+georgeandmary Posted June 27, 2002 Share Posted June 27, 2002 I just posted a new cache under the inspiration of caches like http://www.geocaching.com/seek/cache_details.asp?ID=24907 which was in spired by http://www.geocaching.com/seek/cache_details.asp?ID=16760 but I added a twist that I haven't seen in a cache before. I know many of you are not in a position to hunt it but you may want to try to see if you can solve for the final coordinates. If you send me the coordinates, I'll tell you if you right or wrong. I tried to make the numbers so that very litte rounding is involved, and of course, I made some assumptions to make the problem do able. http://www.geocaching.com/seek/cache_details.asp?ID=26935 There are all kinds of things I could have done to make the problem more 'realistic' but maybe I'll save those for later. BTW, Fizzy has already sent me the correct coordinates. I don't think it took him very long to solve it. george Remember: Half the people you meet are below average. Quote Link to comment
+Markwell Posted June 27, 2002 Share Posted June 27, 2002 Only 4 stars on the difficulty? Man, I wish I could find that one for real. I may have to see if I can get the coordinates... Markwell Chicago Geocachers Quote Link to comment
+parkrrrr Posted June 27, 2002 Share Posted June 27, 2002 I like it. I'm not sure about the 4 stars, though; I think maybe you could have gotten away with 5, since it requires specialized knowledge. My guess as to the coordinates is on its way. Quote Link to comment
+parkrrrr Posted June 27, 2002 Share Posted June 27, 2002 I like it. I'm not sure about the 4 stars, though; I think maybe you could have gotten away with 5, since it requires specialized knowledge. My guess as to the coordinates is on its way. Quote Link to comment
+fizzymagic Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by Warm Fuzzies - Fuzzy: I like it. I'm not sure about the 4 stars, though; I think maybe you could have gotten away with 5, since it requires specialized knowledge. The 4 stars is probably because that's what I rated my math-based cache, which is what DisQuoi rated his in the DC area that inspired mine that inspired George. This one took me about 30 minutes to solve; much of the time was in using a website to do UTM translation. I need to write a program to do it more quickly! Anyway, great job, George! Quote Link to comment
+fizzymagic Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by Warm Fuzzies - Fuzzy: I like it. I'm not sure about the 4 stars, though; I think maybe you could have gotten away with 5, since it requires specialized knowledge. The 4 stars is probably because that's what I rated my math-based cache, which is what DisQuoi rated his in the DC area that inspired mine that inspired George. This one took me about 30 minutes to solve; much of the time was in using a website to do UTM translation. I need to write a program to do it more quickly! Anyway, great job, George! Quote Link to comment
+parkrrrr Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by fizzymagic:The 4 stars is probably because that's what I rated my math-based cache, which is what DisQuoi rated his in the DC area that inspired mine that inspired George. But neither of yours required the finder to know how to calculate the time it takes a body to fall a certain distance in freefall, or to know about conservation of momentum. Truth be told, I wish the GCRS had more questions for "difficulty." I had a lot of trouble deciding how difficult was too difficult on Perfectly Perplexing Puzzles (and I may have erred on the side of "too difficult"; we'll see.) My puzzles don't require specialized knowledge, but they do require a degree of puzzle-solving ability and some intuition. Quote Link to comment
+parkrrrr Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by fizzymagic:The 4 stars is probably because that's what I rated my math-based cache, which is what DisQuoi rated his in the DC area that inspired mine that inspired George. But neither of yours required the finder to know how to calculate the time it takes a body to fall a certain distance in freefall, or to know about conservation of momentum. Truth be told, I wish the GCRS had more questions for "difficulty." I had a lot of trouble deciding how difficult was too difficult on Perfectly Perplexing Puzzles (and I may have erred on the side of "too difficult"; we'll see.) My puzzles don't require specialized knowledge, but they do require a degree of puzzle-solving ability and some intuition. Quote Link to comment
+MrGigabyte Posted June 27, 2002 Share Posted June 27, 2002 Your hint is in error. The s should be squared. Quote Link to comment
+georgeandmary Posted June 27, 2002 Author Share Posted June 27, 2002 quote:Originally posted by MrGigabyte: Your hint is in error. The s should be squared. Yeah, I know, I just didn't know if I should write 9.8m/ss or 9.8 m/s^2 or if I should just leave it 9.8 and leave the units off. george Remember: Half the people you meet are below average. Quote Link to comment
scooterj Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by MrGigabyte: Your hint is in error. The s should be squared. This cache reminds of me all the reasons I decided I hate math. Fortunately for me, though, I remember enough physics to also notice the error in the clue. Quote Link to comment
