tag:blogger.com,1999:blog-4912713243046142041.post1625761853129836469..comments2024-03-28T10:42:53.231-05:00Comments on TYWKIWDBI ("Tai-Wiki-Widbee"): What is the distance between these poles?Minnesotastanhttp://www.blogger.com/profile/01382888179579245181noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-4912713243046142041.post-49876195462943294652014-02-01T08:36:55.487-06:002014-02-01T08:36:55.487-06:00What others have said. Using similar triangles, i...What others have said. Using similar triangles, it's easy enough to figure out the 3:2 ratio between B:A and 15:10 and the parts of L and R. Couldn't progress much from there, however, until I tried using Cartesian equations. Any line is definable by using y = mx + b, where m is the slope and b is the y-intercept. Using that, and setting the origin at the bottom left edge, we get<br /><br />f(x) = 15/(A+B) + 0 and<br /><br />g(x) = -10/(A+B) + 10<br /><br />We also know that f(A) = 6 and g(A) = 6<br /><br />From f(A) we get 6A + 6B = 15A and<br />from g(A) we get 4A + 4B = 10A<br /><br />which gives us no new information. The implication is that we don't have enough information<br />in the problem. So I tried A = 2 and B = 3 (keeping the 3:2 ratio), and f(2) = g(2) = 6.<br />Then I tried A = 4 and B = 6, and still f(4) = g(4) = 6. The intersection point will always be 6 high.<br /><br />Weird.<br /><br />Lurker111<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-66016071466931213082014-01-29T06:45:49.058-06:002014-01-29T06:45:49.058-06:00Also, if instead of using 30 as the denominator yo...Also, if instead of using 30 as the denominator you use 150, you get<br /><br />10XC/150 + 15XC/150 = C or (10 + 15)XC / (10*15)= C<br /><br />Divide both sides by XC to get<br /><br />(10+15) / (10*15) = 1/X<br /><br />Now flip numerator and denominator to get the Futilityn Closet's equation:<br /><br />(10*15)/(10+15) = 6<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-4281107978774207862014-01-28T22:27:14.786-06:002014-01-28T22:27:14.786-06:00Thank you! Now I can understand.
I've inco...Thank you! Now I can understand. <br /><br />I've incorporated your dropbox diagram into the body of the post. Minnesotastanhttps://www.blogger.com/profile/01382888179579245181noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-57260093600679921772014-01-28T20:28:01.561-06:002014-01-28T20:28:01.561-06:00I can try, using (https://www.dropbox.com/s/womlsy...I can try, using (https://www.dropbox.com/s/womlsye6da89t02/pole_position.jpg) as a reference.<br /><br />For starters, the slope of a line is figured as rise / run. So the slope of line L is 10/C, and the slope of line R is 15/C.<br /><br />To get to X, we can see that the rise of line L over distance B = rise of line R over distance A.<br /><br />The run of a line over a certain distance is the rise / slope, so for line L over distance B, it's X / (10/C). Since (X/1) / (10/C) can be said (X/1) * (C/10), we get to the run for line L over distance B, or simply distance B = XC/10.<br /><br />Going through the same exercise with the distance A and slope of line R (15/C), we get that A = XC/15.<br /><br />A + B = C, so by swapping out terms, we get XC/15 + XC/10 = C. To further simplify, let's make those denominators common. The least common multiple of 10 and 15 is 30, so by multiplying the first term by 2/2 and the second by 3/3, we get 2XC/30 + 3XC/30 = C. Add 'em up to get 5XC/30 = C.<br /><br />Factor the left side (divide top and bottom by 5) to arrive at XC/6 = C. Looks like we can divide both sides by C to get X/6 = 1. We want to see the X stand alone, so multiply both sides by 6, finally getting X = 6.<br /><br />Hope that helps!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-826877027513680122014-01-28T14:51:35.274-06:002014-01-28T14:51:35.274-06:00Why? Because... math, sure. But can you ELI5?Why? Because... math, sure. But can you ELI5?Minnesotastanhttps://www.blogger.com/profile/01382888179579245181noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-84004762406775286582014-01-28T13:33:41.241-06:002014-01-28T13:33:41.241-06:00Those lines will always intersect at 6' up unl...Those lines will always intersect at 6' up unless far enough apart for the curvature of the surface of the world (or simple topography) to matterJDJarvishttps://www.blogger.com/profile/07691101939920824546noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-34602995384536080582014-01-28T07:00:01.332-06:002014-01-28T07:00:01.332-06:00Are you figuring the tangents of equivalent angles...Are you figuring the tangents of equivalent angles? If so, I don't think that the angles you're equating are equal. I'd say:<br />4/x1 = 6/x2 = 10/x<br />9/x2 = 6/x1 = 15/x<br />which is also a way of showing the slopes of the two diagonals.<br /><br />I think JPS above is right; the intersection of the lines, regardless of x1 + x2, will always be at 6 high, just as x1:x2 will always be 2:3. Try figuring x1 and x2 for total x of 10 and 100 separately and see what that gives you.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-81239866600642224082014-01-28T01:25:03.262-06:002014-01-28T01:25:03.262-06:00This comment has been removed by the author.Ina Hillevihttps://www.blogger.com/profile/10073567350228611152noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-6537675302828565902014-01-27T20:39:07.599-06:002014-01-27T20:39:07.599-06:00The Futility Closet home page is also inaccessible...The Futility Closet home page is also inaccessible tonight. Someone must have locked the door.Minnesotastanhttps://www.blogger.com/profile/01382888179579245181noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-63236442246293681342014-01-27T20:38:07.578-06:002014-01-27T20:38:07.578-06:00Can you explain why? True, 15x10/15+10 = 6, but h...Can you explain why? True, 15x10/15+10 = 6, but how does that apply?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-1309076156835737402014-01-27T18:59:50.498-06:002014-01-27T18:59:50.498-06:00Stared at this for 10 minutes. Stared some more -...Stared at this for 10 minutes. Stared some more - math, obviously, wasn't my subject. Finally gave up and clicked the link,and what do I get? "502 Bad Gateway"<br /><br />You're a cruel man, Stan. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-6548886775691682752014-01-27T17:16:43.203-06:002014-01-27T17:16:43.203-06:00Can't tell. No matter what the spacing betwee...Can't tell. No matter what the spacing between the poles, the intersection of the top-to-bottom lines will always be 6 feet.JPSnoreply@blogger.com