Yes. But the reason I like the guillotine example is that you don't need to do any calculations. It is obvious that if the blade angle is shallower than 45°, the vertex moves faster than the blade, and for constant velocity of the blade you have constant velocity of the vertex. With scissors, everything is changing during the process, you need to consider average velocity to deal with causality issues, and it's just harder to figure out what's actually happening without resorting to calculations.bystander wrote:Yes but the inverse tangential relationship means that the shallower the angle of the blade, the faster the vertex moves, respectively.
GRED Answer: Scissor vertex speed
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Re: GRED Answer: Scissor vertex speed
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Re: GRED Answer: Scissor vertex speed
It just occurred to me that in real scissors, contact forces between the blades at the vertex is what guides the blades further on to come down on the right side of each other. This guidance obviously is not available in the FTL closing scenarios. The tiniest inaccuracies would lead to either the blades trying to pass through each other, or not forming a vertex at all because the two edges do not touch.
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Re: GRED Answer: Scissor vertex speed
Looks good to me. I said it could be modeled mathematically. http://asterisk.apod.com/vie ... 38#p125438Henning Makholm wrote:With scissors, the tangent of the vertex angle is only one factor in the speed; the other is the transverse speed of the blades which varies with the distance from the pivot.
To be precise, assuming that we can get each blade to move at a constant angular speed ω, the vertex point will move at speed bω/(sinθ·tanθ) where b is the perpendicular distance from the pivot to either blade's edge and 2θ is the angle between the blades. The tips of the scissors cannot move faster than light, so the length of the scissors is at most c/ω and θ must be at least asin(bω/c), or we have run out of blade! Noting that tan(asin(x))~x for small x, the theoretical maximal vertex speed becomes c²/bω. (Which is greater than c because bω, being the speed of the point on the blade edge right beside the pivot, is less than c).
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Re: GRED Answer: Scissor vertex speed
I wasn't disagreeing, I was just saying, even with a guillotine, a shallower angle would increase the tranverse speed of the vertex.Chris Peterson wrote:Yes. But the reason I like the guillotine example is that you don't need to do any calculations. It is obvious that if the blade angle is shallower than 45°, the vertex moves faster than the blade, and for constant velocity of the blade you have constant velocity of the vertex. With scissors, everything is changing during the process, you need to consider average velocity to deal with causality issues, and it's just harder to figure out what's actually happening without resorting to calculations.
Re: GRED Answer: Scissor vertex speed
the crossing of the vertex + mass+ acceleration+speed= no cigar.bystander wrote:I wasn't disagreeing, I was just saying, even with a guillotine, a shallower angle would increase the tranverse speed of the vertex.
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Re: GRED Answer: Scissor vertex speed
???swainy wrote:the crossing of the vertex + mass+ acceleration+speed= no cigar.
The vertex has no mass.
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Re: GRED Answer: Scissor vertex speed
Correct, but unless your crossing it with nothing, then you may as well say a million times the speed of light. it does not matter how shallow your inverted guillotine is. one seconds worth of blade, is a long blade. but that long blade is not long enough due to acceleration. And to get to the speed of light you need all the energy in the universe. And it does not matter how you put it.Chris Peterson wrote:???swainy wrote:the crossing of the vertex + mass+ acceleration+speed= no cigar.
The vertex has no mass.
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Re: GRED Answer: Scissor vertex speed
swainy wrote:the crossing of the vertex + mass+ acceleration+speed= no cigar.
You can at least try to make sense. None of this does.swainy wrote:Correct, but unless your crossing it with nothing, then you may as well say a million times the speed of light. it does not matter how shallow your inverted guillotine is. one seconds worth of blade, is a long blade. but that long blade is not long enough due to acceleration. And to get to the speed of light you need all the energy in the universe. And it does not matter how you put it.
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Re: GRED Answer: Scissor vertex speed
Interesting note for guillotine example: The vertex travel 2 distances, the bottom length of right triangle (width of device) and hypotenuse of right triangle (length of blade edge) at the same time, further lending to fact that it is not a physical object.
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Re: GRED Answer: Scissor vertex speed
I don't follow that.swainy wrote:Correct, but unless your crossing it with nothing, then you may as well say a million times the speed of light. it does not matter how shallow your inverted guillotine is. one seconds worth of blade, is a long blade. but that long blade is not long enough due to acceleration. And to get to the speed of light you need all the energy in the universe. And it does not matter how you put it.
