single side is a circle  (note a circle is the perimeter only of a disc so it is the circle that has a circumference but no area and the disc that has the area).
 

A lens or lenticular has a two sided perimeter

Of course both lenses and circles have curved sides.

A polygon is a many sided figure, with all the sides straight.  (poly means many)

So a rectangle is a four sided figure.

If all the sides are the same length the figure is called a regular polygon and these are the sorts of figures we are considering.

There is more but that's the bare bones of it.

 

[danielj]

13 hours ago, studiot said:

I know its harder to study just out of a book or online because you can't ask questions of a book or pdf, when you come across something you are not sure of.

So you are doing exactly the right thing because SF is exactly the right place to ask.

+1

 

OK so there are no polygons with 1 side or 2 sides (can you say why ?)

So in the expression (1-2/n) n must be 3 or more, that is greater than 2.

so 2/n is always less than 1 and so the whole expression is always positive.

Discussing this helps us since we now know you can understand the notation (brackets and algebraic expression) so keep it up.

This was an example of mathematical reasoning. Butit does not answer you question.

The answer comes from the proof of the formula.

Here is a chatty version. We can do a formal proof if you like but you need understanding rather than formal instruction.

Incidentally do you understand X - Y graphs  ie plots with x and y axes on graph paper ?

 

Is the zero degrees with 1 and 2 correct? And if so, what does that mean?

does it prove that there are no regular polygons with 1 or 2 sides?

Edited February 12 by danielj

[Genady]

1 hour ago, danielj said:

Is the zero degrees with 1 and 2 correct?

For n=2 we get the angle 180(1-2/2) = 180(1-1) = 180(0) = 0 degrees.

For n=1 we get the angle 180(1-2/1) = 180(1-2) = 180(-1) = -180 degrees.

For n=2, I imagine a degenerate, flattened polygon with two sides and two vertices where the sides are on top of each other (Digon - Wikipedia.) The angle between sides in this case is indeed zero degrees.

For n=1, there is no angle to measure.

Edited February 12 by Genady

[danielj]

1 hour ago, Genady said:

For n=2 we get the angle 180(1-2/2) = 180(1-1) = 180(0) = 0 degrees.

For n=1 we get the angle 180(1-2/1) = 180(1-2) = 180(-1) = -180 degrees.

For n=2, I imagine a degenerate, flattened polygon with two sides and two vertices where the sides are on top of each other (Digon - Wikipedia.) The angle between sides in this case is indeed zero degrees.

For n=1, there is no angle to measure.

Ah, yes, got it now. And when I used 1 for n, I think I accidentally swapped the n to the wrong side. That’s cleared that mess up!, thank you

Edited February 12 by danielj

[danielj]

5 hours ago, danielj said:

Ah, yes, got it now. And when I used 1 for n, I think I accidentally swapped the n to the wrong side. That’s cleared that mess up!, thank you

No, I think I still got that wrong. Got a bit confused there somewhere along the line, swapping the n still wouldn’t give that answer. I’ll start writing it down, tough for me to do it in my head.

[Genady]

1 minute ago, danielj said:

No, I think I still got that wrong. Got a bit confused there somewhere along the line, swapping the n still wouldn’t give that answer. I’ll start writing it down, tough for me to do it in my head.

Some of the best advice my best math teacher gave me was to always write down your work as you go, to use one side of paper so you don't need to turn the page to see what you've written, to use pen and not pencil and never erase even mistakes.

[danielj]

Could anyone tell me what the proper name and meaning of the mathematical symbol that is a simple arrow pointing to the right please? 
for instance

a²=c²-b² arrow here a=square root c²-b²

 

 

 

[Genady]

58 minutes ago, danielj said:

Could anyone tell me what the proper name and meaning of the mathematical symbol that is a simple arrow pointing to the right please? 
for instance

a²=c²-b² arrow here a=square root c²-b²

 

 

 

Implies. If ... then ... .

