What does a rod passing through a hole look like in the inertial frame?

[phyti]

This is the question everyone is asking, what are the initial conditions.
If it's what BT described, then the rod and the surface are moving at the approx. the same speed. Because the hole is shorter than the rod (as stated originally), the rod cannot translate sideways through the hole.

 

[Phrak]

Phrak, is this how the question goes? There's still the obvious confusion with what's happening here. Is this what is happening?:

So what we have is a rod not unlike a broomstick. This rod is horizontal and moving at near-light speeds. There is also a horizontal surface the rod is quickly moving and levitating over that is parallel to the rod. There's a hole in the horizontal surface that's roughly the same shape as the rod (thus the hole is more like a slit rather than a hole) but a bit shorter than the rod when the length measurements are taken at rest.
This rod then approaches the slit. As it does so, an external force acts on the rod in such a way that the rod may fall into this slit. This means that the external force acts on the center of the rod, causing it to fall toward the slit while remaining always parallel to the horizontal surface.

You've got it for the most part. But ignore gravity; no there's no 'falling' going on. Also, don't think of the force that's pushing it through the hole, as acting at it's center of mass. Whatever force is used is distributed in such a way that the rod, or broom stick is always parallel to the surface in the inertial frame of the hole.

It's best not to think of it as a rod, but a soft stick of taffy. The bead idea I gave earlier is actually very appropriate, but no one seems to take it up.

(Sigmoidal is the term used to describe the transfer function of an artificial neuro network neuron--soft saturating logic, idealized as the arctangent because the derivative is easy to deal with.)

 

[Fredrik]

The bead idea I gave earlier is actually very appropriate, but no one seems to take it up.

It's a good idea, but even better is to just imagine a rod that's been sliced up into thin slices that are perpendicular to the wall. That's how I think about it. The reason why this is better is that when you consider circular beads,you won't clearly see what the shape of the rod is going to be in the inertial frame that's co-moving with the rod before the push.

The shape of the rod's cross-section is a parallelogram that isn't a rectangle (assuming an instantaneous velocity change and that we're looking at the rod after every part of it has changed its velocity). Note that the slices are still perpendicular to the wall.

 

[Bible Thumper]

A gentle push is a gentle push, you guys. I understand that in reality, the increase in mass the rod takes up will mean far more than a gentle push will be needed for our rod. However, we have to presume the OP meant 'gentle push' when he said 'gentle push.'

And if he meant gentle push, I can't understand why we have to be regarding the rod as necessarily flexible. I mean, can't we make it rigid by forging it from neutron starstuff? Then it will be almost infinitely dense, thus infinitely rigid.

Please explain why it has to be flexible. Don't mention kinematics, either, because we already agreed the OP meant 'gentle push' when he said 'gentle push.'

 

[Fredrik]

I can't understand why we have to be regarding the rod as necessarily flexible. I mean, can't we make it rigid by forging it from neutron starstuff? Then it will be almost infinitely dense, thus infinitely rigid.

No. This was answered earlier in the thread. If it's rigid in one inertial frame, it isn't rigid in others. Suppose that it's rigid in a particular frame. When we give it a push, all the different parts will begin their acceleration at the same time. That means that in other frames, the different parts will begin their acceleration at different times. That makes it "not rigid" in those frames.

So it doesn't matter what it's made of. The properties of spacetime make it impossible for rigid bodies to exist.

The scenario we're discussing in this thread is keeping the rod parallel to the wall at all times in the wall's rest frame. This is only possible if the different parts are all pushed at the same time. We can't just push one part of the rod and let the force propagate through the rod, because the speed of that propagation would have to be infinite. (Otherwise the rod doesn't remain straight, and we have already said that it does).

 

[Bible Thumper]

No. This was answered earlier in the thread. If it's rigid in one inertial frame, it isn't rigid in others. Suppose that it's rigid in a particular frame. When we give it a push, all the different parts will begin their acceleration at the same time. That means that in other frames, the different parts will begin their acceleration at different times. That makes it "not rigid" in those frames.

So it doesn't matter what it's made of. The properties of spacetime make it impossible for rigid bodies to exist.

The scenario we're discussing in this thread is keeping the rod parallel to the wall at all times in the wall's rest frame. This is only possible if the different parts are all pushed at the same time. We can't just push one part of the rod and let the force propagate through the rod, because the speed of that propagation would have to be infinite. (Otherwise the rod doesn't remain straight, and we have already said that it does).

