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# Maths Question.

 Posts: 983 Add as FriendChallenge to a DebateSend a Message 12/11/2010 3:00:47 AMPosted: 7 years agoHow many golf balls does one need to go around the equator while placing one golf ball besides another to make a long line and ultimately a circle? Lets assume that these golf balls are unmovable and they can float. This is just am estimation. Personally I have solved only half of this... Yeah I bet some of you find this ridiculously easy."When he awoke in a tomb three days later he would actually have believed that he rose from the dead" FREEDO about the resurrection of Jesus Christ
 Posts: 16,992 Add as FriendChallenge to a DebateSend a Message 12/11/2010 4:55:01 AMPosted: 7 years agoI looked up, and the diameter of a golf ball is 42.67mm.42.67mm to km = 0.00004267kmThe length of equator: 40,075kmYou divide 40,075 with 0.00004267 =approximately 939,184,439 golf balls
 Posts: 11,793 Add as FriendChallenge to a DebateSend a Message 12/11/2010 7:06:56 AMPosted: 7 years agohow can something be unmoveable and still able to float?signature
 Posts: 983 Add as FriendChallenge to a DebateSend a Message 12/11/2010 10:11:31 AMPosted: 7 years agoAt 12/11/2010 7:06:56 AM, badger wrote:how can something be unmoveable and still able to float?When I said unmovable I meant that once the golf ball is placed you can't move it, whatever on ground or water."When he awoke in a tomb three days later he would actually have believed that he rose from the dead" FREEDO about the resurrection of Jesus Christ
 Posts: 989 Add as FriendChallenge to a DebateSend a Message 1/13/2011 5:36:24 PMPosted: 7 years agoAt 12/11/2010 4:55:01 AM, Mirza wrote:I looked up, and the diameter of a golf ball is 42.67mm.42.67mm to km = 0.00004267kmThe length of equator: 40,075kmYou divide 40,075 with 0.00004267 =approximately 939,184,439 golf ballsI agree with Mirza's calculation (I found the same values for the golf ball and Earth's equatorial circumference and I assume his division was correct). If the significance of the digits matters to you than you should change that to 939,200,000 golf balls. The remaining digits cannot be known precisely because the size of the golf ball and the planet are not given exactly.
 Posts: 4,194 Add as FriendChallenge to a DebateSend a Message 1/13/2011 6:33:55 PMPosted: 7 years agoAt 1/13/2011 5:36:24 PM, Grape wrote:At 12/11/2010 4:55:01 AM, Mirza wrote:I looked up, and the diameter of a golf ball is 42.67mm.42.67mm to km = 0.00004267kmThe length of equator: 40,075kmYou divide 40,075 with 0.00004267 =approximately 939,184,439 golf ballsI agree with Mirza's calculation (I found the same values for the golf ball and Earth's equatorial circumference and I assume his division was correct). If the significance of the digits matters to you than you should change that to 939,200,000 golf balls. The remaining digits cannot be known precisely because the size of the golf ball and the planet are not given exactly.Incorrect. You're assuming that the diameter of each golf ball will be laid end to end. If you could imagine the line circumscribed around the earth at the equator laid out flat on a plane, Mirza's calculation would be correct, however, because the earth is spherical, it'll be different. Try to visualize 1 ball with 3 balls surrounding it. The balls would have to be huge, right? The sum of their diameters would massively exceed the the circumference of the ball in the middle.Where 'r' is radius of the earth and 'g' is the diameter of a golf ball:180/Tan[g/(r+g)]At least I THINK that's what it is. Someone good at math will probably have to correct me.Omar comin' yo https://www.youtube.com...
