tag:blogger.com,1999:blog-14769935.comments2016-12-29T16:07:58.262-08:00Questions Nobody Asked..Oatzyhttp://www.blogger.com/profile/07766533850640317524noreply@blogger.comBlogger56125tag:blogger.com,1999:blog-14769935.post-60824399375836354242016-12-29T14:12:26.474-08:002016-12-29T14:12:26.474-08:00[UPDATE] - actually the code needs to be
#DHH#.$#...[UPDATE] - actually the code needs to be<br /><br /><i>#DHH#.$#Dmm#<6?0$$(round(10*#Dmm#/6))$</i><br /><br />That way, you get the leading zero for for fractions less than 0.1 - e.g. 16:03 -> 16.05, rather than 16.5<br /><br />I've updated the download version as well.Oatzyhttps://www.blogger.com/profile/07766533850640317524noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-20752459912887098432016-12-29T13:59:11.277-08:002016-12-29T13:59:11.277-08:00To do this in Zooper, create a new widget, and go ...To do this in Zooper, create a new widget, and go to<br /><br /><i>Layout -> Add -> Text</i><br /><br />Then select <i>'Edit text manually'</i> and enter<br /><br /><i>#DHH#.$(round(10*#Dmm#/6))$</i><br /><br />I think that's the output you want.<br /><br />Alternatively, you can download <a href="https://dl.dropboxusercontent.com/u/4635169/zooper-templates/fraction-time.zw" rel="nofollow">this one</a> I made earlier - though you'll probably want to play with the formatting to make it look better.<br /><br />Hope this helps.Oatzyhttps://www.blogger.com/profile/07766533850640317524noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-79674661279084421272016-12-29T06:19:24.536-08:002016-12-29T06:19:24.536-08:00I have also been searching for a specific clock fo...I have also been searching for a specific clock for Android. It seems to be a half-breed of several different types.<br />I need a clock that displays the hour (preferably 24 hour style) and the minutes in 100th's of a minute (:15 minutes = .25, :30 = .50, :45 = .75, and :00 = .00).<br />I'm not a developer or programmer, just a PDU (POOR DUMB USER).<br />I would appreciate any assistance you could provide.<br /><br />-rfryeryrfryeryhttps://www.blogger.com/profile/11797174966715093910noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-90240951739791437352016-05-21T05:47:14.018-07:002016-05-21T05:47:14.018-07:00(sorry, bad english)
Well, that's a really goo...(sorry, bad english)<br />Well, that's a really good idea. I was hoping that, at the end, someone would actually give the link to the final game. When someone finally makes it, we will have an interesting tool to introduce general relativity even to kids... this will be awesome! Erichhttps://www.blogger.com/profile/15648423103960265792noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-83899476263964550732016-01-31T18:36:04.680-08:002016-01-31T18:36:04.680-08:00Sorry for some minor grammatical mistake(s) if the...Sorry for some minor grammatical mistake(s) if there is/are one(s). Anyway, I hope you enjoy if not, sorry.<br />nguyen stevenhttps://www.blogger.com/profile/03558212696194321855noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-12530067140357161512016-01-31T18:34:37.053-08:002016-01-31T18:34:37.053-08:00doll kin tem, thokt nid fon ti, ascbe yoi ulule ji...doll kin tem, thokt nid fon ti, ascbe yoi ulule jiku sted hahahahhah <br /><br /> First, we assume that every digits in the number have a score entitled to them, and same number has the same score, and that the number on the the left of the equal sign is the total score of all the digits' score of the number that is on the right side of the equal sign. For instance,: 2 has the score of zero so 2222= 0+0+0+0=0<br /> <br /> Using this method, we can figure out that:<br /> 5555=0 Since same number has same score, the point of 5 is 0/4=0.