[danielj]

Thank you

[danielj]

Could you clarify for me please: 

the statement ‘(the faster you go, the more slowly you age)’, quoted from World Treasury, T.Ferris. 
If it were me travelling faster ( and faster) would time  be passing more slowly for me? (Would I feel it) or only from the viewpoint of some one else not in the same location taking measurements? Or, if I were taking measurements of them, would I record time speeding up for them. Do you see what I’m getting at?
also, what would the difference be between me in an accelerating rocket going away from a planet, and measuring from the rocket, and standing on the planet and measuring the rocket receding into the distance? Are they equivalent?

Simple as you can please

thank you

 

 

Or in other words, does the fact that the rocket is accelerating have a bearing on the measurement (the dilation effect)?

I’m rambling a bit , I know but just trying to think it through. One question leads to another! 
does the orbiting planet also use energy (as the rocket does to accelerate) to stay in orbit, rather than just fly off in a straight line into space?

[Genady]

36 minutes ago, danielj said:

Simple as you can

I don't know if my answers are simple, but they are certainly short and leave a lot of room for more questions. Here it goes.

 

36 minutes ago, danielj said:

If it were me travelling faster ( and faster) would time  be passing more slowly for me? (Would I feel it)

You would not feel it.

 

36 minutes ago, danielj said:

only from the viewpoint of some one else not in the same location taking measurements?

As measured by an observer relative to which you are moving.

 

36 minutes ago, danielj said:

if I were taking measurements of them, would I record time speeding up for them

You would record time slowing down for them.

 

36 minutes ago, danielj said:

Do you see what I’m getting at?

I don't.

 

36 minutes ago, danielj said:

does the fact that the rocket is accelerating have a bearing on the measurement

It does.

36 minutes ago, danielj said:

does the orbiting planet also use energy (as the rocket does to accelerate) to stay in orbit

It does not.

Edited Saturday at 03:15 PM by Genady

[KJW]

2 minutes ago, Genady said:

35 minutes ago, danielj said:

if I were taking measurements of them, would I record time speeding up for them.

You would.

No. You would record time slowing down for them, just as they would record time slowing down for you.

 

[Genady]

Just now, KJW said:

No. You would record time slowing down for them, just as they would record time slowing down for you.

 

Sorry, I have already fixed it a second ago. But thanks for the catch anyway.

[KJW]

55 minutes ago, danielj said:

does the fact that the rocket is accelerating have a bearing on the measurement (the dilation effect)?

The relativistic effects of being in an accelerated frame of reference can be derived from special relativity. That is, acceleration doesn't produce relativistic effects separate from that of velocity. However, the relativistic effects of an accelerated frame of reference are nevertheless different from that of relative velocity. Specifically, clocks that are below you are slower, and clocks that are above you are faster, with the amount by which the clocks are slower or faster depending on the distance of the clock from your location in your accelerated frame of reference.

 

[danielj]

26 minutes ago, Genady said:

58 minutes ago, danielj said:

does the fact that the rocket is accelerating have a bearing on the measurement

Expand  

It does.

Sorry, I didn’t mean it the way it came across. Obviously there would be an effect of time slowing at an accelerated rate from the (let’s say) constant acceleration. It was more the ‘are they equivalent’. I guess they are from your other answers and from the answer that time would slow for both observers from the others point of view.

thank you

[Genady]

1 minute ago, danielj said:

are they equivalent

The measurement by an accelerating observer and by a non-accelerating observer will be different.

[danielj]

14 minutes ago, KJW said:

1 hour ago, danielj said:

does the fact that the rocket is accelerating have a bearing on the measurement (the dilation effect)?

Expand  

The relativistic effects of being in an accelerated frame of reference can be derived from special relativity. That is, acceleration doesn't produce relativistic effects separate from that of velocity. However, the relativistic effects of an accelerated frame of reference are nevertheless different from that of relative velocity. Specifically, clocks that are below you are slower, and clocks that are above you are faster, with the amount by which the clocks are slower or faster depending on the distance of the clock from your location in your accelerated frame of reference.