 Posts: 746 Add as FriendChallenge to a DebateSend a Message 1/13/2011 7:07:48 PMPosted: 7 years agoAssuming the earth is spherical (which it actually isn't, but its close enough for our purposes), imagine looking straight down on the earth from directly above it. You would essentially see a 2-dimensional circle with a circumference equal to that of the equator.Mirza was on the right track, but as J.Kenyon pointed out, his answer is valid only for the case were the equator is laid out in a straight line. Since this isn't actually the case, you have to account for the extra radius added on to the equator by each golf ball, calculate the new circumference and then do the division.The equatorial radius of the earth is 6378.1370 km http://en.wikipedia.org...The radius of a golf ball is 21.335 mm = 0.000021335 kmSo the new radius of the earth at the equator with the golf balls laid across it is 6378.137021335 kmNow, the circumference of the circle through the center of each golf ball is equal to 2*PI*R (where R is the new radius we calculated): C = 40075.01681963024468 kmTherefore the # of golf balls is C/(diameter of each ball) = 939,184,832
 Posts: 14,629 Add as FriendChallenge to a DebateSend a Message 1/13/2011 7:32:32 PMPosted: 7 years agoAt 1/13/2011 7:07:48 PM, MikeLoviN wrote:Assuming the earth is spherical (which it actually isn't, but its close enough for our purposes), imagine looking straight down on the earth from directly above it. You would essentially see a 2-dimensional circle with a circumference equal to that of the equator.Mirza was on the right track, but as J.Kenyon pointed out, his answer is valid only for the case were the equator is laid out in a straight line. Since this isn't actually the case, you have to account for the extra radius added on to the equator by each golf ball, calculate the new circumference and then do the division.The equatorial radius of the earth is 6378.1370 km http://en.wikipedia.org...The radius of a golf ball is 21.335 mm = 0.000021335 kmSo the new radius of the earth at the equator with the golf balls laid across it is 6378.137021335 kmNow, the circumference of the circle through the center of each golf ball is equal to 2*PI*R (where R is the new radius we calculated): C = 40075.01681963024468 kmTherefore the # of golf balls is C/(diameter of each ball) = 939,184,832So it would be about 400 more balls?Without Royal there is a hole inside of me, I have no choice but to leave
 Posts: 3,906 Add as FriendChallenge to a DebateSend a Message 1/13/2011 8:29:10 PMPosted: 7 years agoHow many golf balls would it take to equal the mass of the moon?Minister Of Trolling : At 12/6/2011 2:21:41 PM, badger wrote: : ugly people should beat beautiful people ugly. simple! you'd be killing two birds with the one stone... women like violent men and you're making yourself more attractive, relatively. i met a blonde dude who was prettier than me not so long ago. he's not so pretty now! ha! : : ...and well, he wasn't really prettier than me. he just had nice hair.
 Posts: 4,113 Add as FriendChallenge to a DebateSend a Message 1/13/2011 9:29:41 PMPosted: 7 years agoAt 1/13/2011 7:07:48 PM, MikeLoviN wrote:Assuming the earth is spherical (which it actually isn't, but its close enough for our purposes), imagine looking straight down on the earth from directly above it. You would essentially see a 2-dimensional circle with a circumference equal to that of the equator.Mirza was on the right track, but as J.Kenyon pointed out, his answer is valid only for the case were the equator is laid out in a straight line. Since this isn't actually the case, you have to account for the extra radius added on to the equator by each golf ball, calculate the new circumference and then do the division.The equatorial radius of the earth is 6378.1370 km http://en.wikipedia.org...The radius of a golf ball is 21.335 mm = 0.000021335 kmSo the new radius of the earth at the equator with the golf balls laid across it is 6378.137021335 kmNow, the circumference of the circle through the center of each golf ball is equal to 2*PI*R (where R is the new radius we calculated): C = 40075.01681963024468 kmTherefore the # of golf balls is C/(diameter of each ball) = 939,184,832which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...evidently i only come to ddo to avoid doing homework...
 Posts: 3,906 Add as FriendChallenge to a DebateSend a Message 1/13/2011 9:42:35 PMPosted: 7 years agoAt 1/13/2011 8:29:10 PM, SuperRobotWars wrote:How many golf balls would it take to equal the mass of the moon?A golf balls mass is around 45.93 grams and the moons mass is around 7.36 × 1022 kilograms so . . . it is about 163 769.214 golf balls . . . [?]Minister Of Trolling : At 12/6/2011 2:21:41 PM, badger wrote: : ugly people should beat beautiful people ugly. simple! you'd be killing two birds with the one stone... women like violent men and you're making yourself more attractive, relatively. i met a blonde dude who was prettier than me not so long ago. he's not so pretty now! ha! : : ...and well, he wasn't really prettier than me. he just had nice hair.