I wasn't regarding this kind of independence. I mean, I know things may appear nonrigid to one person but completely rigid to another.
Phrack mentioned kinematics, and that the application of force on the rod will turn the problem into a kinematic one. He supplements this with his bead-on-a-string thought experiment. The bead-on-a-string thinking can go both ways: with different observations or with the introduction of kinematic ideas (which would complicate things greatly and unnecessarily).

Since an external force ("gentle push") acts on our rod; and must necessarily on the center of the rod if we wish to keep the rod parallel to the plane it's moving upon, then I propose the rod will fall directly into the hole, but front-end first, relative to the person on the rod.
The man beside the slit OTOH will see the event taking place with the rod always parallel to the slit.
The rod, according to the person riding it OTOH, will seem to enter nose-first, then the remaining of the rod in a parabolic fashion, until the end of the rod falls in.
The person on the rod will be scared at first, as the slit will look radically shortened--way too short for his rod--right before it enters the slit. But as his rod appears to bend into it, he is relieved.
The person standing beside the slit sees a very short rod (much shorter in appearance than when he took his measurements with the rod at rest) entering effortlessly into the slit without incident... :)

 

[Fredrik]

Since an external force ("gentle push") acts on our rod; and must necessarily on the center of the rod if we wish to keep the rod parallel to the plane it's moving upon,...

Again, no. If a push applied to the center of the rod keeps the rod parallel to the wall in the wall's rest frame, then the rigidity is a property of the rod, not of the push. If it's a property of the rod, then it's rigid in all frames, and that contradicts special relativity.

At the very least, this problem becomes much more complicated when we consider a push applied to the center of the rod. Just think about the fact that the deformation of the rod will propagate at the speed of sound in the material the rod's made of, in the rod's rest frame. That speed, transformed to the wall's rest frame, must be high enough to reach the front of the rod before the front of the rod reaches the front of the hole. And you can't push the rod until at least half of it is over the hole. If the propagation speed in the wall's frame is greater than the rod's speed, then you have to wait even longer before you make the push.

There's a lot to think about here. If we do the math, we may find that even if we assume that the rod is infinitesimally thin and that its rest length is only infinitesimally greater than the hole's rest length, the speed of sound needs to be greater than the speed of light for this to work. I don't know if that's what we'll find. My point is that this is something we would have to check.

I'm not saying it's not an interesting problem, but you should try to understand the simple problem first.

 

[Phrak]

A gentle push is a gentle push, you guys. I understand that in reality, the increase in mass the rod takes up will mean far more than a gentle push will be needed for our rod. However, we have to presume the OP meant 'gentle push' when he said 'gentle push.'

Point taken. I was being flipant. Think of the rod as very thin and the surface as well--and then give it a good push.

 

[atyy]

A gentle push is a gentle push, you guys. I understand that in reality, the increase in mass the rod takes up will mean far more than a gentle push will be needed for our rod. However, we have to presume the OP meant 'gentle push' when he said 'gentle push.'

Point taken. I was being flipant. Think of the rod as very thin and the surface as well--and then give it a good push.

I don't understand why your bead idea isn't correct. In anyone (global) inertial frame, there is no rigid body if we push at only one point. So even in one inertial frame, we have to push at all points simultaneously in order to achieve rigidity. Since rigidity is achieved by simultaneous pushing and not the internal structure of the rod, it seems reasonable and equivalent to make the problem as simple as possible and consider the rod as independent beads.

 

[Bible Thumper]

The rod is its own intertial reference frame. For this reason the gentle push affects the entire rod at the same time, relative to anyone on the rod.
The effect of this pushing, however, is to see the rod enter the slit, front-end first, for the person on the rod.

 

[Fredrik]

The rod is its own intertial reference frame. For this reason the gentle push affects the entire rod at the same time, relative to anyone on the rod.
The effect of this pushing, however, is to see the rod enter the slit, front-end first, for the person on the rod.

Everything in this post is wrong.

* The rod's motion before the push defines an inertial frame. (If that's what you meant, you weren't wrong about that, but you stated your claim a bit carelessly).

* A force applied to a point doesn't propagate at infinite speed through the material. It propagates at the speed of sound, which can't be greater than the speed of light.

* Even if it did, so that the rod remains parallel to the wall in the rod's original rest frame, then this would contradict both what we have said before (parallel in the other frame) and what you said next.