scooterj Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by MrGigabyte: Your hint is in error. The s should be squared. This cache reminds of me all the reasons I decided I hate math. Fortunately for me, though, I remember enough physics to also notice the error in the clue. Quote Link to comment
chloew Posted June 27, 2002 Share Posted June 27, 2002 This cache is another good reason why we should be using the metric system. Now how many meter per foot, and how many feet per mile? And how can I easily find the distance between two given coordinates? Quote Link to comment
chloew Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by mdmax371: Why is this a cache? Because it can be. Seriously though, because with a little work you get a set of coordinates, and you then run to those coordinates guided by your GPS, where hopefully there will be a tupperware box (or black box) filled with trinkets. Quote Link to comment
+georgeandmary Posted June 27, 2002 Author Share Posted June 27, 2002 quote:Originally posted by mdmax371: Why is this a cache? Because there is actaully a box with tradeable items in it. You just have to find the coordinates from the information given. george Remember: Half the people you meet are below average. Quote Link to comment
+fizzymagic Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by georgeandmary: Yeah, I know, I just didn't know if I should write 9.8m/ss or 9.8 m/s^2 or if I should just leave it 9.8 and leave the units off. How about 9.8 m/s/s? That is technically right, and probably less confusing than s^2. Quote Link to comment
Us 5 Camp Posted June 27, 2002 Share Posted June 27, 2002 This was a fun little exercise.... I am sending you via e-mail my coordinates. I just read the cache page again, didn't realize it was a "traditional" cache. I won't be going to CA anytime soon but would appreciate a reply to verify my problem solving was correct. I will leave my log on your page as a "note" only. Thanks for the exercise! [This message was edited by Us 5 Camp on June 27, 2002 at 10:24 AM.] Quote Link to comment
chloew Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by fizzymagic: quote:Originally posted by georgeandmary: Yeah, I know, I just didn't know if I should write 9.8m/ss or 9.8 m/s^2 or if I should just leave it 9.8 and leave the units off. How about 9.8 m/s/s? That is technically right, and probably less confusing than s^2. If you enter the following string (HTML) in the hint it will come out with the normal superscript 2 (you must remove the space between "sup" and the "2", I added it here so that it would not interpret it as HTML): g = 9.8m/s⊃ 2; will be g = 9.8m/s² It will look a bit strange when not decrypted however and might confuse people decrypting by hand, but I don't think so. Will send in my answer by e-mail as well. [This message was edited by chloew on June 27, 2002 at 10:42 AM.] Quote Link to comment
+Jamie Z Posted June 27, 2002 Share Posted June 27, 2002 Fizzy... I'm surprised to read in your profile that you are a physicist, but you say that m/s/s is more correct than m/s^2. The two terms are mathematically identical, but m/s^2 is written in an acceptable form. (i.e., you never see 1/4 written as 1/2/2) I understand that in the past, m/s/s was used more often, but the current standard, I believe, is to use the supersript. In any case, I find m/s^2 to be much easier to use in calculations, since the units can be easily canceled or combined that way. Jamie Quote Link to comment
chloew Posted June 27, 2002 Share Posted June 27, 2002 Nsgre fbyivat gur ceboyrz (ubcrshyyl pbeerpgyl) V ernyvmrq gung ol frcnengvat gur ceboyrz vagb vgf cebcre pbzcbaragf (yngvghqr naq ybatvghqr) vg pna or fbyirq eryngvir rnfvyl jvgu srj pbairefvbaf naq jvgubhg univat gb qrny jvgu gur pbbeqvangrf bgure guna n bar gvzr pbairefvba bs fcrrq/qvfgnapr gb pbbeqvangrf. Qvq abg unir gb hfr HGZ be natyrf. Quote Link to comment
+fizzymagic Posted June 27, 2002 Share Posted June 27, 2002 quote:Qvq abg unir gb hfr HGZ be natyrf. I don't know how close you will be with that approach. I think that the scale of the problem is large enough that it may introduce significant errors. Quote Link to comment
+fizzymagic Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by Jamie Z: Fizzy... I'm surprised to read in your profile that you are a physicist, but you say that m/s/s is more correct than m/s^2. That's not at all what I meant. The notation you prefer is in fact the best one; I was simply suggesting a notation that wouldn't require any HTML and would be understandable by non-experts. In any case, it is something easily looked up by anyone. Quote Link to comment
+MrGigabyte Posted June 27, 2002 Share Posted June 27, 2002 Something does not look right here. You say the explosion occurred at N 37° 02.485 and the aircraft was heading north. You state that part was found about 5km north at N37 5.187. You also go on to state that another part was found at N37 1.063, which is about 3km south of the explosion and in entirely the wrong direction. Is this correct? Quote Link to comment
