Attacking Darius:

penny wrote:And yes, that fact can yield wildly divergent microjumps. If the scalar is large enough then the negative result of that scalar can place us quite some distance away. But the hypotenuse of the scalar will always be the same.

That knowledge still does not solve our problem. If as you said the wall is as wildly chaotic as you claim it to be. In which case, the fluctuations could vary wildly from one jump to another. But I personally do not think the wall is fluctuating so much in a small timeframe. Rather than wildly varying... in time measured in hours, days, weeks or even months. Or the equation will have changed before the second jump in a dogleg. Therefore, one reason a ship might hyper in and out is to measure the degree of fluctuations currently operating on the wall.

For my edification, please explain to me what the "hypotenuse of the scalar" is? Before now I have only heard "hypotenuse" used as the longest side of a right triangle, the one opposite the right angle.

The thing is, if the solution to the problem of microjumps were as simple as you thought; then why haven't some of the brightest people in the Honorverse (and we know there are some with enhanced intelligence) already come up with the same solution?

tlb wrote:

penny wrote:And yes, that fact can yield wildly divergent microjumps. If the scalar is large enough then the negative result of that scalar can place us quite some distance away. But the hypotenuse of the scalar will always be the same.

That knowledge still does not solve our problem. If as you said the wall is as wildly chaotic as you claim it to be. In which case, the fluctuations could vary wildly from one jump to another. But I personally do not think the wall is fluctuating so much in a small timeframe. Rather than wildly varying... in time measured in hours, days, weeks or even months. Or the equation will have changed before the second jump in a dogleg. Therefore, one reason a ship might hyper in and out is to measure the degree of fluctuations currently operating on the wall.

For my edification, please explain to me what the "hypotenuse of the scalar" is? Before now I have only heard "hypotenuse" used as the longest side of a right triangle, the one opposite the right angle. 1.

The thing is, if the solution to the problem of microjumps were as simple as you thought; then why haven't some of the brightest people in the Honorverse (and we know there are some with enhanced intelligence) already come up with the same solution? 2.

To be taken in the context of Jonathan's post and my response.

1. The origin of the microjump is C.

The two possibly varying results of the microjump will always be equidistant from the origin. Represented as the vertices of an equidistant triangle and labeled as A, and B.

CA = CB since they represent the hypotenuse of the two right triangles formed by the isosceles triangle.

2. I would put money on my own translation of the difficulties if I had to bet. Since it also isn't as haphazard as you and Jonathan make it out to be. Or Honor's several different astrogator's would not have been able to consistently give her what she asks. Especially considering the difficult dogleg right inside the resonance zone??? If memory serves.

If it is as wildly chaotic as you make it out to be, then Honor was irresponsible taking that risk which could have put her anywhere in the system ... and dead. Along with the system.

penny wrote:1. The origin of the microjump is C

The two possibly varying results of the microjump will always be equidistant from the origin. Represented as the vertices of an equidistant triangle and labeled as A, and B.

CA = CB since they represent the hypotenuse of the two right triangles formed by the isosceles triangle.

There is NO reason to assume that there are only two possible results, NOR that two successive jumps would end up with the same displacement from the origin. Instead they will end up somewhere in the circle defined by the origin and what we hope will be the maximum possible displacement from that origin. From the text in chapter 38 of Echoes of Honor (given earlier in this thread), we know that could be on the order of light seconds.

Again, if the answer were easy, then would already have been solved in the Honorverse.

tlb wrote:

penny wrote:1. The origin of the microjump is C

The two possibly varying results of the microjump will always be equidistant from the origin. Represented as the vertices of an equidistant triangle and labeled as A, and B.

CA = CB since they represent the hypotenuse of the two right triangles formed by the isosceles triangle.

There is NO reason to assume that there are only two possible results, NOR that two successive jumps would end up with the same displacement from the origin. Instead they will end up somewhere in the circle defined by the origin and what we hope will be the maximum possible displacement from that origin. From the text in chapter 38 of Echoes of Honor (given earlier in this thread), we know that could be on the order of light seconds.