Thank you , that is a very useful explanation that I can (just) grasp. 

[KJW]

34 minutes ago, danielj said:

Thank you , that is a very useful explanation that I can (just) grasp. 

Note that when you are standing on the ground, you are being accelerated upward. Also, if you are in an accelerated rocket, you are being accelerated upward. And it is always upward due to perception of the vertical direction in response to acceleration by the vestibular system of the inner ear. If you are on the fifth floor of a ten-storey building, a clock on the ground floor would tick slower relative to your clock, and a clock on the tenth floor would tick faster relative to your clock. It is important to note that the clocks themselves are not actually affected, and all are ticking at the same intrinsic rate of "one second per second".
 

[danielj]

3 hours ago, KJW said:

Note that when you are standing on the ground, you are being accelerated upward.

Would you mind just expanding that a little for me please? I can see that human perception skews reality somewhat, but I can’t work out that bit for myself. 
thank you

[swansont]

1 minute ago, danielj said:

Would you mind just expanding that a little for me please? I can see that human perception skews reality somewhat, but I can’t work out that bit for myself. 
thank you

You can only feel a force when standing because the earth is pushing up on you. In General Relativity, freefall is inertial motion, and being stationary requires an acceleration.

[studiot]

On 2/11/2025 at 10:06 PM, danielj said:

 

I understand reasonably well x,y and z axes, positive and negative as I have to occasionally use a laser cnc machine at work. I can visualise it.

That's really good news. I think we can build on this mechanical experience.

I have been busy these last week but I see other members have been looking after you.

I would, however like to bring your feet fimly back to ground to feel the gravity.

I think you could benefit from this book. It is one of a kind where the author tries to use physics to explain well known mathematics, rather than the other way round.

So rather than dry mathematical general theorems about polygons try the polygon of least area (p35)

The Mathematical Mechanic  :  Mark Levi :  Princeton University Press.

 

But remember to keep asking questions as you will find much unfamiliar as well as some familiar stuff.

 

[KJW]

15 hours ago, danielj said:

Would you mind just expanding that a little for me please?

The Equivalence Principle basically says that over distances that are sufficiently small for the tidal effect to be negligible, being in a gravitational field is indistinguishable from being in an accelerated frame of reference. This is illustrated by the following diagram:
 

 

However, over larger distances, the gravitational field does differ from being in an accelerated frame of reference due to the tidal effect, which is a manifestation of spacetime curvature that is absence from being in an accelerated frame of reference in flat spacetime.

 

 

15 hours ago, danielj said:

I can see that human perception skews reality somewhat, but I can’t work out that bit for myself.

What I said about perception of the vertical direction was not about skewing reality, but a consequence of the equivalence principle. I personally discovered the perception of the vertical direction when I was a teenager in an amusement park ride called the "Rotor". I noticed that the people directly opposite me before the start of the ride were very much above me during the ride. I immediately realised that what we regard as up or down is actually a perception that we don't normally notice unless we are in an environment that challenges the notion of up or down.

 

 

[Genady]

18 hours ago, KJW said:

perception of the vertical direction in response to acceleration by the vestibular system of the inner ear

The vestibular system input is an important one but there are other factors that determine our perception of the vertical direction. One such factor is tactile. When we stand, sit, or lie on the ground, we feel how the ground pushes at us. This feel disappears when we are submerged and buoyant underwater. The perception of vertical direction is not so sharp anymore.

We still respond to a visual clue: it is lighter above and darker below. But in night diving or when diving in very murky water this factor disappears too. It is very easy then to lose the vertical orientation completely. To make sure where is up and where is down in such case, we look at where the bubbles go.

[Genady]

1 hour ago, Genady said:

we look at where the bubbles go

BTW, the bubbles may be also misleading. When diving Molokini Crater we got caught in a down current once. We held to a rock, but our bubbles were pulled down. The visibility was perfect thought, so we did not have a problem to see where the surface is.