Math Problem ...
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Chris Losinger wrote:
.2 * 4 = .8
Chris Losinger wrote:
.2 * 3 = .6
Last time I checked, these are the correct answers.
yup. he was right up until the last little bit of arithmetic. Cleek | Image Toolkits | Thumbnail maker
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Ed.Poore wrote:
the "complex" formula is the correct one
The solution by Jun Du is also correct. Tried both your solution and his for several different points and I get the same answers from both. Consider if this problem was being programmed on some weird machine where no square root functions or trig functions were available via hardware, then Jun Du's solution would be best.
Paul Conrad wrote:
The solution by Jun Du is also correct.
:wtf: How? Can you provide some proof!
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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Paul Conrad wrote:
The solution by Jun Du is also correct.
:wtf: How? Can you provide some proof!
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
Can't you just accept the fact someone else came up with an easier solution than yours (and yours is correct, too)? You can't even get horizontal and vertical lines correct. Division by zero occurs when you have a vertical line. I knew this BEFORE you were even born. Mr. Losinger even pointed out math errors in your one post.
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Can't you just accept the fact someone else came up with an easier solution than yours (and yours is correct, too)? You can't even get horizontal and vertical lines correct. Division by zero occurs when you have a vertical line. I knew this BEFORE you were even born. Mr. Losinger even pointed out math errors in your one post.
If I've understood the post correctly he wants to get 20% of the line between the two points. Agree? If so then how can you simply take 20% of the width of the "triangle" formed and "20%" of the height to get a hypotenuse that is 20% of the original triangle? If you use Pythagoras it'll show you that if you do that you'll actually get a hypotenuse that's ~28% of the original rather than 20%.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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Can't you just accept the fact someone else came up with an easier solution than yours (and yours is correct, too)? You can't even get horizontal and vertical lines correct. Division by zero occurs when you have a vertical line. I knew this BEFORE you were even born. Mr. Losinger even pointed out math errors in your one post.
Oooooooooopppppppppppssssssssssssssss :doh::doh::doh::doh::doh: :-O:-O:-O:-O:-O Now I know I should go to bed, sorry I've completely ballsed this up havn't I, just got a pencil and paper out to work it out, forgot to substitute for cos and sin :doh: Combination of too much revision and a late night.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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Hey Guys, I have a math issue, I know how you like brain teasers so I thought I would post it here, say I have two points ab & cd, how do I find the point 20% of distance between these two points on the same slope ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
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If I've understood the post correctly he wants to get 20% of the line between the two points. Agree? If so then how can you simply take 20% of the width of the "triangle" formed and "20%" of the height to get a hypotenuse that is 20% of the original triangle? If you use Pythagoras it'll show you that if you do that you'll actually get a hypotenuse that's ~28% of the original rather than 20%.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
Though it does not qualify as an actual mathematical proof, I've tested your approach and Jun Du's approach in Excel with 10-15 different values and I get the same results. Remember that: θ = arctan( y / x ) x = cos( θ ) y = sin( θ ) from Inverse Tangent[^] You wrote: θ = arctan((d - b) / (c - a)) x = a + (0.2*|AB|*cos(θ) y = b + (0.2*|AB|*sin(θ) and based on the link from MathWorld, (c-a) = cos( θ ) and (d-b) = sin( θ ) PJC
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yup. he was right up until the last little bit of arithmetic. Cleek | Image Toolkits | Thumbnail maker
Chris Losinger wrote:
he was right up until the last little bit of arithmetic.
Yep, I've now confirmed that I've been a complete and utter nincompoop, see http://www.codeproject.com/lounge.asp?msg=1530321#xx1530321xx[^] for my explanation and excuses. I now realise I was barking up the wrong tree :doh:
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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Though it does not qualify as an actual mathematical proof, I've tested your approach and Jun Du's approach in Excel with 10-15 different values and I get the same results. Remember that: θ = arctan( y / x ) x = cos( θ ) y = sin( θ ) from Inverse Tangent[^] You wrote: θ = arctan((d - b) / (c - a)) x = a + (0.2*|AB|*cos(θ) y = b + (0.2*|AB|*sin(θ) and based on the link from MathWorld, (c-a) = cos( θ ) and (d-b) = sin( θ ) PJC
I just wrote it out on paper, basically if you substitute adjacent / hypotenuse in place of cos then the root will cancel out and you'll get what Jun Du gave. I'm just too exhausted from 12 hours revision plus a 2 hour shooting walk with my father and the dog. I stand corrected and apologise if I was dogmatic :-O.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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I just wrote it out on paper, basically if you substitute adjacent / hypotenuse in place of cos then the root will cancel out and you'll get what Jun Du gave. I'm just too exhausted from 12 hours revision plus a 2 hour shooting walk with my father and the dog. I stand corrected and apologise if I was dogmatic :-O.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
Ed.Poore wrote:
I stand corrected and apologise if I was dogmatic .