+fizzymagic Posted June 27, 2002 Share Posted June 27, 2002 It was a very powerful explosion! Quote Link to comment
jfitzpat Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by fizzymagic: It was a _very_ powerful explosion! And a bit magic as well Kuddos to G&M for a job well done, but I've always disliked that aspect of story problems. -jjf Quote Link to comment
jfitzpat Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by fizzymagic: It was a _very_ powerful explosion! And a bit magic as well Kuddos to G&M for a job well done, but I've always disliked that aspect of story problems. -jjf Quote Link to comment
+Jamie Z Posted June 27, 2002 Share Posted June 27, 2002 quote:Originally posted by fizzymagic:I was simply suggesting a notation that wouldn't require any HTML and would be understandable by non-experts. Ok.. Got it. Jamie Quote Link to comment
Greg Nelson Posted June 27, 2002 Share Posted June 27, 2002 Well, I'm being picky here, but that's my job as a physics teacher. Also, I might be putting my foot in my mouth, as I haven't actually worked out the problem. To get an answer here, don't you have to assume that all 3 parts of the plane have the same free-fall time? That's the same as assuming that the explosion sent the three parts of the plane apart in a plane parallel to the earth's surface, rather than blowing them in 3 random directions (such that the total momentum is conserved). In simpler terms, it looks like you have to assume that all three parts have zero starting velocity in the z direction (the straight line from the plane to the ground). This explosion becomes ever more special. Quote Link to comment
+georgeandmary Posted June 27, 2002 Author Share Posted June 27, 2002 quote:Originally posted by Greg Nelson: Well, I'm being picky here, but that's my job as a physics teacher. Also, I might be putting my foot in my mouth, as I haven't actually worked out the problem. To get an answer here, don't you have to assume that all 3 parts of the plane have the same free-fall time? That's the same as assuming that the explosion sent the three parts of the plane apart in a plane parallel to the earth's surface, rather than blowing them in 3 random directions (such that the total momentum is conserved). In simpler terms, it looks like you have to assume that all three parts have zero starting velocity in the z direction (the straight line from the plane to the ground). This explosion becomes ever more special. This is basically the same problem out of my general physics book, Physics for scientists and Engineers, Fourth Edition, by Serway. Conservation of momentum problem Example 9.22 pg 259, The numbers are similar but the situation is the same. The problem only requires that you find the velocity for the third piece given that each piece is 1/3 of the total mass... to quote "the total momentum just before the explosion must equal the total momentum of the framents after the explosion since the forces of the explosion are internal to the sytem and cannot affect its total momentum" Now i did not include wind resistance and I assumed all pieces landed at the same elvation so each piece had the same free fall time. I could have spent the extra time account for the various elevation changes at plotted a more accurate landing spot but I wanted to keep it easy. The second pieces actual elevation is a few hundred feet higher than the first but I didn't think this added anything to the problem. The center of mass of the system should still be moving forward at 300m/s. george Remember: Half the people you meet are below average. Quote Link to comment
+fizzymagic Posted June 28, 2002 Share Posted June 28, 2002 quote:Originally posted by fizzymagic: I don't know how close you will be with that approach. I think that the scale of the problem is large enough that it may introduce significant errors. I was wrong. The scale is small enough that Lat/Long works fine. Quote Link to comment
DisQuoi Posted June 28, 2002 Share Posted June 28, 2002 As I sat down with pencil and paper, I came to a stop when I had the same question as Greg. The fact that all of the pieces ended up at elev=200' does not mean that they all had the same fall time (i.e., that they all hit the ground at the same moment). I'm pretty sure that this problem is not designed to require a three-dimensional analysis of the explosion. Simply add the word "simulateously" to the statement that all of the pieces landed at the same elevation. By the way, I think this is a great way to lay out a cache. While some people see this as merely a "word problem", it is very much unlike many caches that require that you solve a problem that has nothing to do with navigation or geometry which is inherent in global positioning. "Why is this a cache?" ... I'm sure I can guess what type of caches appeal to persons who would ask this ... the kind that I find too simple and not very challenging. Well done! Quote Link to comment