 Posts: 4,194 Add as FriendChallenge to a DebateSend a Message 1/13/2011 10:14:34 PMPosted: 7 years agoAt 1/13/2011 10:05:13 PM, MikeLoviN wrote:At 1/13/2011 9:29:41 PM, belle wrote:which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...I was assuming it was the process and not the final answer that matters. If the scale difference wasn't so drastic, this would have a very noticeable affect. There's a reason you'll never hear a competent engineer or mathematician say "well that looks about right, lets go with it"...No, it's not that the figures don't matter, they're not significant in the sense that they are meaningless; they fall within the margin of error given the imprecise measurements. http://en.wikipedia.org... That's why I only posted an equation: 180/[tan/(r+g)] instead of a solution.Btw, belle, was my equation correct?Omar comin' yo https://www.youtube.com...
 Posts: 4,113 Add as FriendChallenge to a DebateSend a Message 1/13/2011 10:20:46 PMPosted: 7 years agonot sure where you got the tangent from... i think you're thinking about it too hard.it would just be PI(g+e)/g where g is the radius of the golf ball and e is the radius of the earthevidently i only come to ddo to avoid doing homework...
 Posts: 11,204 Add as FriendChallenge to a DebateSend a Message 1/13/2011 10:50:50 PMPosted: 7 years agoOh boy math. We need more mathematical questions to add to the geekiness of DDO. Now J.Kenyon is correct, that you have to account for the curvature, if you want to do this calculation. However the curvature would be so small it would be insignificant since the earth's radius is so large compared to the golf balls. If we were to look at the world outside, it looks roughly flat. The circumfrence is so large, that the curvature is very small for thousands of meter.The fact that that the earth is not a perfect circle but has ridges (ex. mountains, holes) would have a greater effect then the curvature of earth.Open borders debate: http://www.debate.org...
 Posts: 746 Add as FriendChallenge to a DebateSend a Message 1/14/2011 12:22:00 AMPosted: 7 years agoAt 1/13/2011 10:14:34 PM, J.Kenyon wrote:At 1/13/2011 10:05:13 PM, MikeLoviN wrote:At 1/13/2011 9:29:41 PM, belle wrote:which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...I was assuming it was the process and not the final answer that matters. If the scale difference wasn't so drastic, this would have a very noticeable affect. There's a reason you'll never hear a competent engineer or mathematician say "well that looks about right, lets go with it"...No, it's not that the figures don't matter, they're not significant in the sense that they are meaningless; they fall within the margin of error given the imprecise measurements. http://en.wikipedia.org... That's why I only posted an equation: 180/[tan/(r+g)] instead of a solution.Btw, belle, was my equation correct?Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).BTW, I understand the concept of sig figs. I purposely tossed the extra numbers in there to give as much a difference as possible in the final answer in an attempt to demonstrate that accounting for the curvature (normally) matters. It just doesn't seem that way here because of the scale difference. If we were talking about more practical scenarios, you wouldn't be able to get away with that straight line estimation, at least not outside of high school.
 Posts: 19,305 Add as FriendChallenge to a DebateSend a Message 1/14/2011 12:35:28 AMPosted: 7 years agoHow can there be a plural of math?It came to be at its height. It was commanded to command. It was a capital before its first stone was laid. It was a monument to the spirit of man.
