CraigD Posted June 15, 2007 Author Report Posted June 15, 2007 I forgot the minus sign. Let's try it again... Ball mass=1.7x=8.5v=(-1.5,3)Very interesting – no ball contact with the 1 pin (nor, unfortunately between anything and the 1 pin). Left the 1, 8, and 10. c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 f 2 3 4 5 6 7 8 9 a e Sim T:1.220494506428849876 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.520657330500272825 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.83928483584012753 c d 1 2 3 f 4 5 7 8 9 6 a e Sim T:2.110007673355126901 c d 1 2 3 f 4 7 8 5 9 a e Sim T:2.331230756545607822 c d 2 1 3 f 4 7 8 5 9 a e Sim T:2.462843202122834408 c d 2 1 3 4 f 8 9 a 7 e Sim T:2.799182380347328782 c d 2 1 3 4 f 8 9 a e Sim T:2.935230846449298945 c d 2 1 3 4 f 8 9 a e Sim T:3.160594880905789525 c d 2 3 1 f 8 a 9 e Sim T:4.009338243279255224 c d 2 3 1 f 8 a e Sim T:4.533009777017878759 c d 2 1 f 8 a e Sim T:5.828071327319384421 c d 1 8 f a e Sim T:7.677107475593590691 c d 1 8 a e Sim T:10.64620043929137057 Quote
freeztar Posted June 15, 2007 Report Posted June 15, 2007 OT, I was daydreaming and just thought of this type of game, but with billiards! It would be great fun to have a game in progress where whoever,whenever, can come along and take a shot. The winner would be the one who hits the most balls in the pockets. Of course, I have no idea how difficult this would be to simulate, but it doesn't seem like that far of a stretch from the bowling. Just an idea. And with my favorite ball in hand (I call it "Force") I step up to the line, throw my arm back and then forward, and release... Ball mass=1.7x=8.5v=(-1.25,3) Quote
CraigD Posted June 16, 2007 Author Report Posted June 16, 2007 OT, I was daydreaming and just thought of this type of game, but with billiards! It would be great fun to have a game in progress where whoever,whenever, can come along and take a shot. The winner would be the one who hits the most balls in the pockets. Of course, I have no idea how difficult this would be to simulate, but it doesn't seem like that far of a stretch from the bowling. Just an idea. :shade:That shouldn’t be too hard. It would require a couple a new “bumper” body type, an easy addition. The trick would be adding friction to the “ball” body type. An essential element of a pool game is that the balls eventually stop moving, which, of course, will never happen in a frictionless simulation. A simple uniform acceleration in the opposite direction of the ball’s velocity with a “stop” event at zero velocity should be enough for a reasonably realistic-looking simulation… Sounds like fun – I’ll see if I can do it tonight, or later this weekend And with my favorite ball in hand (I call it "Force") I step up to the line, throw my arm back and then forward, and release... Ball mass=1.7x=8.5v=(-1.25,3)Good adjustment – it took out the 1 pin. Left the 4, 7, 8, and 10 c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.4369955981006127 c d 1 3 2 f 4 5 6 7 8 9 a e Sim T:1.454533916821456982 c d 1 3 2 4 5 f 7 8 9 6 a e Sim T:1.960168411703652255 c d 1 3 2 4 5 f 7 8 a 9 e Sim T:2.201176293766480755 c d 1 3 2 4 5 f 7 8 a 9 e Sim T:2.294675071470479173 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.34579719391154908 c d 1 3 2 5 4 f 7 8 a e Sim T:2.380138884856690461 c d 1 3 2 5 4 f 7 8 a e Sim T:3.070475142422318752 c d 1 3 2 5 4 7 8 f a e Sim T:3.37438285889799684 c d 1 3 2 5 4 7 8 a e Sim T:4.631445773150469158 c d 1 2 4 5 7 8 a e Sim T:5.779230158156360029 c d 2 4 7 5 8 a e Sim T:8.900663434423030335 c d 4 7 8 a 5 e Sim T:9.192619205512951801 c d 4 7 8 a e Sim T:10.56154117855066968 Quote