* If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first.

 

[Bible Thumper]

Everything in this post is wrong.

* The rod's motion before the push defines an inertial frame. (If that's what you meant, you weren't wrong about that, but you stated your claim a bit carelessly).

* A force applied to a point doesn't propagate at infinite speed through the material. It propagates at the speed of sound, which can't be greater than the speed of light.

* Even if it did, so that the rod remains parallel to the wall in the rod's original rest frame, then this would contradict both what we have said before (parallel in the other frame) and what you said next.

* If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first.

What about atomospheric resistance on the rod? You forgot that. If the rod is going at nearly the speed of light, atmospheric resistance will heat the rod up to the point it will melt.

 

[Fredrik]

What about atomospheric resistance on the rod? You forgot that. If the rod is going at nearly the speed of light, atmospheric resistance will heat the rod up to the point it will melt.

Can you really not see the difference between pointing out that what you're saying is wrong and being too concerned with practical matters in a discussion about a thought experiment?

 

[Bible Thumper]

Can you really not see the difference between pointing out that what you're saying is wrong and being too concerned with practical matters in a discussion about a thought experiment?

Fredrik, this little guy: <--- he means, "sarcasm". Knowing that, realize I was being sarcastic when I brought up air resistance. Why was I being sarcastic? Because you were bringing up the rate at which force is distributed thru the rod, that's why. If you want to bring up the rate at which the force runs thru the rod, why not other matters, such as air resistance?

• A 'gentle push' can imply an accelerating force on the rod, at which consideration, the rate at which the force runs thru the rod will depend on the rate at which the 'gentle push' acceleration takes place.
• Since the rate of the external force (assuming an impact force) runs thru the rod will depend on factors, such as the material of which the rod is made from.

Thus I identify two scenarios that alone tell us they weren't meant to be taken into consideration in the initial question. In other words, factors such as the rate at which the force is transmitted thru the rod (from a 'gentle push' which means the force goes thru the rod very slowly--as slowly as the gentle push itself--to an impact force, which goes thru the rod much faster and is material-dependent) were never meant to be taken into consideration.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/polebarn.html" link helps show that the pole will remain always parallel to the surface relative to the slit, but goes into the slit nose-first relative to someone on the pole.

 

[atyy]

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/polebarn.html" link helps show that the pole will remain always parallel to the surface relative to the slit, but goes into the slit nose-first relative to someone on the pole.

There is no gentle push in the pole-barn paradox, because the pole is moving at constant velocity and does not accelerate.

 

[Bible Thumper]

There is no gentle push in the pole-barn paradox.

I thought his main concern was with contraction, as evidenced by this:

"If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first."

For this reason and only this reason I posted the barn-pole paradox link. Regard the barn-pole paradox after the 'gentle pushing' is done.

 

[atyy]

I thought his main concern was with contraction, as evidenced by this:

"If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first."

For this reason and only this reason I posted the barn-pole paradox link. Regard the barn-pole paradox after the 'gentle pushing' is done.

I see. I thought Phrak was interested in the acceleration. I'm a bit confused what the latest incarnation of this problem is.

 

[Bible Thumper]

I see. I thought Phrak was interested in the acceleration. I'm a bit confused what the latest incarnation of this problem is.

I was under the impression the only thing he cared about was what the slit looks like from the pole's perspective.

 

[Fredrik]

If you want to bring up the rate at which the force runs thru the rod, why not other matters, such as air resistance?

Because an infinite propagation speed contradicts the theory that we're trying use to solve this problem. That makes it absolutely necessary to avoid it, even in the most mathematically idealized thought experiment we can think of. Air resistance on the other hand is only a problem in an actual experiment performed by actual physicists.

The difference between the two is enormous. That's why I asked if you really don't see it, and apparently you don't. It's like the difference between being able to eat a trillion hamburgers and being able to eat oneself*. One of them is impossible in practice, one of them is impossible in principle. I wouldn't have said anything if the scenario you described involved a person who eats a trillion hamburgers.

*) Yes, I'm aware of how that can be interpreted.

the pole will remain always parallel to the surface relative to the slit, but goes into the slit nose-first relative to someone on the pole.

Yes, that's what I've been saying in every post in this thread, but it's not what you've been saying.

 

[Fredrik]

I thought his main concern was with contraction, as evidenced by this:

"If the rod stays parallel to the wall in the rod's original rest frame, then the front and back are obviously doing everything at the same time in that frame, so the front can't go through first."