<br /> 1111=0 Since same number has same score, the point of 1 is 0/4=0.<br /> <br /> Now, here arrives the more challenging part, finding the score of 8.<br /> To begin with, let's find the account for 9 and 3.<br /> Since 9999=4 and same digit has same score, 9 has the point of:<br /> 1 (Check: 1+1+1+1=4).<br /> Since 3333=, 3's score is 0.<br /> <br /> Next, I gonna use the number 8193 (u can use some others numbers too, pal) to discover the point of 8 .It was given that 8193=3 and we found out that 1 has the score of 0<br /> 9 has the score of 1 <br /> 3 has the score of 0<br /> let x present the score of 8<br /> <br /> (I expressed the equation 8193=3 into another one based on the score of the digits)<br /> 8193= x + 0 + 1 + 0= x + 1 = 3<br /> - 1 -1<br /> x = 2<br /> <br /> In conclusion to our objective mentioned, the score of 8 is 2.<br /> <br /> So, the answer to the question 2581 is:<br /> 2581= 0 + 0 + 2 + 0 = 2<br /><br /> Yassss, finished,the answer is 2.<br /><br /> P.S: This is original and it works!<br /><br /><br /> 3213= 0 + 0 + 0 + 0 = 0.<br /> <br /> P.S.S: Bish, still don't believe me? Chek the math urself like yeah imma get some orange juice rn. Anyway,you are awesome, do not doubt that ever!nguyen stevenhttps://www.blogger.com/profile/03558212696194321855noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-5628219399446224792015-12-08T14:33:12.110-08:002015-12-08T14:33:12.110-08:00You know, I didn't think to take the average c...You know, I didn't think to take the average circumference to find the number of turns - that's brilliantly simple (and simply brilliant).<br /><br />I was looking at the formula for the arc length of a <a href="http://mathworld.wolfram.com/ConicalSpiral.html" rel="nofollow">conic spiral</a>. To find the vertical separation between rows of lights, you would have to solve<br /><br />\[ L = \frac{1}{2} H \sqrt{1 +r^2 + \left(\frac{2\pi r H}{\delta h}\right)^2} + \delta h \frac{1+r^2}{4\pi r} \sinh^{-1}\left(\frac{1}{\delta h}\frac{2\pi r H}{\sqrt{1 + r^2}}\right) \]<br /><br />for dh - where L is the length of lights and r is the radius of the tree divided by the height (just to confuse you). I don't know if it's possible to solve analytically, but if you put your numbers in (using <a href="http://www.wolframalpha.com/input/?i=6.1*sqrt%281%2B0.042*%281%2B5876%2Fx%5E2%29%29+%2B+0.4046*x*sinh%5E-1%2815.39%2Fx%29+%3D+700" rel="nofollow">wolfram alpha</a>) you get dh~13.7cm OR 89.05 turns(!)<br /><br />Okay, so I've done a 'proof' that having the strings evenly spaced leads to a uniform light density, but I don't know if Tex will render in comments, so you can see it <a href="http://www.texpaste.com/n/4y3aid8i" rel="nofollow">here</a>.<br /><br />Having said that, I think you're right that the light's will be more sparse at the top of the tree - especially while the circumference is roughly equal to the space between lights on the string, in which case you'll only have one or two lights per row. So maybe you want to bunch them up a little at the top. But once the circumference is much bigger than the light separation, you can just space the rows out evenly.<br /><br />But, also, as I mentioned in my last comment, looking up at the top of the tree from ground level, the lights will appear closer together anyway because of the perspective.<br /><br />Anyway, good luck with your tree! I'm sure it'll turn out great.Oatzyhttps://www.blogger.com/profile/07766533850640317524noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-74277163570505571902015-12-07T17:25:26.825-08:002015-12-07T17:25:26.825-08:00Hi Oatsy!
