Solve block triangle question

[Lea-disabled]

Hi Everyone,
Please see the enclosed MathCad worksheet & plot for an explanation of what I am trying to do.
I also enclose a PDF file with Figure 1 & Figure 2 to help explain as well.
Notes: -
#1 - Given the Fixed Hypotenuse ( =25.837) & Fixed Opposite (=8) I am trying to arrange a slope from top to bottom of the Opposite line marked in the centre of the red triangle in figure 1.
#2 - There are 3 'legs'/lines down which are HYP1, HYP2 & HYP3 which HYP1+ HYP2+ HYP3 = Hypotenuse = 25.83723
#3 - All the slopes of the legs / lines down are the same.
#4 - I have setup the current configuration for ease of explanation of what I am trying to do.
#5 - Note also that each HYP / leg meets the edge of the red triangle & the other HYP / leg start or finishes from that horizontal point.
#6 - The Combination of the vertical heights OPP1+OPP2+OPP3 = Opposite = 8
#7 - And lastly / obviously the ADJ1+ADJ2+ADJ3 = Adjacent = SQRT(Hypotenuse^2-Opposite^2) = (24.567508)
QUESTION: - What I am trying to do is this :-
- Adjust the base (& therefore the overall dimensions of the red triangle), say by reducing the value of PHI, in the MathCad worksheet.
(The dimensions of the red triangle are the height is half the base.)
- The OVERALL OBJECTIVE is to confine the Fixed lengths of Opposite, ( always in centre of red triangle), & the three legs of the Hypotenuse within a ,say smaller red triangle, ( for reduced value of PHI) but maintaining a similar configuration as outlined in figure 1 where the legs of the Hypotenuse begin & end at the top & bottom of the Opposite line respectively. As the value of PHI changes the size of the red triangle changes & then the slope of all 3 HYP/legs (which all have the same slope) increase or decrease accordingly.
I have been looking at this problem for quite some time & I am not sure if it can be done - Is my objective above possible? Please advise.
- I think it should be possible within a certain small range of values for PHI / changes in the size of the red triangle base.
If so, please could someone show this in a working MathCad worksheet, perhaps with an edited up date of mine enclosed.
Many thanks for everyone's help & attention.
Best regards - Lea...
Ps. I am version 14 of MathCAd.

[RichardJ]

On 2/1/2010 8:04:14 AM, Lea wrote:

>I have been looking at this
>problem for quite some time &
>I am not sure if it can be
>done - Is my objective above
>possible? Please advise.

It is not possible unless you relax at least one of your constraints. As you have defined the problem, all the equations in your solve block are correct. That means PHI is defined by those six equations and the given values of Opposite and Hypotenuse.

Richard

[RichardJ]

Your pieces aren't the same size.

Richard

[RichardJ]

On 2/1/2010 11:18:00 AM, jmG wrote:

>Blue here or blue there is same, isn't ?
>Pink here or pink there is same, isn't ?

The way you have drawn it the blue and the pink are both different sizes. In the original problem they are the same sizes, but the sloping line is not quite straight.

Richard

[RichardJ]

On 2/1/2010 11:38:39 AM, jmG wrote:

>Just blowing smoke about my imperfect
>drawing.

Yes.

>Do it better with Corel, collabs will
>appreciate.

I'll go one better. This is done in CAD software.



Richard

[RichardJ]

On 2/1/2010 3:28:41 PM, jmG wrote:

>Just redo mine in CAD, flat

Sorry. Too boring.

+ squares

OK. Squares you can have.




Richard

[IRstuff]

That's a trivial problem even for high school. The two triangles are not similar. One's slope is 2/5, while the other's is 3/8. Neither are similar to the larger triangle whose slope is 5/13.


TTFN,
Eden

[Lea-disabled]

Hi Richard & Everyone,

Many thanks for all the responses.

Thanks Richard for pointing out that this problem is not possible with 2 constrains Hypotenuse AND Opposite.

OK... Now the Question is:-
How would the setup look like in a MathCad worksheet if just the Hypotenuse was fixed but difference values for the Opposite could be entered thus changing the size of the red triangle, but maintaining the a similar configuration as in the diagram. I.E. 3 'legs' down touching the sides of the red triangle.
- the slope of all 3 legs down being the same
would presumably the increase & decrease accordingly depending on the value of the Opposite value.

Any help you could offer for this setup would be greatly received.

Many thanks for you help & attention.
Best regards,
Lea...

[RichardJ]

If you get rid of the defined value of Opposite you can change the value of PHI, and the triangle will change.

To get what I think you want you actually need to change the equations in the solve block too. The equations you had before were all correct, but when I looked at it closer I realized they did not constrain the three triangles to be similar. I rearranged a few other things in the worksheet too.

Richard

[Lea-disabled]

Hi Richard & jmG,

Thank you so much for all your extensive help & advice & links etc....Absolutely excellent.

Richard the MathCad worksheet you provided with adjustable PHI is great - Many thanks.

jmG - I liked your picture of solution of adjustable isosceles triangle. Is it possible to post the MathCad worksheet version of this so I can study more closely - Many thanks.

Warm regards,
Lea...

[ptc-1368288]

On 2/3/2010 9:16:32 AM, Lea wrote:
>Hi Richard & jmG,
>
>Thank you so much for all your
>extensive help & advice &
>links etc....Absolutely
>excellent.
>
>Richard the MathCad worksheet
>you provided with adjustable
>PHI is great - Many thanks.
>
>jmG - I liked your picture of
>solution of adjustable
>isosceles triangle. Is it
>possible to post the MathCad
>worksheet version of this so I
>can study more closely - Many
>thanks.
>
>Warm regards,
>Lea...
______________________________

Lea,

You must be able to spare a fraction of the time I spent. It will take you few minutes and so simple. If you don't reconstruct, you will forget and won't appreciate, eventually miss to make your project more comprehensive/visible/adaptable.
The other point is that any version can take or leave.

jmG

[Lea-disabled]

Sure OK. Many thanks for your excellent help on this.
Best regards - Lea...

[ptc-1368288]

On 2/4/2010 12:56:13 AM, Lea wrote:
>Sure OK. Many thanks for your
>excellent help on this.
>Best regards - Lea...
______________________________

Few more things 2d plots

jmG

[Lea-disabled]

On 2/2/2010 2:51:57 PM, rijackson wrote:
>If you get rid of the defined
>value of Opposite you can
>change the value of PHI, and
>the triangle will change.
>
>To get what I think you want
>you actually need to change
>the equations in the solve
>block too. The equations you
>had before were all correct,
>but when I looked at it closer
>I realized they did not
>constrain the three triangles
>to be similar. I rearranged a
>few other things in the
>worksheet too.
>
>Richard

Hi Richard,

I have been studying your excellent worksheet provided above( which is great)
QUESTION - But now I am trying to create a vector output for Opposite by input certain range values for PHI. Unfortunately I have forgotten how to transform your solve block so the that the range input for PHI can be added ( as a vector) => then calculated => to give output values of Opposite ( as a vector) , so I can plot this...

Obviously Range:-0.5,0.6...5.0

Please could you show me how to do this one.

Many thanks for your help & attention.
Best regards,
Lea ...

[RichardJ]

On 2/11/2010 8:48:50 AM, Lea wrote:

>QUESTION - But now I am trying to create
>a vector output for Opposite by input
>certain range values for PHI.
>Unfortunately I have forgotten how to
>transform your solve block so the that
>the range input for PHI can be added (
>as a vector) => then calculated => to
>give output values of Opposite ( as a
>vector) , so I can plot this...

You need to convert the solve block to a function of PHI. See the attached.

Richard

[Lea-disabled]

On 2/11/2010 10:28:53 AM, rijackson wrote:
>On 2/11/2010 8:48:50 AM, Lea wrote:
>
>>QUESTION - But now I am trying to create
>>a vector output for Opposite by input
>>certain range values for PHI.
>>Unfortunately I have forgotten how to
>>transform your solve block so the that
>>the range input for PHI can be added (
>>as a vector) => then calculated => to
>>give output values of Opposite ( as a
>>vector) , so I can plot this...
>
>You need to convert the solve block to a
>function of PHI. See the attached.
>
>Richard
>
>
Hi Richard,

Many thanks for the enclosed adjustment for PHI.
Here I have defined the range of values for PHI so I can produce output vector values for each O1, O2, O3, A1, A2, A3. Obviously the function does not accept the PHI range as an input.
Please could you advise & show me the correct setup to get the desired vector output.

Many thanks for your help & attention.
Best regards,
Lea...

[RichardJ]

See the attached.

Richard

[RichardJ]

On 2/1/2010 8:04:14 AM, Lea wrote:

>#7 - And lastly / obviously
>the ADJ1+ADJ2+ADJ3 = Adjacent
>=
>SQRT(Hypotenuse^2-Opposite^2)
>= (24.567508)

I pointed out to you last time that this is not correct. You cannot apply Pythagoras theorem to the sums of the triangle sides.

Take

H1^2 = O1^2+A1^2
H2^2 = O2^2+A2^2

Then

H1^2+H2^2 = O1^2+A1^2+O2^2+A2^2

That is NOT equal to

(O1+O2)^2 + (A1+A2)^2

Richard

[ptc-1368288]

On 2/1/2010 8:04:14 AM, Lea wrote:
........

>QUESTION: - What I am trying to do is this :-
- Adjust the base (& therefore the overall dimensions of the red triangle), say by reducing the value of PHI, in the MathCad worksheet.
(The dimensions of the red triangle are the height is half the base.)<<br>
......
......

>Many thanks for everyone's
>help & attention.
>Best regards - Lea...
>Ps. I am version 14 of
>MathCAd.
___________________________

I have ignored "say by reducing the value of PHI" for the construct of the isocele triangle as you have stated.



jmG

[ptc-1368288]

... drawn with the "Calipers"

jmG