Here are some simple numbers. If you have a guillotine blade that is 1 meter long, and slanted 89° from its direction of travel (that is, the edge is one degree from being parallel with the cross bar), the intersection point of the blade and the cross bar (the vertex) will appear to move at the speed of light if the blade is moving about 5200 km/s. There is nothing in physics that makes that speed unreasonable- it isn't even relativistic. And there is no reason that you couldn't accelerate a perfectly ordinary material like steel to that speed without damaging it or requiring a physically unreasonable amount of energy. It's nothing more than an engineering problem. Double that perfectly reasonable speed, and you'll see the vertex moving at 2c. Make the angle shallower, and you'll see the vertex move faster even with a slower blade speed.
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Re: GRED Answer: Scissor vertex speed
Chris Peterson wrote:I don't follow that.swainy wrote:Correct, but unless your crossing it with nothing, then you may as well say a million times the speed of light. it does not matter how shallow your inverted guillotine is. one seconds worth of blade, is a long blade. but that long blade is not long enough due to acceleration. And to get to the speed of light you need all the energy in the universe. And it does not matter how you put it.
Here are some simple numbers. If you have a guillotine blade that is 1 meter long, and slanted 89° from its direction of travel (that is, the edge is one degree from being parallel with the cross bar), the intersection point of the blade and the cross bar (the vertex) will appear to move at the speed of light if the blade is moving about 5200 km/s. There is nothing in physics that makes that speed unreasonable- it isn't even relativistic. And there is no reason that you couldn't accelerate a perfectly ordinary material like steel to that speed without damaging it or requiring a physically unreasonable amount of energy. It's nothing more than an engineering problem. Double that perfectly reasonable speed, and you'll see the vertex moving at 2c. Make the angle shallower, and you'll see the vertex move faster even with a slower blade speed.
And the opposite force is Chris? No, Mass and acceleration says NO. The mass you want to use will weigh as much as the whole universe, but will still need all the mass in the universe to convert to energy, to get to light speed. and you still can,t do it.
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Re: GRED Answer: Scissor vertex speed
No part of either blade is moving at relativistic speeds, I'm not sure your analysis holds. The vertex is not a physical entity and its speed is just an illusion, or a mathematical construct.Henning Makholm wrote:It just occurred to me that in real scissors, contact forces between the blades at the vertex is what guides the blades further on to come down on the right side of each other. This guidance obviously is not available in the FTL closing scenarios. The tiniest inaccuracies would lead to either the blades trying to pass through each other, or not forming a vertex at all because the two edges do not touch.
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Re: GRED Answer: Scissor vertex speed
Nothing physical is moving at relativistic speeds. The vertex is not a physical entity, it does not have any mass. It is not bound by SR or GR.swainy wrote:And the opposite force is Chris? No, Mass and acceleration says NO. The mass you want to use will weigh as much as the whole universe, but will still need all the mass in the universe to convert to energy, to get to light speed. and you still can,t do it.
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Re: GRED Answer: Scissor vertex speed
What are you talking about? I gave you a blade that is one meter wide. It might reasonably weigh a kilogram. Its kinetic energy when moving at 5200 km/s is 1.4e13 J - less than 20% of the energy of the bomb used on Hiroshima. The energy requirements for moving this blade at the required speed are well inside our current technology. There is nothing moving at anything close to the speed of light, except for a virtual point marking the intersection of the blade with the cross bar. You might as well hold your hand out and move your finger between Polaris and Arcturus in one second, and then let your brain explode when you realize your finger just moved at a huge multiple of c!swainy wrote:And the opposite force is Chris? No, Mass and acceleration says NO. The mass you want to use will weigh as much as the whole universe, but will still need all the mass in the universe to convert to energy, to get to light speed. and you still can,t do it.
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Re: GRED Answer: Scissor vertex speed
My god, is that it? I used to think that 30 years ago. Yeah i just moved billions of light years at the end of my finger in seconds, whats the point?
tc
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Re: GRED Answer: Scissor vertex speed
So, is our fun meter pegged yet?
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Re: GRED Answer: Scissor vertex speed
now, i have taken time to read this. Your one meter blade is not long enough, and there is not enough energy in the universe to complete what you want to. you really should know this Chris. your Da Man.Chris Peterson wrote:What are you talking about? I gave you a blade that is one meter wide. It might reasonably weigh a kilogram. Its kinetic energy when moving at 5200 km/s is 1.4e13 J - less than 20% of the energy of the bomb used on Hiroshima. The energy requirements for moving this blade at the required speed are well inside our current technology. There is nothing moving at anything close to the speed of light, except for a virtual point marking the intersection of the blade with the cross bar. You might as well hold your hand out and move your finger between Polaris and Arcturus in one second, and then let your brain explode when you realize your finger just moved at a huge multiple of c!swainy wrote:And the opposite force is Chris? No, Mass and acceleration says NO. The mass you want to use will weigh as much as the whole universe, but will still need all the mass in the universe to convert to energy, to get to light speed. and you still can,t do it.
tc
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Re: GRED Answer: Scissor vertex speed
What's the length of the blade have to do with anything? I gave you some numbers based on simple calculations. I'd welcome you providing your own analysis explaining where I might have gone wrong. In the absence of such, I can't place much weight on your opinion.swainy wrote:now, i have taken time to read this. Your one meter blade is not long enough, and there is not enough energy in the universe to complete what you want to. you really should know this Chris. your Da Man.
