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Posted by Grant Erwin on May 6, 2006, 9:59 pm
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This is actually a little worse than shop math. I have solved this; will confirm
the first correct answer.
Find an equation of a conic (2nd order function) which goes through the points
(0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is nearly
vertical at the point (2.5,5). Put it in the form y = F(x); in other words with
y alone on the left hand side of the equation.
GWE
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Posted by Dave on May 7, 2006, 9:48 am
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Grant Erwin wrote:
> This is actually a little worse than shop math. I have solved this; will
confirm
> the first correct answer.
>
> Find an equation of a conic (2nd order function) which goes through the points
> (0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is nearly
> vertical at the point (2.5,5). Put it in the form y = F(x); in other words with
> y alone on the left hand side of the equation.
Got any background to make this interesting? Looks pretty dull.
y= ax^2 + bx + c
0= 0 + 0 + c so c = 0
0.75 = 4a + 2b
5 = 6.25a + 2.5b
Two equations in two unknowns so put a or b in terms of the other and
plug it in and grind out an answer. Not sure if a "conic" might mean
you'd really need to start with the form y = ax^2 + bxy + cy^2 + dx + e
but I guess you'd get the same thing.
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Posted by Grant Erwin on May 7, 2006, 10:40 am
Please log in for more thread options Dave wrote:
> Grant Erwin wrote:
>
>>This is actually a little worse than shop math. I have solved this; will
confirm
>>the first correct answer.
>>
>>Find an equation of a conic (2nd order function) which goes through the points
>>(0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is nearly
>>vertical at the point (2.5,5). Put it in the form y = F(x); in other words with
>>y alone on the left hand side of the equation.
>
>
> Got any background to make this interesting? Looks pretty dull.
>
> y= ax^2 + bx + c
> 0= 0 + 0 + c so c = 0
> 0.75 = 4a + 2b
> 5 = 6.25a + 2.5b
>
> Two equations in two unknowns so put a or b in terms of the other and
> plug it in and grind out an answer. Not sure if a "conic" might mean
> you'd really need to start with the form y = ax^2 + bxy + cy^2 + dx + e
> but I guess you'd get the same thing.
>
Background? Well, I'm musing on ways to make the center piece of a vibratory
polishing bowl.
It's a little more complicated than your analysis, but here 'tis:
In my solution, the final equation looks like
y = Px + Q + SQRT(R*X^2 + S*X + T)
and for a curve going through (0,0) (2,0.75) and (2.5,5) and tangent to the x
axis at (0,0) and very nearly vertical at (2.5,5) I got my 5 constants as:
P = -12.083
Q = 35.208
R = 142.007 ;; there may be slight roundoff or truncation errors in these
S = -850.868
T = 1239.627
It took me an amazing amount of time flailing at this problem (which I didn't at
all need to solve; I just like doing math sometimes) before I resorted to
looking up how aircraft designers used to lay out airframes, and found that back
in the 1950s those guys really knew about 2nd order equations and how to whip
them into shape.
GWE
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Posted by Dave on May 7, 2006, 11:04 am
Please log in for more thread options Grant Erwin wrote:
> Dave wrote:
> > Grant Erwin wrote:
> >
> >>This is actually a little worse than shop math. I have solved this; will
confirm
> >>the first correct answer.
> >>
> >>Find an equation of a conic (2nd order function) which goes through the
points
> >>(0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is
nearly
> >>vertical at the point (2.5,5). Put it in the form y = F(x); in other words
with
> >>y alone on the left hand side of the equation.
> >
> >
> > Got any background to make this interesting? Looks pretty dull.
> >
> > y= ax^2 + bx + c
> > 0= 0 + 0 + c so c = 0
> > 0.75 = 4a + 2b
> > 5 = 6.25a + 2.5b
> >
> > Two equations in two unknowns so put a or b in terms of the other and
> > plug it in and grind out an answer. Not sure if a "conic" might mean
> > you'd really need to start with the form y = ax^2 + bxy + cy^2 + dx + e
> > but I guess you'd get the same thing.
> >
>
> Background? Well, I'm musing on ways to make the center piece of a vibratory
> polishing bowl.
>
> It's a little more complicated than your analysis, but here 'tis:
>
> In my solution, the final equation looks like
>
> y = Px + Q + SQRT(R*X^2 + S*X + T)
>
> and for a curve going through (0,0) (2,0.75) and (2.5,5) and tangent to the x
> axis at (0,0) and very nearly vertical at (2.5,5) I got my 5 constants as:
>
> P = -12.083
> Q = 35.208
> R = 142.007 ;; there may be slight roundoff or truncation errors in these
> S = -850.868
> T = 1239.627
>
> It took me an amazing amount of time flailing at this problem (which I didn't
at
> all need to solve; I just like doing math sometimes) before I resorted to
> looking up how aircraft designers used to lay out airframes, and found that
back
> in the 1950s those guys really knew about 2nd order equations and how to whip
> them into shape.
>
> GWE
Your proposed solution seems to fail the (0,0) test...
0 = 0 + Q + SQRT(0 + 0 + T)
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Posted by mike on May 7, 2006, 11:21 am
Please log in for more thread options
>Grant Erwin wrote:
>> Dave wrote:
>> > Grant Erwin wrote:
>> >
>> >>This is actually a little worse than shop math. I have solved this; will
confirm
>> >>the first correct answer.
>> >>
>> >>Find an equation of a conic (2nd order function) which goes through the
points
>> >>(0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is
nearly
>> >>vertical at the point (2.5,5). Put it in the form y = F(x); in other words
with
>> >>y alone on the left hand side of the equation.
>> >
>> >
>> > Got any background to make this interesting? Looks pretty dull.
>> >
>> > y= ax^2 + bx + c
>> > 0= 0 + 0 + c so c = 0
>> > 0.75 = 4a + 2b
>> > 5 = 6.25a + 2.5b
>> >
>> > Two equations in two unknowns so put a or b in terms of the other and
>> > plug it in and grind out an answer. Not sure if a "conic" might mean
>> > you'd really need to start with the form y = ax^2 + bxy + cy^2 + dx + e
>> > but I guess you'd get the same thing.
>> >
>>
>> Background? Well, I'm musing on ways to make the center piece of a vibratory
>> polishing bowl.
>>
>> It's a little more complicated than your analysis, but here 'tis:
>>
>> In my solution, the final equation looks like
>>
>> y = Px + Q + SQRT(R*X^2 + S*X + T)
>>
>> and for a curve going through (0,0) (2,0.75) and (2.5,5) and tangent to the x
>> axis at (0,0) and very nearly vertical at (2.5,5) I got my 5 constants as:
>>
>> P = -12.083
>> Q = 35.208
>> R = 142.007 ;; there may be slight roundoff or truncation errors in these
>> S = -850.868
>> T = 1239.627
>>
>> It took me an amazing amount of time flailing at this problem (which I didn't
at
>> all need to solve; I just like doing math sometimes) before I resorted to
>> looking up how aircraft designers used to lay out airframes, and found that
back
>> in the 1950s those guys really knew about 2nd order equations and how to whip
>> them into shape.
>>
>> GWE
>
>Your proposed solution seems to fail the (0,0) test...
>
>0 = 0 + Q + SQRT(0 + 0 + T)
Dave, remember Square roots have a + and a - value.
Mike in BC
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