jfitzpat Posted June 28, 2002 Share Posted June 28, 2002 quote:Originally posted by DisQuoi: I'm pretty sure that this problem is not designed to require a three-dimensional analysis of the explosion... Precisely. What if the explosion occured above the black box, or was a missile that detonated below the rocket? Regardless of the location, there is no way to know that the force was evenly distributed to the rocket's total mass (ex. crumple zones in a car). FWIW, I wasn't trying to put G&M's cache down above. I think it is very clever, and, based on the log, well done. I just, in general, have a problem with story problems. Too often, the student isn't solving a real world problem with math, he/she is trying to find the math problem hidden in the story. -jjf Quote Link to comment
+fizzymagic Posted June 28, 2002 Share Posted June 28, 2002 quote:Originally posted by jfitzpat:What if the explosion occured above the black box, or was a missile that detonated below the rocket? Regardless of the location, there is no way to know that the force was evenly distributed to the rocket's total mass (ex. crumple zones in a car). These questions are actually kinda interesting. The answer to the above concern is that it makes no difference. Since there are only 3 pieces, their geometry at the time of the explosion must be coplanar. In fact, I've had an amusing time considering how the problem as stated constrains the initial plane. It does constrain it, but not quite enough. I think, however, that one simple change can remove the ambiguity. Simply specify the speeds (magnitudes of velocities) of Pieces #1 and #2 when they hit the ground. I'm pretty happy with this solution, since it can easily be incorporated into the problem statement something like this: from the depth and characteristics of the craters formed by Pieces #1 and #2, their speeds when they hit were V1 and V2. Am I correct here? Doesn't this change completely specify the initial plane of the pieces? [This message was edited by fizzymagic on June 28, 2002 at 10:10 AM.] Quote Link to comment
+georgeandmary Posted June 28, 2002 Author Share Posted June 28, 2002 Now it's someone else turn to improve on my first attempt. I just wanted a simple little problem so more people could solve it. I don't want to go three dimensional just yet, maybe the next one. Initially the cache started out as a collision but I decided against it. george Remember: Half the people you meet are below average. Quote Link to comment
+jthorson Posted June 28, 2002 Share Posted June 28, 2002 We've got one of these in Wisconsin, very close to Chicago. No plane crash but similar game. http://www.geocaching.com/seek/cache_details.asp?ID=23267 Quote Link to comment
jfitzpat Posted June 28, 2002 Share Posted June 28, 2002 quote:Originally posted by fizzymagic: Since there are only 3 pieces, their geometry at the time of the explosion _must_ be coplanar. Really? It would seem that the forces have to add up, but the actual vectors would depend on angle to the original combustion point. I mentioned crumple zones because of a mental image. The rocket is blown in half, one half tumbles in one direction (because the end nearest the explosion...) then, weakened, seperates again. But, again, I think it is great as it is. And, I hope G&M does a follow-up. -jjf Quote Link to comment
chloew Posted June 28, 2002 Share Posted June 28, 2002 I assume the solution does not take into account that the surface of the earth is curved. Then again given that the straight line flight of a plane is actually a very low near-circular orbit and the gravitational force is perpendicular, it may cancel out the curvature, approximating a flat planar surface. I may have to write a complete simulation of the problem including all 3-D physics and see what initial velocities of the parts come up with the solution. Of course this will assume that the earth is a perfect sphere so that I can more easily map the xyz coordinates to the LatLong coordinates. BTW, is there a defined standard for how many meters or feet in a minute of latitude? I have found a number of different answers based on different definitions, ranging from ~1851.5-~1853.5 meters per minute. ------------ About the planar initial situation, in general any three objects are coplanar in position and their velocity vectors will also be coplanar due to conservation of momentum. (It's the same as assuming you start with an object at rest and have it explode from an internal source. If you look at any two velocity vectors originating from the same point, the third vector originating from the same point will have to in the same plane as the other two to have the sum of the three be zero.) Therefore for this simplified problem we assume that at the time of the explosion the three velocity vectors are coplanar and parallel to the surface of the earth. Quote Link to comment
chloew Posted June 28, 2002 Share Posted June 28, 2002 G&M, I you have not already noticed, I really like this cache and math/physics problems in general. Excellent work on your part, and keep 'em coming. BTW, I left out a factor of two in my previous calculations which threw off my answer. I'll send my updated numbers soon. Thanks to one of your previous posts I noticed that the numbers come out much nicer when done in metric. I used imperial in my initial calculations. Quote Link to comment
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