Hey it's cool. It happens.
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yup. he was right up until the last little bit of arithmetic. Cleek | Image Toolkits | Thumbnail maker
Chris Losinger wrote:
he was right up until the last little bit of arithmetic.
I had that happen when I was in Calculus 15 years ago. It happens. The teacher dinged everyone on arithmetic mistakes even if we displayed knowledge of the material. Now I look back at those old midterm exams and laugh. PJC
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I just wrote it out on paper, basically if you substitute adjacent / hypotenuse in place of cos then the root will cancel out and you'll get what Jun Du gave. I'm just too exhausted from 12 hours revision plus a 2 hour shooting walk with my father and the dog. I stand corrected and apologise if I was dogmatic :-O.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
happens to the best of us... and the least :) without sitting down and working it through, i don't see how DU's solution works. maybe i'll give it some time later tonight. Cleek | Image Toolkits | Thumbnail maker
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happens to the best of us... and the least :) without sitting down and working it through, i don't see how DU's solution works. maybe i'll give it some time later tonight. Cleek | Image Toolkits | Thumbnail maker
When substituting cos θ by
x2 - x1
√(y2 - y1)² + (x2 - x1)²
the magnitude of the two vectors cancel out and you are left with
x2 - x1With the equation for the y component, sin θ is substituted by
y2 - y1
√(y2 - y1)² + (x2 - x1)²
the magnitude of the two vectors cancelled out again and you are left with
y2 - y1So in Jun Du's method, the magnitude of the two vectors and the trig functions drop out. PJC
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I just wrote it out on paper, basically if you substitute adjacent / hypotenuse in place of cos then the root will cancel out and you'll get what Jun Du gave. I'm just too exhausted from 12 hours revision plus a 2 hour shooting walk with my father and the dog. I stand corrected and apologise if I was dogmatic :-O.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
As a friendly suggestion, I would recommend bookmarking the www.mathworld.com[^] link. I use it often as a good web reference :) Paul
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Hey Guys, I have a math issue, I know how you like brain teasers so I thought I would post it here, say I have two points ab & cd, how do I find the point 20% of distance between these two points on the same slope ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
here it is suppose a and b are points.. I want point C 20% from point a then c.x= a.x+ 0.2 * (distnce between a & b ) * cos (angle between a and b ); c.Y= a.y+ 0.2 * (distnce between a & b ) * sin (angle between a and b ); Leya -- modified at 3:04 Wednesday 14th June, 2006
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Hey Guys, I have a math issue, I know how you like brain teasers so I thought I would post it here, say I have two points ab & cd, how do I find the point 20% of distance between these two points on the same slope ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
Use a tape measure :) I still remember having to write your own code in FORTRAN rather than be a cut and paste merchant being pampered by colour coded Intellisense - ahh proper programming - those were the days :)
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As a friendly suggestion, I would recommend bookmarking the www.mathworld.com[^] link. I use it often as a good web reference :) Paul
I know about MathWorld since I've got a copy of Mathematica, it's just I was being blind to the obvious yesterday, was feeling a bit more awake today but then sat down outside to read a book in the sun, big mistake.. Fell alseep.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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happens to the best of us... and the least :) without sitting down and working it through, i don't see how DU's solution works. maybe i'll give it some time later tonight. Cleek | Image Toolkits | Thumbnail maker
Chris Losinger wrote:
happens to the best of us
God I hope you're not implying I'm one of the best :laugh: Far from it.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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Use a tape measure :) I still remember having to write your own code in FORTRAN rather than be a cut and paste merchant being pampered by colour coded Intellisense - ahh proper programming - those were the days :)
I was told I was making things too complicated (which I was by the way), I think you're oversimplifying a bit here :laugh: But it's the most realiable method, I've got to give you that.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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I know about MathWorld since I've got a copy of Mathematica, it's just I was being blind to the obvious yesterday, was feeling a bit more awake today but then sat down outside to read a book in the sun, big mistake.. Fell alseep.
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
Ed.Poore wrote:
Mathematica
I've always liked mathematica. Haven't worked with it in some time :sigh: