admin Posted September 7, 2009 Report Share Posted September 7, 2009 What did I say a few weeks ago? No need to roll your eyes. I've had this on my to-do list for some time. Finally got to it today. Quote Link to comment Share on other sites More sharing options...
Overlord Shinnra Posted September 7, 2009 Report Share Posted September 7, 2009 shortest path on the Surface of the moonWell, to be accurate, I checked angles between the lines connecting the center of the moon to these locations, which are proportional to the shortest distance on the surface of the moon. Shinnra, is your guess based on some data? just an educated guess based on distances between the percentages. It really seems the the hotspot is so very small while the other percentage areas are large. Quote Link to comment Share on other sites More sharing options...
Echostorm Posted September 7, 2009 Report Share Posted September 7, 2009 Quote Link to comment Share on other sites More sharing options...
jonnygozy Posted September 7, 2009 Report Share Posted September 7, 2009 Maybe it's a 2D Gaussian-type surface given how small the hotspot seems to be. I'll input all the data points later today and try some different 2D surface fits to see if any of them will work. Does that seem like a decent idea? Quote Link to comment Share on other sites More sharing options...
poach Posted September 7, 2009 Report Share Posted September 7, 2009 (edited) I predict the hotspot to be around (-70.4165, 71.333). I'm concerned that elevation also affects this, so I was trying to keep on a 0km elevation point while searching for the coordinates when I 'submitted' prematurely. Even so, (-73.603, 69.16992) got me 96% efficiency. My method was to plot all of the locations in this topic (in radians), and had enough approximate points to construct a circle with my prediction as the center. A couple of data points (which I consider reliable) don't quite match which is why I'm thinking there is a modifier with the elevation. This was evident if you calculate the distance from my predicted center point using the great circle equation (which assumes a sphere) 1400.71 67.00 1152.55 72.00 976.12 76.00 813.58 80.00 768.80 84.00 585.87 85.00 589.40 88.00 333.91 90.00 (Golan's point on an 8KM rise) 473.23 90.00 307.46 91.00 (OS's 91% point on a 2-4 KM+ rise) 424.25 91.00 443.23 91.00 391.77 92.00 163.31 94.00 98.71 96.00 (distance in KM vs percent. Unless otherwise noted, the 85%+ points are all on a 0 +/- 2 KM rise) Edited September 7, 2009 by poach Quote Link to comment Share on other sites More sharing options...
Ron Paul Posted September 7, 2009 Report Share Posted September 7, 2009 Poach, I got 98% with your prediction- "I predict the hotspot to be around (-70.4165, 71.333)." Quote Link to comment Share on other sites More sharing options...
poach Posted September 7, 2009 Report Share Posted September 7, 2009 (edited) Poach, I got 98% with your prediction- "I predict the hotspot to be around (-70.4165, 71.333)." Considering the accuracy of the data (since every percentage has a certain range around it), I'll take that as a victory Edited September 7, 2009 by poach Quote Link to comment Share on other sites More sharing options...
JoshuaR Posted September 7, 2009 Report Share Posted September 7, 2009 Poach, could you elaborate? You are converting Lat,Long into theta,psi from the center of the moon, and then you are using the percent values to construct a sphere from a predicted point on the surface of the moon, so that the percentages match? (thus distance from hotspots goes through the moon, not along the surface.) And to predict a point, you are looking at the plotted points and applying a formula that calculates the center (assuming linear correspondence between distance and percentages, or something non-linear?, or are you using a guess and check method (computer assisted or otherwise) in order to predict a center of the sphere where you think the hot spot is?) 1. Are those km vs percentage plots km along the surface of the moon or going through the moon? 2. Guess and check to predict hot spot, or equation? Quote Link to comment Share on other sites More sharing options...
poach Posted September 7, 2009 Report Share Posted September 7, 2009 Poach, could you elaborate? You are converting Lat,Long into theta,psi from the center of the moon, and then you are using the percent values to construct a sphere from a predicted point on the surface of the moon, so that the percentages match? (thus distance from hotspots goes through the moon, not along the surface.)And to predict a point, you are looking at the plotted points and applying a formula that calculates the center (assuming linear correspondence between distance and percentages, or something non-linear?, or are you using a guess and check method (computer assisted or otherwise) in order to predict a center of the sphere where you think the hot spot is?) 1. Are those km vs percentage plots km along the surface of the moon or going through the moon? 2. Guess and check to predict hot spot, or equation? Well, at first I just wanted to get a feel of the distance between the points, so I used the great circle equation to come up with distances ( r * acos[sin(lat1) * sin(lat2) + cos(lat1) * cos(lat2) * cos(lon2 - lon1)] ). This equation uses distance across the surface. Second, I decided to plot all of the closer points at the 90+ percent range. I found that I had a few data points, and using a compass, was able to connect the points with a circle. The center point of the compass was my prediction. I then drew a circle around every data point I had, and everything (except as noted) seemed to match up. A couple of the data points should have had different percentages though. So there is something else at work here, which is why I suspect elevation is a factor. And before you start on the 2D doesn't translate into 3D (Golan already gave me this argument), remember the plot I did was in radians, so each given fraction of a radian had a different distance associated with it, making the plot "good enough for government work". Quote Link to comment Share on other sites More sharing options...
JoshuaR Posted September 7, 2009 Report Share Posted September 7, 2009 Ah good. So the distances are across the surface. And from the description, the percentages actually seem linear? This means we can figure out the equation that converts distances to percentages. Perhaps you can plot your data in excel and see if you can fit a curve to the data? Quick request: Using the predicted/realish hot spot, could you plot surface distances to a couple points from the hotspot, using points that seem accurate. Knowing that there is a radius R that would be equal to 50% efficiency, would you be able to make a prediction for that radius R? ^I.E. Say 95% is 100km, 79% is 1000k (along the surface), so the difference between these is 900km and the efficiency changed by 16%. Then you could take that slope and solve for when the efficiency equals 50% to find the distance 9along the surface). We could even have someone test the prediction by moving a base just inside the prediction, hoping to get 50-51% efficiency. BY THE WAY: I thought I had posted in this thread, but it looks like I hadn't. I moved my base on 9/03: -65.36684, 80.00000 = 97% efficiency. <---So this can add to your data points. Quote Link to comment Share on other sites More sharing options...
Ron Paul Posted September 7, 2009 Report Share Posted September 7, 2009 Considering the accuracy of the data (since every percentage has a certain range around it), I'll take that as a victory No complaints at all; just providing data to help out. Good show. Thanks. Quote Link to comment Share on other sites More sharing options...
poach Posted September 7, 2009 Report Share Posted September 7, 2009 (edited) Ah good. So the distances are across the surface. And from the description, the percentages actually seem linear? This means we can figure out the equation that converts distances to percentages. Perhaps you can plot your data in excel and see if you can fit a curve to the data?Quick request: Using the predicted/realish hot spot, could you plot surface distances to a couple points from the hotspot, using points that seem accurate. Knowing that there is a radius R that would be equal to 50% efficiency, would you be able to make a prediction for that radius R? ^I.E. Say 95% is 100km, 79% is 1000k (along the surface), so the difference between these is 900km and the efficiency changed by 16%. Then you could take that slope and solve for when the efficiency equals 50% to find the distance 9along the surface). We could even have someone test the prediction by moving a base just inside the prediction, hoping to get 50-51% efficiency. BY THE WAY: I thought I had posted in this thread, but it looks like I hadn't. I moved my base on 9/03: -65.36684, 80.00000 = 97% efficiency. <---So this can add to your data points. I'm at work atm, so I won't be able to do anything with data. The problem I have is that I'm using open office, and I couldn't get it to do a decent plot. Anyone know any free software that will do a x,y plot and that you can draw circles on? Also, message me and I'll email you the open office file with the data points I have. Also, for the percent ranges, I don't think they are linear, but I don't have enough data points yet to really decide either way. Having said that though, I think we're looking at around 25km or less for 100, going to at least 200km for the 91% (which I had the most data points for). Then if elevation is indeed a factor, it'll make it harder to figure. Best thing to do: More data points. Edited September 7, 2009 by poach Quote Link to comment Share on other sites More sharing options...
JoshuaR Posted September 7, 2009 Report Share Posted September 7, 2009 K. Funny how before these hot spot hunts... we thought we'd be averaging 75% efficiency even if we weren't moving our bases! Quote Link to comment Share on other sites More sharing options...
enderland Posted September 8, 2009 Report Share Posted September 8, 2009 K. Funny how before these hot spot hunts... we thought we'd be averaging 75% efficiency even if we weren't moving our bases! I was pretty sure this would happen the moment hotspots were released. Quote Link to comment Share on other sites More sharing options...
JoshuaR Posted September 8, 2009 Report Share Posted September 8, 2009 Yep, and yet we were wrong. Quote Link to comment Share on other sites More sharing options...
enderland Posted September 8, 2009 Report Share Posted September 8, 2009 Yep, and yet we were wrong. I should have clarified - because of the "once per week move" it was obvious that what will happen is that someone finds it, and then it is trivial to get over 90% efficiency, meaning that it doesn't really matter if you have it right away or not, because someone else will find it and you can just copy them (basically ^^). Quote Link to comment Share on other sites More sharing options...
jonnygozy Posted September 8, 2009 Report Share Posted September 8, 2009 (edited) I decided to guess in between the 97% and 98% points recently posted, and got 99%. Thanks guys! Edited September 8, 2009 by jonnygozy Quote Link to comment Share on other sites More sharing options...