 Posts: 4,194 Add as FriendChallenge to a DebateSend a Message 1/14/2011 12:38:10 AMPosted: 7 years agoAt 1/14/2011 12:22:00 AM, MikeLoviN wrote:At 1/13/2011 10:14:34 PM, J.Kenyon wrote:At 1/13/2011 10:05:13 PM, MikeLoviN wrote:At 1/13/2011 9:29:41 PM, belle wrote:which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...I was assuming it was the process and not the final answer that matters. If the scale difference wasn't so drastic, this would have a very noticeable affect. There's a reason you'll never hear a competent engineer or mathematician say "well that looks about right, lets go with it"...No, it's not that the figures don't matter, they're not significant in the sense that they are meaningless; they fall within the margin of error given the imprecise measurements. http://en.wikipedia.org... That's why I only posted an equation: 180/[tan/(r+g)] instead of a solution.Btw, belle, was my equation correct?Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).I used trigonometry; my equation gave me roughly the same answer yours did. It's kind of difficult to explain without a diagram, but it works. Fairly ingenious if I do say so myself :POmar comin' yo https://www.youtube.com...
 Posts: 4,194 Add as FriendChallenge to a DebateSend a Message 1/14/2011 12:46:22 AMPosted: 7 years agoAt 1/14/2011 12:38:10 AM, J.Kenyon wrote:At 1/14/2011 12:22:00 AM, MikeLoviN wrote:Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).I used trigonometry; my equation gave me roughly the same answer yours did. It's kind of difficult to explain without a diagram, but it works. Fairly ingenious if I do say so myself :PBasically, imagine a triangle circumscribed around a circle, then a square, then a pentagon, then a hexagon, then an octagon, etc.Hint: what's the interior angle of an octagon?Omar comin' yo https://www.youtube.com...
 Posts: 10,806 Add as FriendChallenge to a DebateSend a Message 1/14/2011 2:08:48 AMPosted: 7 years agoAt 12/11/2010 3:00:47 AM, vardas0antras wrote:How many golf balls does one need to go around the equator while placing one golf ball besides another to make a long line and ultimately a circle? Lets assume that these golf balls are unmovable and they can float. This is just am estimation. Personally I have solved only half of this... Yeah I bet some of you find this ridiculously easy.At what height are these golf balls floating above the earth, are they all at uniform height?Otherwise it's very simple to work out.I am voting for Innomen because of his intelligence, common sense, humility and the fact that Juggle appears to listen to him. Any other Presidential style would have a large sub-section of the site up in arms. If I was President I would destroy the site though elitism, others would let it run riot. Innomen represents a middle way that works, neither draconian nor anarchic and that is the only way things can work. Plus he does it all without ego trips.
 Posts: 4,597 Add as FriendChallenge to a DebateSend a Message 1/14/2011 2:16:35 AMPosted: 7 years agoAre you supposed to factor in the geography of the land at the equator? I doubt it but its something to consider.When large numbers of otherwise-law abiding people break specific laws en masse, it's usually a fault that lies with the law. - Unknown
 Posts: 746 Add as FriendChallenge to a DebateSend a Message 1/14/2011 10:56:44 AMPosted: 7 years agoAt 1/14/2011 12:46:22 AM, J.Kenyon wrote:At 1/14/2011 12:38:10 AM, J.Kenyon wrote:At 1/14/2011 12:22:00 AM, MikeLoviN wrote:Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).I used trigonometry; my equation gave me roughly the same answer yours did. It's kind of difficult to explain without a diagram, but it works. Fairly ingenious if I do say so myself :PBasically, imagine a triangle circumscribed around a circle, then a square, then a pentagon, then a hexagon, then an octagon, etc.So basically you approximated the circumference as an infinite sum of straight lines / tangents?Hint: what's the interior angle of an octagon?135 degreesI do understand you're method though. Interesting way of doing it... though you are assuming that the distance from the center of one ball to the point of contact with an adjacent ball forms a right angle with the (r+g) radius (ie. it's tangent to the circle). Since the balls are so small comparatively, it works perfectly fine here and you do infact get the right answer. But this is still technically an approximation and would not be that accurate if the size difference wasn't so huge (that right triangle you used wouldn't actually be a right triangle).BTW, You're equation is actually 180/[arctan{g/(r+g)}] ...correct?Good job though, I'll give credit where credit is due... It would have taken me a while before I even considered using trig to do this, but that's because I've been conditioned to always find the easy way out.