Contraction? No, not at all. What I was concerned about there is a contradiction.

You said that the rod is parallel to the wall in the rod's rest frame. That's the same thing as saying that every part of it does everything simultaneously in the rod's frame. Immediately after saying that, you said that the front went in first in the rod's frame. That's a contradiction.

 

[Fredrik]

I see. I thought Phrak was interested in the acceleration. I'm a bit confused what the latest incarnation of this problem is.

There is no new incarnation of the problem. Just about everything he (Bible Thumper) said contradicts SR or something else he said.

 

[Bible Thumper]

Contraction? No, not at all. What I was concerned about there is a contradiction.

You said that the rod is parallel to the wall in the rod's rest frame. That's the same thing as saying that every part of it does everything simultaneously in the rod's frame. Immediately after saying that, you said that the front went in first in the rod's frame. That's a contradiction.

Lulz...
This was my first proposal. I posted it yesterday:

"Since an external force ("gentle push") acts on our rod; and must necessarily on the center of the rod if we wish to keep the rod parallel to the plane it's moving upon, then I propose the rod will fall directly into the hole, but front-end first, relative to the person on the rod.
The man beside the slit OTOH will see the event taking place with the rod always parallel to the slit.
The rod, according to the person riding it OTOH, will seem to enter nose-first, then the remaining of the rod in a parabolic fashion, until the end of the rod falls in.
The person on the rod will be scared at first, as the slit will look radically shortened--way too short for his rod--right before it enters the slit. But as his rod appears to bend into it, he is relieved.
The person standing beside the slit sees a very short rod (much shorter in appearance than when he took his measurements with the rod at rest) entering effortlessly into the slit without incident... :) "​

So as you can see, I made no such assertion that the rod appears parallel to the surface while the rod is moving; in fact, I made it clear that the rod sees its front-end enter into the slit first thing.
The man standing on the surface, however, sees the rod moving into the slit while remaining always parallel to thwe surface the man is standing on.
I placed, in bold, the observed reference frames.
Period.

 

[Bible Thumper]

There is no new incarnation of the problem. Just about everything he (Bible Thumper) said contradicts SR or something else he said.

Lulz (x2)...
Just because I read and thump the Bible and just because skeptics think the Bible (G-d's holy Word) is full of contradictions, doesn't mean everything I type or say has to contradict necessarily...

 

[Phrak]

I was under the impression the only thing he cared about was what the slit looks like from the pole's perspective.

It doen't matter what I care about; the problem should stand on it's own, right?

As a reminder, I should say I didn't pose the problem as well as could be, and the improved problem statement is in post #7. Maybe Frederik could clean it up, so the misleading stuff, like "a gentle push" is gone.

 

[Fredrik]

So as you can see, I made no such assertion that the rod appears parallel to the surface while the rod is moving

I didn't say that you made that assertion in every post. I said that you made it in #60, and you did. What you said there implies that the rod is parallel to the wall in the rod's original rest frame, even if you don't understand that:

The rod is its own intertial reference frame. For this reason the gentle push affects the entire rod at the same time, relative to anyone on the rod.

But OK, we don't have to debate what you said for ever. We apparently agree that this thread is about a rod that remains parallel to the wall in the wall's frame.

 

[Fredrik]

Maybe Frederik could clean it up, so the misleading stuff, like "a gentle push" is gone.

What we're interested in a scenario where we push the rod through the hole while keeping it parallel to the wall in the wall's rest frame. As long as we all agree on that, and that "the contracted length of the rod" < "the rest length of the hole" < "the rest length of the rod", the other details don't really matter.

I prefer to leave the acceleration unspecified because it's interesting to see how the shape of the rod in the rod's original rest frame depends on the details of the velocity change.

The fact that the front of the rod goes through the hole before the rear of the rod in the rod's original rest frame is however independent of the details of the velocity change.

 

[DrGreg]

It doen't matter what I care about; the problem should stand on it's own, right?

As a reminder, I should say I didn't pose the problem as well as could be, and the improved problem statement is in post #7. Maybe Frederik could clean it up, so the misleading stuff, like "a gentle push" is gone.

This is my precise interpretion of the problem. Phrak, is my interpretation correct?

There is a flat two-dimensional surface S and a one-dimensional rod R.

Initially R is straight and parallel to S and R moves inertially relative to S in the direction of its own length.