Again, if the answer were easy, then would already have been solved in the Honorverse.

To be taken in the context of the exchange with Jonathan.

penny wrote:1. The origin of the microjump is C

The two possibly varying results of the microjump will always be equidistant from the origin. Represented as the vertices of an equidistant triangle and labeled as A, and B.

CA = CB since they represent the hypotenuse of the two right triangles formed by the isosceles triangle.

tlb wrote:There is NO reason to assume that there are only two possible results, NOR that two successive jumps would end up with the same displacement from the origin. Instead they will end up somewhere in the circle defined by the origin and what we hope will be the maximum possible displacement from that origin. From the text in chapter 38 of Echoes of Honor (given earlier in this thread), we know that could be on the order of light seconds.

Again, if the answer were easy, then would already have been solved in the Honorverse.

penny wrote:To be taken in the context of the exchange with Jonathan.

Since Jonathan_S was agreeing with me and my answer was in that context, I do not think this last post answers or explains anything.

tlb wrote:

penny wrote:1. The origin of the microjump is C

The two possibly varying results of the microjump will always be equidistant from the origin. Represented as the vertices of an equidistant triangle and labeled as A, and B.

CA = CB since they represent the hypotenuse of the two right triangles formed by the isosceles triangle.

tlb wrote:There is NO reason to assume that there are only two possible results, NOR that two successive jumps would end up with the same displacement from the origin. Instead they will end up somewhere in the circle defined by the origin and what we hope will be the maximum possible displacement from that origin. From the text in chapter 38 of Echoes of Honor (given earlier in this thread), we know that could be on the order of light seconds.

Again, if the answer were easy, then would already have been solved in the Honorverse.

penny wrote:To be taken in the context of the exchange with Jonathan.

Since Jonathan_S was agreeing with me and my answer was in that context, I do not think this last post answers or explains anything.

What you mean is it doesn't coincidence with your own translation. Agreed. But it definitely explains my own take on it. And I am entitled to that translation.

penny wrote:What you mean is it doesn't coincidence with your own translation. Agreed. But it definitely explains my own take on it. And I am entitled to that translation.

No one denies that you are entitled to an opinion and if you want to call it an interpretation or translation instead, that is certainly fine also.

But saying I am entitled to my opinion is not the same thing as defending it. However in 1862, Emily Dickinson wrote, "The heart wants what it wants, or else it does not care"; certainly it does not care about logic or evidence. So if that is what your heart wants, then go ahead and cling to it.

As for me, I am only trying to understand the problem as described in the books. The only reason that I do not accept your interpretation is that your proposed solution is so simple that it should already be implemented.

tlb wrote:There is NO reason to assume that there are only two possible results, NOR that two successive jumps would end up with the same displacement from the origin. Instead they will end up somewhere in the circle defined by the origin and what we hope will be the maximum possible displacement from that origin. From the text in chapter 38 of Echoes of Honor (given earlier in this thread), we know that could be on the order of light seconds.

Again, if the answer were easy, then would already have been solved in the Honorverse.

Yeah, I would expect a spherical error margin; unless something goes especially wrong you're going to appear within X of your target point, but you're equally likely to be displaced in any direction from it.

If the variations were purely quantum then it'd probably be an evenly weighted error (you're just as likely to appear at any point within that probability sphere) - but if there's some classical component then it seems more likely it'd be a weighted distribution (for example, you're more likely to emerge closer to the center of the target than right out at the edges of the probability sphere).

Jonathan_S wrote:

tlb wrote:There is NO reason to assume that there are only two possible results, NOR that two successive jumps would end up with the same displacement from the origin. Instead they will end up somewhere in the circle defined by the origin and what we hope will be the maximum possible displacement from that origin. From the text in chapter 38 of Echoes of Honor (given earlier in this thread), we know that could be on the order of light seconds.