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Re: GRED Answer: Scissor vertex speed
But you did, and i thought it. I am saying nothing from now on. Good Job Chris. Let some others have a go.Chris Peterson wrote:What's the length of the blade have to do with anything? I gave you some numbers based on simple calculations. I'd welcome you providing your own analysis explaining where I might have gone wrong. In the absence of such, I can't place much weight on your opinion.swainy wrote:now, i have taken time to read this. Your one meter blade is not long enough, and there is not enough energy in the universe to complete what you want to. you really should know this Chris. your Da Man.
tc
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Re: GRED Answer: Scissor vertex speed
Hah... you had me thinking about this for a minute. But really, this doesn't suggest the point is non-physical. It just means the point moves at a different velocity along the baseline than along the blade. This could be accomplished just as well with a physical object. Make your blade edge and baseline out of stiff wires, and put a bead at their intersection. If you now move the blade edge down, the bead moves horizontally along the baseline wire, marking the intersection of the two. It will travel different distances on each wire, in exactly the same time.hstarbuck wrote:Interesting note for guillotine example: The vertex travel 2 distances, the bottom length of right triangle (width of device) and hypotenuse of right triangle (length of blade edge) at the same time, further lending to fact that it is not a physical object.
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Re: GRED Answer: Scissor vertex speed
The parts of your quote I bolded are absolutely correct. I just don't see what either of them has to do with my remark.bystander wrote:No part of either blade is moving at relativistic speeds, I'm not sure your analysis holds. The vertex is not a physical entity and its speed is just an illusion, or a mathematical construct.Henning Makholm wrote:It just occurred to me that in real scissors, contact forces between the blades at the vertex is what guides the blades further on to come down on the right side of each other. This guidance obviously is not available in the FTL closing scenarios. The tiniest inaccuracies would lead to either the blades trying to pass through each other, or not forming a vertex at all because the two edges do not touch.
It's not really a question of relativity. The same thing applies if we attempt to close two straight, parallel blades against-and-past each other at a millimeter per second. Either you'll get a gap between the blades (in the z-direction), and no vertex events at all, or you'll be virtually certain to have the leading edge bounce into each other at multiple places instead of slicing cleanly.
Imagine you're an atom right on the edge of one of the blades coming down on another. The opposite edge is approaching moderately fast. You need to pass it either to the left or to the right, but close enough that a vertex will actually form between your point on the blade and the opposite one (because if a vertex doesn't form, it certainly won't be moving at any speed, faster than light or not). You have temporarily forgotten whether the blade you're part of is supposed to be the left blade or the right one (atoms have notoriously bad memory). How are you going to decide which side of the opposite blade to shoot for?
Under ordinary scissor conditions you would just observe what the atom next to you in the direction of the pivot does, and then do the same. However, in the experiment we're contemplating, you need to make a decision before information about what any other points on the edge chose can reach you.
If some parts of the blade decide to try to be the right blade and other parts try to be the left one, there will be a gnashing sound and a ruined scissors, not any moving vertex.
It would be quite a different thing if we had a single blade (a line) slicing through a plane of paper. Then it would not need any particular z-axis guidance in order to form a vertex.
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Re: GRED Answer: Scissor vertex speed
Ignoring any relativistic effects, of course. :–)Chris Peterson wrote:Make your blade edge and baseline out of stiff wires, and put a bead at their intersection. If you now move the blade edge down, the bead moves horizontally along the baseline wire, marking the intersection of the two. It will travel different distances on each wire, in exactly the same time.
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Re: GRED Answer: Scissor vertex speed
Yes, obviously this "vertex" is a physical object and cannot exceed c. Unlike the virtual vertex, it requires energy pumped into the blade movement to drive the bead. We'll just run my little bead experiment at low speed to demonstrate that the bead can simultaneously be moving a different velocities along the blade and baseline.Henning Makholm wrote:Ignoring any relativistic effects, of course. :–)Chris Peterson wrote:Make your blade edge and baseline out of stiff wires, and put a bead at their intersection. If you now move the blade edge down, the bead moves horizontally along the baseline wire, marking the intersection of the two. It will travel different distances on each wire, in exactly the same time.
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