freeztar Posted June 16, 2007 Report Posted June 16, 2007 That shouldn’t be too hard. It would require a couple a new “bumper” body type, an easy addition. The trick would be adding friction to the “ball” body type. An essential element of a pool game is that the balls eventually stop moving, which, of course, will never happen in a frictionless simulation. A simple uniform acceleration in the opposite direction of the ball’s velocity with a “stop” event at zero velocity should be enough for a reasonably realistic-looking simulation… Sounds like fun – I’ll see if I can do it tonight, or later this weekend :) Somehow I knew you'd be into the idea. ;)I have some other ideas for it as well (friction is a must of course). Since numbered balls are not needed (assuming it's just a game of try to hit a ball in), perhaps the symbology can be changed a bit in the ASCII? Perhaps an "x" for the cue ball and an "o" for all the other balls. It would be easier to estimate angles, imo. It might also be easier to create a table with only four pockets (one in each corner). Cheers for your efforts! :) Good adjustment – it took out the 1 pin. Left the 4, 7, 8, and 10 Well my goal was accomplished because I wanted that 1 pin. Unfortunately, I'm still not getting it quite right. I think my ball mass is too low so it's not having much movement after a few encounters with "pins" (at least from what I can judge by the snapshot positions). So this time: ball mass=1.8x=8.5v=(-1.25,3) Quote
CraigD Posted June 16, 2007 Author Report Posted June 16, 2007 Somehow I knew you'd be into the idea. :)I have some other ideas for it as well (friction is a must of course). Since numbered balls are not needed (assuming it's just a game of try to hit a ball in), perhaps the symbology can be changed a bit in the ASCII? Perhaps an "x" for the cue ball and an "o" for all the other balls.It’s very easy to add different interfaces to the core simulator code, so minor changes to the ASCII are no problem. With a bit of html ingenuity, it should be possible to render some simple graphics. One of my key ambitions is to write a generic GIF or similar graphics format renderer in ANSI-standard MUMPS – I’ve been getting “rumors from the top” that I may soon get a chance to do this at work :)Well my goal was accomplished because I wanted that 1 pin. Unfortunately, I'm still not getting it quite right. I think my ball mass is too low so it's not having much movement after a few encounters with "pins" (at least from what I can judge by the snapshot positions). So this time: ball mass=1.8x=8.5v=(-1.25,3)The slight ball mass increase was significant. Only the 10 pin left – and the 6 came close to getting the 10 after its collision with the ball. c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.4369955981006127 c d 1 3 2 f 4 5 6 7 8 9 a e Sim T:1.45533533558412979 c d 1 3 2 4 5 f 7 8 9 6 a e Sim T:1.951896949917645616 c d 1 3 2 4 5 f 7 8 a 9 e Sim T:2.185452411139611041 c d 1 3 2 4 5 f 7 8 a 9 e Sim T:2.316420653730639343 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.322727638320321595 c d 1 3 2 5 4 f 7 8 a e Sim T:2.359013726995956065 c d 1 3 2 5 4 7 8 f a e Sim T:3.083658749841490086 c d 1 3 2 5 4 7 8 f a e Sim T:3.334724309435983317 c d 1 3 2 5 4 7 8 f a e Sim T:3.61339115351040725 c d 1 3 2 5 4 8 7 a e Sim T:4.324471978605371914 c d 1 2 4 5 8 7 a e Sim T:6.130134771817362993 c d 1 2 8 5 7 a e Sim T:6.776911083059362577 c d 1 4 8 5 7 a e Sim T:7.529689810447353354 c d 4 8 7 5 a e Sim T:8.891296845722150753 c d 4 8 7 a e Sim T:10.54110253531142023 c d 4 8 7 a e Sim T:10.75490125692885145 c d 4 8 a e Sim T:17.00727985720945931 c d 8 a e Sim T:32.2849296346853025 c d a e Sim T:63.71596461652395272 Quote
freeztar Posted June 16, 2007 Report Posted June 16, 2007 That was close! Now it is a matter of determining whether small changes will be able to hit the ten, or whether it will offset the other interactions in a negative manner. I'm hoping that an increase in vertical velocity will create a slightly different angle (and of course speed) resulting in a guttered 10 ("a") with no significant differences from last throw expressed throughout the rest of the pins' fates. ball mass=1.8x=8.5v=(-1.25,3.1) Quote
CraigD Posted June 16, 2007 Author Report Posted June 16, 2007 ball mass=1.8x=8.5v=(-1.25,3.1)The ball hit the 6 before the 3, but the 6 again just missed the 10. Left the 8 and 10. c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.382420723513313731 c d 1 3 2 f 4 5 6 7 8 9 a e Sim T:1.451156207415546746 c d 1 3 2 4 5 f 7 8 9 a 6 e Sim T:1.974323021892850737 c d 1 3 2 4 5 f 7 8 9 a e Sim T:2.016980570221942208 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.238004309934766647 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.351017001507023179 c d 1 3 2 5 4 f 7 8 a e Sim T:2.447970225807951226 c d 1 2 3 5 4 7 8 f a e Sim T:3.158469896573131361 c d 1 2 3 5 4 7 8 f a e Sim T:3.254586589817747876 c d 1 3 2 5 4 7 8 a f e Sim T:4.138297877879919264 c d 1 3 2 5 4 7 8 a e Sim T:4.250500669640843155 c d 1 3 2 5 4 7 8 a e Sim T:5.450141905671206687 c d 1 3 5 4 7 8 a e Sim T:6.098209174294177906 c 1 d 3 5 4 8 a e Sim T:7.272800897774767571 c d 3 5 4 8 a e Sim T:7.574588574503165316 c d 3 4 8 a e Sim T:14.66193799137168496 c 3 d 8 a e Sim T:16.16738866365658776 c d 8 a e Sim T:16.75527393239548141 Quote
CraigD Posted June 16, 2007 Author Report Posted June 16, 2007 ball mass=1.8x=8.5v=(-1.25,3.05)Closer. The 6 was within about .09 of colliding with the 10. Left the 4, 7, 8, and 10: c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.408815120008463065 c d 1 3 2 f 4 5 6 7 8 9 a e Sim T:1.452187156519894873 c d 1 3 2 4 5 f 7 8 9 a 6 e Sim T:1.961763199009984793 c d 1 3 2 4 5 f 7 8 9 a e Sim T:2.0932546113106836 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.271230516020480831 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.321987739277099583 c d 1 3 2 5 4 f 7 8 a e Sim T:2.399662861283214819 c d 1 3 2 5 4 7 8 f a e Sim T:3.11700776848938123 c d 1 3 25 4 7 8 f a e Sim T:3.211072781747479976 c d 1 3 2 4 5 7 8 a e Sim T:4.323908975210361986 c d 1 3 4 5 7 8 a e Sim T:6.230310749445818001 c d 3 4 7 8 a 5 e Sim T:7.865789120536657338 c d 3 4 7 8 a e Sim T:8.34037373498238144 c d 4 7 8 a e Sim T:9.39795236027726405 Quote
CraigD Posted June 16, 2007 Author Report Posted June 16, 2007 ball mass=1.85x=8.5v=(-1.25,3)Left the 10, by small (about .04) margin: c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.4369955981006127 c d 1 3 2 f 4 5 6 7 8 9 a e Sim T:1.455744190208665279 c d 1 3 2 4 5 f 7 8 9 6 a e Sim T:1.948022485611826304 c d 1 3 2 4 5 f 7 8 a 9 e Sim T:2.178227924527265498 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.31380249633533563 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.328111046088005752 c d 1 3 2 5 4 f 7 8 a e Sim T:2.349151747630366705 c d 1 3 2 5 4 7 8 f a e Sim T:3.098279080004641694 c d 1 3 2 5 4 7 8 f a e Sim T:3.331914154121467859 c d 1 3 2 5 4 7 8 f a e Sim T:3.500973533513430129 c d 1 3 2 5 4 8 7 a e Sim T:4.220506030306238759 c d 1 3 2 4 8 5 7 a e Sim T:5.957956444064893964 c d 1 2 4 5 8 7 a e Sim T:6.322290839601223487 c d 1 4 8 5 7 a e Sim T:7.185503707579171382 c 1 d 4 8 7 5 a e Sim T:8.217234634624083366 c d 4 8 7 5 a e Sim T:8.897132889517921104 c d 4 8 a 7 e Sim T:11.04163734399184514 c d 4 8 a e Sim T:13.32282255261854727 c d 4 a e Sim T:15.5165879938316908 c d a e Sim T:29.29289503343805665 Quote
freeztar Posted June 16, 2007 Report Posted June 16, 2007 Left the 10, by small (about .04) margin That pesky ten!!If anything is going to take it out, then it will be the 6 (in this scenario, hopefully). So I'll increase the angle a slight amount to hopefully cause a 6-10 collision. The question I have now is: Will a 0.01 increase in horizontal velocity create the necessary circumstances for a collision with the ten, needing a 0.04 margin of difference? ball mass=1.85x=8.5v=(-1.26,3) Quote