Thanks for your quick response - actu...Hi Oatsy! <br /><br />Thanks for your quick response - actually we are trying to work this out tonight, since the lights go up tomorrow.<br /><br />Yes the tree is tall, its for the downtown square. It is artificial, so a perfect cone in shape<br /><br />I think our problem is a bit different - i will try to explain. Here are the actual numbers (all in meters, i have learned that the tree is 12.2 meters (40ft).<br /><br />string of lights=700m ... the problem is we don't want to test out how to put the lights on, then take them back off... etc, so the question becomes how much space should be put in between each spiral of light string to make the tree look even<br /><br />Here are the numbers<br /><br />height= 12.2m<br />diameter at bottom 5m<br />radius at bottom 2.5 m<br />hypotenus (outside length 12.5m (calculated)<br /><br />Circumference at bottom = 15.7m<br /><br />since circumference is a linear function, we can average the circumference at each end to determine the average circumference of the cone = 15.7m + 0m = 7.85m<br /><br />Thus 700m of light will go around the cone 89.1 times<br /><br />Easy answer would be: Since the hypotenuse is 12.5 m long, the space between each spiral would be 14 cm (12.5m/89).<br /><br />However I'm not sure it is the right answer. Here is why.<br /><br />the horizontal distance at the top of the cone is very small horizontally, resulting in a steep decent angle for the spiral. As the spiral continues lower, each decent angle is reduced. Thus it seems (i believe) the result is lower density of lights on the top (per surface area), and higher density on the bottom. The resulting impression will be a bright bottom and less bright top...<br /><br />I have some imagination that an integral function may solve the problem, but i have not been able to figure it out... hahaha<br /><br />Does this detail spark any ideas?<br /><br />Really appreciate your help! We are string this tree with a crane tomorrow, doing it once will be enough! hahaha! <br /><br />Stuart Tillhttps://www.blogger.com/profile/06104145982800492378noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-31583486724486051812015-12-07T15:41:06.283-08:002015-12-07T15:41:06.283-08:00How do you practically hang lights on a 30ft tree?...How do you practically hang lights on a 30ft tree? With a big ladder XD<br /><br />Sorry.<br /><br />So wait, do you have an actual 30ft tree you want to hang lights on? I don't know if I can offer much in the way of practical advice. Some thought though...<br /><br />With a tree 30ft tall, perspective becomes important. For example, if you space the lights out perfectly, the rows at the top of the tree will appear closer together than the rows at the bottom, when viewed from below (and vice versa). You can do some trigonometry to figure out how to compensate for that, but the answer depends on how far away and from what height people will be viewing the tree.<br /><br />Okay, practical advice. So as I mention in the blog, the distance between lights on the string is fixed - that's your guide. Presumably you'll be hanging them from top to bottom. Wing it for the first couple of rows while it's still too narrow to really space things out. Then you want to make sure each new light is roughly the same distance from the nearest light(s) on the previous row as the distance between lights on the string. If that makes sense.<br /><br />As a rough guide, if you do that, then the distance between rows will be ~sqrt(3)/2 * S ~ 0.87*S, where S is the distance between light on your string.<br /><br />Of course, that assumes the layout is locally flat (or near enough). If you wanted to be exact you'd have to take into account the curvature / base-to-height ratio of the tree. But working that out would be a nightmare (see the link under 'tinsel' for the length of a conic spiral). It'd be interesting to try and calculate, but from a practical perspective, you'd be better just estimating. <br /><br />I don't know if any of this helps...Oatzyhttps://www.blogger.com/profile/07766533850640317524noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-27789457696948621492015-12-07T10:39:34.270-08:002015-12-07T10:39:34.270-08:00This is agreat discussion - but it still leaves a ...This is agreat discussion - but it still leaves a key question uncovered - on a large tree (30 ft.) this question becomes key<br /><br />How to practically hange lights with the same density - so on a conical surface - what is the distance between each string would give the same density of lights. The answer will yield rows of lights that are parallel - The answer will also start a a small number (e.g. 3in between the rows) and at the bottom be much larger (e.g. 10 cm) Any ideas?<br /><br />Help?Stuart Tillhttps://www.blogger.com/profile/06104145982800492378noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-57645195555052815892015-11-05T15:48:18.959-08:002015-11-05T15:48:18.959-08:00Ah, I see what you're doing now. Neat!
So at ...Ah, I see what you're doing now. Neat!<br /><br />So at the moment, you have a function that slows to (and accelerates from) a stop at 12. If you don't want that stop, one thing you could do is tweak the function so that the rollercoaster car moves at a constant speed through a small region around 12 - say, +/- 45 degrees<br /><br />So for example you could do<br /><br />{drms}<45 and {drms}/4 or {drms} <180 and ({drms}^2)/180 or {drms}<315 and 360-(((360-{drms})^2)/180) or 270+{drms}/4<br /><br />The difference is subtle, but I think it works. <br /><br />Or if that's not smooth enough, try below, replacing 'x' with a bigger angle (between 45 and 180)<br /><br />{drms}<x and x*{drms}/180 or {drms}<180 and ({drms}^2)/180 or {drms}<(360-x) and 360-(((360-{drms})^2)/180) or (360-2*x) + x*{drms}/180<br /><br />Glad I could help!Oatzyhttps://www.blogger.com/profile/07766533850640317524noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-44455223207538305622015-11-05T01:49:03.027-08:002015-11-05T01:49:03.027-08:00Wow, thanks for your quick and detailed reply!