Let I be the initial inertial rest frame of the rod. The rest-length of the rod, measured in I, is L.

There is a linear slot cut in the surface S, parallel to R, wide enough for R to fit through, but its rest-length (measured in the S-frame) is less than L. However its rest-length is greater than L/\gamma, where \gamma is the relative Lorentz factor between S and I. Thus, due to length contraction of the rod relative to S, the rod is short enough, in the S-frame, to parallel-slide through the hole, but in the I-frame it is not.

The rod now moves towards the hole in such a way that it it remains straight and parallel to the hole according to the S-frame, and then passes through the hole.

Aside: The actual practicalities of how this is achieved are not relevant to the problem. You'd need to pre-arrange for a continuum of forces to be applied along the whole length of the rod, all pre-programmed to exert a specific force at a specific time to achieve the desired effect. It would be difficult to engineer in practice, but, as a thought experiment, no laws of physics need be broken to achieve this.​

The question is then how does this appear in the I-frame, in view of the fact that the rod is longer than the hole is this frame?

The answer which has already been thrashed out in this thread is that the rod does not remain straight in this frame: it curves and passes through the hole at an angle.

(Note that rod curvature --relative to I -- is inevitable during acceleration, but afterwards it will straighten and so may well be straight by the time the rod meets the hole. But it will slide through at an angle, head first, so it doesn't matter that it's too long to fit. See diagrams below.)

 

[Bible Thumper]

W00t! W00t! Finally, someone who acquiesces!
He said:

This is my precise interpretion of the problem.
There is a linear slot cut in the surface S, parallel to R, wide enough for R to fit through, but its rest-length (measured in the S-frame) is less than L.
However its rest-length is greater than L/\gamma, where \gamma is the relative Lorentz factor between S and I. Thus, due to length contraction of the rod relative to S, the rod is short enough, in the S-frame, to parallel-slide through the hole, but in the I-frame it is not.
The rod now moves towards the hole in such a way that it it remains straight and parallel to the hole according to the S-frame, and then passes through the hole.
The answer which has already been thrashed out in this thread is that the rod does not remain straight in this frame: it curves and passes through the hole at an angle.
(Note that rod curvature --relative to I -- is inevitable during acceleration, but afterwards it will straighten and so may well be straight by the time the rod meets the hole. But it will slide through at an angle, head first, so it doesn't matter that it's too long to fit.

And I said:

Originally posted by Bible Thumper:
I propose the rod will fall directly into the hole, but front-end first, relative to the person on the rod.
The man beside the slit OTOH will see the event taking place with the rod always parallel to the slit.
The rod, according to the person riding it OTOH, will seem to enter nose-first, then the remaining of the rod in a parabolic fashion, until the end of the rod falls in.
The person on the rod will be scared at first, as the slit will look radically shortened--way too short for his rod--right before it enters the slit. But as his rod appears to bend into it, he is relieved.
The person standing beside the slit sees a very short rod (much shorter in appearance than when he took his measurements with the rod at rest) entering effortlessly into the slit without incident... :)

 

[Phrak]

Greg. How do you do that? I want to draw pictures. It would make things so much clearer.

I'd draw things a little different, with the forces applied to the rod only when the rod is directly over the hole in the hole's frame.

Your last frame has an error--sort of. Any portion of the rod sticking-out, north of the hole, is parallel to the surface. Recall that when there are no forces applied, the rod is just Lorentz contracted.

If you wished, you could argue that the gentle push has boosted the rod to have a perpendicular component of velocity as large as its velocity along the surface, which is sort of what your picture respresents, but it simpler if you consider that the added velocity in the perpendicular direction is v_p << c and v_p << v_l.

I didn't forsee a force adding as much y velocity as x velocity.

I have a drawing that makes the Bell Inequality visually obvious, but I don't know how to post it. Can you give me a hint on how to make and send pics?

 

[Dale]

Your last frame has an error--sort of. Any portion of the rod sticking-out, north of the hole, is parallel to the surface.

No, that is incorrect. There is only one bend in the worldline. If you wanted two bends in the worldline then you would have to push it twice.

 

[Phrak]

No, that is incorrect. There is only one bend in the worldline. If you wanted two bends in the worldline then you would have to push it twice.

I'll be ding-donged. I am sure you're right.