poach Posted September 8, 2009 Report Share Posted September 8, 2009 With the new data, my 'paper method' has broken down and can't get any more accurate. Measuring distance from 99%, we get this: 90.00 408.6779815523 -1.4198951600 1.3440000000 90.00 389.3700052567 -.9612988548 1.3591069781 91.00 373.3578047334 -1.3993825405 1.4448053714 91.00 359.0929109946 -.9787521473 1.3591069781 91.00 340.2302337869 -.9896803237 1.3575729973 92.00 309.1207733328 -1.0079464224 1.3499030933 94.00 219.8657791893 -1.3114539758 1.3989904786 95.00 198.0383920255 -1.2985626626 1.3621750269 96.00 201.8612005043 -1.2846146893 1.2072428473 97.00 76.6660386086 -1.1408665796 1.3962634016 98.00 117.9273818012 -1.2290000000 1.2450000000 99.00 .0000000000 -1.1849331762 1.3904427285 (percentage, distance in KM from 99% point, location in radians) With the placement of each point, even throwing in the maximum error, plotting no longer works. The only thing I am somewhat certain of is that the hotspot is with a circle with centerpoint (-1.21, 1.335) and a radius of 0.105 radians. That circle contains every point above 95%, but distances from that point look like this: 90.00 432.5076356480 90.00 364.6897724789 91.00 402.2037121581 91.00 383.1757016104 91.00 332.3635721306 92.00 351.2289404303 94.00 179.4165749398 95.00 154.5556758722 96.00 147.3558529350 97.00 126.8655688224 98.00 63.0970559637 99.00 55.9405500987 The distances, on the other hand, seem to match up. (percent, distance from -1.21, 1.335) On paper, it doesn't work, but until we get a couple points at 100% (or a few more at 99% so we can triangulate from there), this is about as accurate as I can do. Already though, We can see some patterns. The range for 98% seems to be about 100km. For 95% seems to be about 150km. and for 90% is in the neighborhood of 500km. The 75% point is around 1000km. The diameter of the moon is around 3475 km. Quote Link to comment Share on other sites More sharing options...
JoshuaR Posted September 9, 2009 Report Share Posted September 9, 2009 Circumference (distance along the surface, as we suspect) is up to 10,921 km. So these hot spots don't have that great a range. Quote Link to comment Share on other sites More sharing options...
Coffee Shock Posted September 9, 2009 Report Share Posted September 9, 2009 (edited) This seems to work for me. moon diameter = 3474 efficiency = max((1 - (distance to hot spot / moon diameter)) * 100, 50) That gives: for efficiency > 50 distance to hot spot = (1 - (efficiency / 100)) * moon diameter Now it's easy to find the hot spot from three coordinates with trilateration. B) Edit: Incorrectly placed parentheses. Edited September 9, 2009 by Coffee Shock Quote Link to comment Share on other sites More sharing options...
poach Posted September 9, 2009 Report Share Posted September 9, 2009 (edited) It looks like that equation is correct. From playing around with a spreadsheet, it looks like 50km per percent. I was playing around with determining the hotspot, but kept getting my butt kicked by the math. Here's what I'm trying to do, maybe one of you math types can tell me what I'm doing wrong From the law of cosines for a spherical triangle C = acos [ (cos(.c) - cos(.a)cos(.b.))/(sin(.a)sin(.b.))] Latitude=asin(sin(lat1)*cos(distance)+cos(lat1)*sin(distance)*cos(.C) dlon = atan2(sin(.C)*sin(distance)*cos(lat1),cos(distance)-sin(lat1)*sin(lat2)) Longitude = mod( lon1-dlon +pi,2*pi )-pi [all decimals above were to defeat the bbcode from changing the variables -- not actual decimals) any takers? If this will work, then from 2 points > 50% efficiency, we should be able to get the position for the hotspot. Edited September 9, 2009 by poach Quote Link to comment Share on other sites More sharing options...
King Irwin Posted September 9, 2009 Report Share Posted September 9, 2009 I've added a location indicator on the popup boxes for the Moon & Mars wonders so that you don't have to go digging through the source code to get your coordinates. Hope that helps. Woohoo! Too bad you hadn't done that a couple days earlier. I was trying to do this from a hotel lobby kiosk last week, but I couldn't see the source code from there. Now if only you could add the abillity to type in new coordinates directly instead of just dragging the icon. That would be ideal! Unfortunately, I have nothing useful to add right now. I actually managed to mess up with moving BOTH of my wonders! The first one I managed to accidentally hit enter before changing the coordinates, and the second one I hadn't realized that longitude was first, so typed the numbers in backwards. I guess the lesson here is that I shouldn't do this when I'm sleep deprived! Quote Link to comment Share on other sites More sharing options...
poach Posted September 9, 2009 Report Share Posted September 9, 2009 It looks like that equation is correct. From playing around with a spreadsheet, it looks like 50km per percent.I was playing around with determining the hotspot, but kept getting my butt kicked by the math. Here's what I'm trying to do, maybe one of you math types can tell me what I'm doing wrong From the law of cosines for a spherical triangle C = acos [ (cos(.c) - cos(.a)cos(.b.))/(sin(.a)sin(.b.))] Just an fyi, I figured out why this doesn't work. With the level of error, you can't construct a true triangle, and thus you can't get an accurate angle for C, making the equation set fall apart. Quote Link to comment Share on other sites More sharing options...
Rich333 Posted September 10, 2009 Report Share Posted September 10, 2009 Efficiency: 99% Latitude: -67.89 Longitude: 75.67 Quote Link to comment Share on other sites More sharing options...
Golan 1st Posted September 11, 2009 Report Share Posted September 11, 2009 I get 99% at (-67.88763, 72.01126) Quote Link to comment Share on other sites More sharing options...
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