I ...Wow, thanks for your quick and detailed reply!<br /><br />I took your first line and changed it a bit:<br /><br />{drms} <180 and ({drms}^2)/180 or 360-(((360-{drms})^2)/180)<br /><br />This does the job pretty good, many thanks!<br />Here's the example: https://www.youtube.com/watch?v=ssgWt4YYtsc<br />I'm not sure if it's possible to make the motion at the top smoother (so that it doesn't stop completely), but I'm very happy with the result already.<br /><br />Thanks again!Lennerthttps://www.blogger.com/profile/01665562935428836794noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-57798180210534646242015-11-04T16:25:59.310-08:002015-11-04T16:25:59.310-08:00For deceleration, you want to change the function ...For deceleration, you want to change the function to ((360-{drms})^2)/360 when you get to six - so something like:<br /><br />{drms} <180 and ({drms}^2)/360 or ((360-{drms})^2)/360<br /><br />Alternatively, I don't know if this is what you're looking for, but you could try :<br /><br />x: 180*math.sin(math.rad(180*math.cos(math.rad({drms}/2))))<br />y: 180*math.cos(math.rad(180*math.cos(math.rad({drms}/2))))<br /><br />I think this is the behaviour you're describing (it's based on the motion of a pendulum/simple harmonic oscillator). <br /><br />Hope this helps. Oatzyhttps://www.blogger.com/profile/07766533850640317524noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-53411865768329512282015-11-04T02:45:39.699-08:002015-11-04T02:45:39.699-08:00Thanks for the useful examples. I'm trying to ...Thanks for the useful examples. I'm trying to figure out how to implement a rotating rollercoaster which accelerates from 12 to 6 and decelerates from 6 to 12, like going through a loop. I came as far as ({drms}^2)/360 to have it accelerate from the top position, but it obviously doesn't decelerate. Any tips would be greatly appreciated.<br />(I used {drms} for the speed, the coaster doesn't have a practical function in the watch)Lennerthttps://www.blogger.com/profile/01665562935428836794noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-32263983354773381142015-09-10T05:26:18.015-07:002015-09-10T05:26:18.015-07:00Find the missing number,your answer is absolutely ...Find the missing number,your answer is absolutely right,below answers were circled ones given in the problem<br />1st one:<br /> 12+6+5+9=32/10=3.2~ (3)<br />2nd one:<br /> 10+10+18+2=40/10= 4<br />3rd one:<br /> 23+12+3+20=58/10= 5.8~ (6)<br /><br />B is the right answer (i.e) '6'sageni karthikhttps://www.blogger.com/profile/09981317253543126064noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-71677086266335534162015-04-06T22:30:53.336-07:002015-04-06T22:30:53.336-07:00Useful information shared. I am very happy to read...Useful information shared. I am very happy to read this article about <a href="http://technogigs.com/" rel="nofollow">android watch</a>. Thanks for giving us nice info. Fantastic walk through. I appreciate this post.robert yarnhttps://www.blogger.com/profile/18272769035515111751noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-44739075967294616162015-02-23T13:21:33.280-08:002015-02-23T13:21:33.280-08:00LOL, you just murdered that for me. Thank you so m...LOL, you just murdered that for me. Thank you so much. You wouldn't believe the amount of XDA and Google searching I did for such a simpler idea. You explained a lot, and gave me usable example code. So, so grateful.thistimearoundhttps://www.blogger.com/profile/03967597936711332069noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-76686459946130007582015-02-23T12:20:06.503-08:002015-02-23T12:20:06.503-08:00The problem is, when you try to put a new expressi...The problem is, when you try to put a new expression in the OR statement, the '$' closes the previous IF. So the code you've posted will be interpreted as 3 IFs. For example, when the condition is 'Mixed Snow and Sleet', the first statement evaluates to nothing, the second evaluates to 'Snow & Sleet:' and the third evaluates to #W0COND#<br /><br />Unfortunately, there's no way around this. You'll have to do it the long way:<br /><br />$#W0COND#=scattered clouds?S-Clouds$<br />$#W0COND#=mixed snow and sleet?Snow & Sleet$<br />$#W0COND#=scattered thunderstorms || #W0COND#=isolated thunderstorms?Thunderstorms$<br />$#W0COND#!=scattered clouds && #W0COND#!=mixed snow and sleet && #W0COND#!=scattered thunderstorms && #W0COND#!=isolated thunderstorms?#W0COND#$<br /><br />Alternatively, you could try using the condition codes #W0CODE#, explained <a href="http://www.zooper.org/wp/archives/1155" rel="nofollow">here</a>.<br /><br />For example, 'Thunderstorms', 'Scattered Thunderstorms', 'Isolated Thunderstorms', etc. all have a code of 2. So you could do something like:<br /><br />$#W0CODE#=13?S-Clouds$<br />$#W0CODE#=7?Snow & Sleet$<br />$#W0CODE#=2?Thunderstorms$<br />$#W0CODE#!=13 && #W0CODE#!=7 && #W0CODE#!=2?#W0COND#$<br /><br />It's more succinct, but I've had to assume, for example, that 'Snow and Sleet' is equivalent to code 7: 'Snow and Rain'.<br /><br />Hope this helps.Oatzyhttps://www.blogger.com/profile/07766533850640317524noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-1203282348596731452015-02-22T22:57:47.759-08:002015-02-22T22:57:47.759-08:00Stumbled upon this blog after lots of searching. Y...Stumbled upon this blog after lots of searching. You definitely seem to know your way around Zooper. I know it's got limited nested conditionals, but I'm trying to set up a string that condenses the weather into less characters. Example:<br /><br />$$#W0COND#=scattered clouds?S-Clouds:$$#W0COND#=mixed snow and sleet?Snow & Sleet:$$#W0COND#=scattered thunderstorms || isolated thunderstorms?Thunderstorms:#W0COND#$<br /><br />If the weather conditions are either of the thunderstorms variables I am greeted with Thunderstorms. However, if the condition is Mixed Snow and Sleet I'm presented with Snow & Sleet:Mixed Snow and Sleet. I don't understand how it's breaking, or where. I suspect it's too many nested conditions, but I'm not well versed enough to come up with an alternative.<br /><br />If you can help, that would be awesome. Good tutorials! Thanks!thistimearoundhttps://www.blogger.com/profile/03967597936711332069noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-17504062692451268142015-01-30T19:27:54.354-08:002015-01-30T19:27:54.354-08:00Hello there!
I had been searching hours on end f...Hello there! <br /><br />I had been searching <i>hours on end</i> for how to convert normal-time into decimal (metric) time. Finally. A snippet of your blog turns up in the results. <br /><br />It is precisely (and better!) what I have been seeking: a math expression, using ISO 8601. Even better: It's <i>written for Zooper Widget</i>. (This had been exactly my end goal.) <br /><br />Better yet: It's compiled and available to download! The cherry on top: There's more where it came from. <br /><br />Please understand that I am not a mathematician or programmer, but only a curious student. So, I copied your code myself, and it works perfectly. I am so grateful for you having posted this just recently, that I'm spreading the joy of Android, humanity, and Zooper Widget. <br /><br />Thank you kindly<i>!</i> Keep it up<i>!</i><br /><br />-katsumii / GG<br /><br />9:35 ESTGGhttps://www.blogger.com/profile/14215247442018986927noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-47887312504205773142013-11-08T07:29:26.656-08:002013-11-08T07:29:26.656-08:00check out http://www.lotteryaxiom.comcheck out http://www.lotteryaxiom.comChristopher Wonghttps://www.blogger.com/profile/18173353141540349936noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-74196317790604611992013-04-24T01:04:42.051-07:002013-04-24T01:04:42.051-07:00It would look nicer if you could remove notificati...It would look nicer if you could remove notification of my removed comments... why do we need to know? For anybody reading, they say the exact same thing as the one you can read with the only difference that it has one or two fewer typos. Now you are enlightened! :)vapourhttps://www.blogger.com/profile/03734004341470017099noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-20410437683977793682013-04-24T00:56:40.642-07:002013-04-24T00:56:40.642-07:00The post you make here is more important than you ...The post you make here is more important than you realise. I have often considered what I think of as 'the programming problem'. I think there is no better way of approaching the task of programmability using games. Human-derived, invented, games have an intrinsic systematic model-ability about them making it a great field for recreational programming and education. So, you inspired me to have a go because I can encode a S+L games using my limited knowledge of C++ without having to know any graphics! :) Thanks. :)vapourhttps://www.blogger.com/profile/03734004341470017099noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-10300136771212368152013-04-24T00:54:27.125-07:002013-04-24T00:54:27.125-07:00This comment has been removed by the author.vapourhttps://www.blogger.com/profile/03734004341470017099noreply@blogger.comtag:blogger.com,1999:blog-14769935.post-53228382717963111072013-04-24T00:54:19.484-07:002013-04-24T00:54:19.484-07:00This comment has been removed by the author.vapourhttps://www.blogger.com/profile/03734004341470017099noreply@blogger.com