CraigD Posted June 17, 2007 Author Report Posted June 17, 2007 That pesky ten!!If anything is going to take it out, then it will be the 6 (in this scenario, hopefully). So I'll increase the angle a slight amount to hopefully cause a 6-10 collision. The question I have now is: Will a 0.01 increase in horizontal velocity create the necessary circumstances for a collision with the ten, needing a 0.04 margin of difference? ball mass=1.85x=8.5v=(-1.26,3)Got the 10. Left the 7. This game’s sensitive enough to initial conditions, it might deserve to be call chaotic. I didn’t recall it being quite this tricky, but I didn’t keep track of how many tries it took me, (at seconds per try, rather than the minutes or hours it takes via the thread) and my notes do look like I got into the 3rd place after the decimal before finding a strike. c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.439619459733645983 c d 1 3 2 f 4 5 6 7 8 9 a e Sim T:1.44597032418443439 c d 1 3 2 4 5 f 7 8 9 6 a e Sim T:1.869223519316237221 c d 1 3 2 4 5 f a 7 8 9 6 e Sim T:1.929272516980551514 c d 1 3 2 4 5 f a 7 8 9 e Sim T:2.195217383929961003 c d 1 3 2 4 5 f a 7 8 e Sim T:2.31421061886395671 c d 1 3 2 4 5 f a 7 8 e Sim T:2.315926972155221472 c d 1 3 2 5 4 f a 7 8 e Sim T:2.333961204436647757 c d 1 3 25 4 a 7 8 f e Sim T:3.450963965146518755 c d 1 3 2 5 4 8 a 7 f e Sim T:3.526749626250119982 c d 1 3 2 5 4 8 7 f e Sim T:3.645262288335542338 c d 1 3 2 5 4 8 7 e Sim T:4.223907286691612795 c d 1 2 5 4 8 7 e Sim T:5.613714563224727914 c d 1 2 5 84 7 e Sim T:5.702573769855143931 c d 1 2 4 5 8 7 e Sim T:6.538405345388079797 c 1 d 2 4 5 7 e Sim T:9.27875024506144916 c d 2 4 5 7 e Sim T:10.10131717878659505 c d 4 5 7 e Sim T:11.48935608493591156 c d 5 7 e Sim T:14.11847360907285196 c d 7 e Sim T:24.40032031919833594 Quote
freeztar Posted June 17, 2007 Report Posted June 17, 2007 I'll keep trying until I get it! :)It may take a while though! :lol: ball mass=1.85x=8.5v=(-1.255,3) Quote
CraigD Posted June 17, 2007 Author Report Posted June 17, 2007 ball mass=1.85x=8.5v=(-1.255,3)Back to leaving the 10: c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.431371817717634314 c d 1 3 2 f 4 5 6 7 8 9 a e Sim T:1.482662446105680168 c d 1 3 2 4 5 f 7 8 9 a 6 e Sim T:1.999424229006156221 c d 1 3 2 4 5 f 7 8 9 a e Sim T:2.110662178624148804 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.294790719389043269 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.394377693341718218 c d 1 3 2 5 4 f 7 8 a e Sim T:2.447952440312601677 c d 1 2 3 5 4 7 8 f a e Sim T:3.25978853262991127 c d 1 2 3 5 4 7 8 f a e Sim T:3.265085948342712363 c d 1 3 2 5 4 7 8 a f e Sim T:3.690457470732009173 c d 1 3 2 5 4 8 7 a e Sim T:4.289679379061931475 c d 1 3 2 5 4 8 7 a e Sim T:4.476343500327442449 c d 1 3 2 5 4 8 7 a e Sim T:6.039579380765562966 c d 1 3 5 4 8 7 a e Sim T:6.269203357357582806 c d 3 5 4 8 a 7 e Sim T:7.961011859216656395 c d 3 5 4 8 a e Sim T:8.151714711755685863 c d 3 5 8 a e Sim T:9.50466568508821571 c d 5 8 a e Sim T:10.65149384018028741 c d 8 a e Sim T:32.46134604171524869 c d a e Sim T:187.2558241222434398 Quote
CraigD Posted June 17, 2007 Author Report Posted June 17, 2007 ball mass=1.75x=8.5v=(-1.464,3.5)Left the 7 and 10. c d f 1 2 3 4 5 6 7 8 9 a e Sim T:0 c d 1 2 3 f 4 5 6 7 8 9 a e Sim T:1.232777917332150592 c d 1 3 2 f 4 5 6 7 8 9 a e Sim T:1.243201986126970166 c d 1 3 2 4 5 f 7 8 9 6 a e Sim T:1.668218114553045979 c d 1 3 2 4 5 f 7 8 a 9 e Sim T:1.893490553906617197 c d 1 3 2 4 5 f 7 8 a 9 e Sim T:1.97149728899151607 c d 1 3 2 5 4 f 7 8 a 9 e Sim T:2.005267559231502918 c d 1 3 2 5 4 f 7 8 a e Sim T:2.015510054694335119 c d 1 3 2 5 4 7 8 f a e Sim T:2.966705202304666561 c d 1 3 2 5 4 7 8 f a e Sim T:3.189780822471465187 c d 1 3 2 5 4 8 7 a e Sim T:3.807803554657139853 c d 1 2 3 5 4 8 7 a e Sim T:4.601319962727099262 c d 1 2 5 4 8 7 a e Sim T:4.83058333837877471 c d 2 5 8 4 7 a e Sim T:8.006152865339078732 c 2 d 5 8 7 a e Sim T:8.904289123135568052 c d 5 8 7 a e Sim T:9.36742444016623971 c d 5 7 a e Sim T:16.22880426161805783 c d 7 a e Sim T:186.2288042616180578 Quote
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