 

[phyti]

The drawing is not accurate. Try drawing it as a Minkowski space-time diagram, and you will see the difference. The scenario requires 2 dimensions, one for the direction of motion and a small offset for the separation of the rod and hole.

 

[Fredrik]

Greg knows what the spacetime diagram would look like. See his post earlier in this thread. A 2+1-dimensional spacetime diagram is precisely what you need to use to figure out what the shape of the rod is going to be at different times in different frames, and I'm pretty sure that's what he did. I don't see anything wrong with the drawing.

 

[DrGreg]

The drawing is not accurate. Try drawing it as a Minkowski space-time diagram, and you will see the difference. The scenario requires 2 dimensions, one for the direction of motion and a small offset for the separation of the rod and hole.

The diagrams I drew were not spacetime diagrams, they were space-only diagrams. Each picture is a 2D spacelike slice through 3D (2+1) spacetime, i.e. a plane of simultaneity relative to the particular frame.

It's hard to draw unambiguous 3D spacetime diagrams on a 2D computer screen. I described such a diagram in words in post #48[/color] of this thread.

I still stand by the accuracy of my pictures (for my interpretation of what the problem is).

 

[DrGreg]

Greg. How do you do that? I want to draw pictures. It would make things so much clearer.

I'd draw things a little different, with the forces applied to the rod only when the rod is directly over the hole in the hole's frame.

Your last frame has an error--sort of. Any portion of the rod sticking-out, north of the hole, is parallel to the surface. Recall that when there are no forces applied, the rod is just Lorentz contracted.

If you wished, you could argue that the gentle push has boosted the rod to have a perpendicular component of velocity as large as its velocity along the surface, which is sort of what your picture respresents, but it simpler if you consider that the added velocity in the perpendicular direction is v_p << c and v_p << v_l.

I didn't forsee a force adding as much y velocity as x velocity.

I have a drawing that makes the Bell Inequality visually obvious, but I don't know how to post it. Can you give me a hint on how to make and send pics?

For the avoidance of any doubt, note the arrows are not forces, they are velocity-directions (relative to the frame).

The "push" has to be applied before the rod gets to the hole, if you wait until it's next to it, it would be too late. As I've drawn it, the force is applied only for a very short period of time, when the rod's direction changes and when it appears bent in the I-frame. For the rest of the time the rod moves inertially.

In the S-frame, the forces are applied at all points of the rod simultaneously. (In my diagram, just before the 3rd blue snapshot only.) In the I-frame, the forces begin at the front and ripple their way through to the back. The rod is bent at the points where there are forces. (In my diagram, between the 2nd and 4th pink snapshots, moving from top to bottom of R.)

The relative x- and y-components of velocity depend on how early you push and how hard and how long -- an early push can be gentle, a late push will have to be much harder.

 

[DrGreg]

How to upload pictures

Greg. How do you do that? I want to draw pictures. It would make things so much clearer.

I have a drawing that makes the Bell Inequality visually obvious, but I don't know how to post it. Can you give me a hint on how to make and send pics?

There are two halves to this.

1. Do you know how to draw diagrams on your computer?

2. Do you know how to upload a picture?

1. There are lots of computer applications that can create diagrams. E.g. Microsoft Powerpoint or Microsoft Word and lots of others. Failing that, you could even draw on a piece of paper and scan it in, if you have access to a scanner.

2A. You need to save the picture in the right format. For computer-generated line diagrams the best formats are PNG or GIF. For a full list of supported types see step 2C below.

2B. If your application cannot save in your preferred format, you need to make a screen dump as follows:
(This assumes you use Microsoft Windows. Other operating systems vary.)
- arrange your picture so it is completely visible in the front window, at the right zoom level
- hold down ALT and press the "Print Screen" (or "Prt Sc" etc) key on your keyboard. This copies a picture of the front window to the clipboard.
- open Microsoft Photo Editor (or any image editor)
- select "Paste as New Image" from the Edit menu
- use the "select tool" to select the area you want to upload (exclude all the window borders).
- Use Image > Crop > OK
- File > Save As... to save as a PNG file.

2C. When you are composing your message using the forum's "advanced edit" option, press the "Manage Attachments" button. This let's you upload a file and includes a list of supported file types and limits on how big files can be. Avoid file types that others may not be able to read.

 

[Phrak]

Thanks, DrGreg. That's quit a post. Nicely written. I can use autoCad for perspective, but my skills at it are no to great. I think MS Word will convert drawings to jpg of gifs when inserted.