Again, if the answer were easy, then would already have been solved in the Honorverse.

Yeah, I would expect a spherical error margin; unless something goes especially wrong you're going to appear within X of your target point, but you're equally likely to be displaced in any direction from it.

If the variations were purely quantum then it'd probably be an evenly weighted error (you're just as likely to appear at any point within that probability sphere) - but if there's some classical component then it seems more likely it'd be a weighted distribution (for example, you're more likely to emerge closer to the center of the target than right out at the edges of the probability sphere).

Jonathan, you also are not staying within the parameters of our exchange.

I posited the ability to duplicate a jump inputting the exact same data under the exact same conditions with the same ship. I do not think the fluctuations are from moment to moment or doglegs would be impossible. Period. It wasn't luck what Theophile Kgari pulled off.

So, under the exact same conditions, a jump should produce nearly the same results. I attempted to explain my interpretations of the differences, and why.

But, let's stop pussyfooting around the bush and get to the nitty-gritty of the argument by... mounting this horse from the other side. As Pavel Young might say.

Regarding kzt's suggested tactic of microjumping in the middle of a GA fleet, is that doable by a navy other than a navy belonging to the GA?

penny wrote:Quantum effects are consistent, therefore quantum fluctuations should also be consistent. Quantum manipulation is consistent as well; or it would be impossible to use it in our next generation quantum computers because these quantum computers would then be impossible.

Jonathan_S wrote:On the contrary quantum fluctuations are about the only source we know of of basically perfect randomness.

Yes, they vary within a definable range - but they are equally likely to be any value within that range; and their value at one instant has zero correlation to their value at the next instant.

penny wrote:You said exactly what I meant. Quantum fluctuations will always vary within a definable range. Consistent. That is why I said all you have to do is add the appropriate weight back in from the tables in the back or front of the book (adjust by the appropriate weight). But of course, you cannot pinpoint a certain value; which would be completely contrary to quantum theory.

tlb wrote:We are trying to say that the wall is chaotic, involving gravity shears, energy flows and fluctuations. Instead you think an "appropriate weight" can be ADDED to account for all effects. What if sometimes that has to be subtracted? What if it manifests itself as a force that can have both a random magnitude (even if it is capped by some value) and a random direction?

Jonathan_S wrote:Exactly - knowing an unmeasurably value might, for example, be any value between 0.5 and 2.5 doesn't really help if that range of fluctuations causes a multiple lightsecond dispersion in your entry point. What can you add in to the unknown to counter that?

tlb wrote:There is NO reason to assume that there are only two possible results, NOR that two successive jumps would end up with the same displacement from the origin. Instead they will end up somewhere in the circle defined by the origin and what we hope will be the maximum possible displacement from that origin. From the text in chapter 38 of Echoes of Honor (given earlier in this thread), we know that could be on the order of light seconds.

Again, if the answer were easy, then would already have been solved in the Honorverse.

Jonathan_S wrote:Yeah, I would expect a spherical error margin; unless something goes especially wrong you're going to appear within X of your target point, but you're equally likely to be displaced in any direction from it.

If the variations were purely quantum then it'd probably be an evenly weighted error (you're just as likely to appear at any point within that probability sphere) - but if there's some classical component then it seems more likely it'd be a weighted distribution (for example, you're more likely to emerge closer to the center of the target than right out at the edges of the probability sphere).

penny wrote:Jonathan, you also are not staying within the parameters of our exchange.

I posited the ability to duplicate a jump inputting the exact same data under the exact same conditions with the same ship. I do not think the fluctuations are from moment to moment or doglegs would be impossible. Period. It wasn't luck what Theophile Kgari pulled off.

So, under the exact same conditions, a jump should produce nearly the same results. I attempted to explain my interpretations of the differences, and why.

Jonathan_S has been consistent and you seem to avoid what he is saying. So what do you mean about "the parameters of the exchange"? Do you mean that he should allow you to incorrectly interpret what he said?