Math/geometry question 1

[Walterke]

I'm actually ashamed asking you guys for help on this.

I've wasted my entire day trying to solve this and haven't gotten any further yet. I guess I'll have to go back to school and freshen up my math skills.

In attachment you'll find an image of an irregular pentagon with given:
angles: a, b
lengths: H, R, J, C

Requested is the angle marked with a "?" in the drawing.

If I draw it out in NX and define these, I can't move anything else, so I'm pretty sure it "should" be solvable.

I really don't know what I'm missing, any help would be appreciated

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

 

[Walterke]

Of course I forgot to mention something:
The angle that seems 90° (near the "?") is, so you know 3 angles.

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

 

[TheTick]

Usually I chop shapes like this into triangles and determine point positions. Eventually that leads to a solution.

 

[zekeman]

You simply write the vector equation

R=He^ja+Ce^j(a+x-180)+y*e^j(a+x-90)+Je^j(a+x+b-270)

where x is the angle sought and y is the link unknown

From this you write the 2 orthogonal equations in in the variables, x and y.

R=Hcosa+Jcos(a+x-180)+ysin(a+x)+Jcos(a+x+b+90)
0=Hsina+Jsin(a+x-180)-ysin(a+x)+Jsin(a+x+b+90)

The 2 equations should yield the answer, maybe with some sweat.

Good luck!

 

[Walterke]

@tick: you're right, it usually does :s

@Zeke, you're also right. It "should" yield the answer, only that's where my problems lie. I can't seem to work it out.


More sweat it is.

Again: All hints are welcome.

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

 

[Wrenchbender]

Looks like 2eqn/2 unk. Starting at what looks like the origin, label the points (going CCW), as O, A, B, C,D.

Did you try this?
Consider triangle OCD, and polygon OABC.

Forget the trig, try vectors
OC + CD + DO = zero
OA + AB + BC +CD +CO = zero

Too lazy to try myself but looks like it might work.

 

[MiketheEngineer]

You could draw it up in CAD and get e very close answer??

 

[rb1957]

internal angles of a pentagon sum to 540deg (i think).

so the two unknown angles sum to something known.

then i think you can find the RH lower unknown angle by changing the shape to be a quadrateral, replacing the "dog leg" side with a straight (yes?).
now you've got a quadrateral with three known sides and one known angle.

does that lead anywhere ?

 

[JohnRBaker]

You could draw it up in CAD and get e very close answer??

Trust me, he already has, and what he has should work (at least I was able to make it work using the same version of NX that Walterke is using). I've attached a file for him to review, but for the rest of you, here's what a sketched solution would look like:

John R. Baker, P.E.
Product 'Evangelist'
Product Engineering Software
Siemens PLM Software Inc.
Industry Sector
Cypress, CA
Siemens PLM:
UG/NX Museum:

To an Engineer, the glass is twice as big as it needs to be.

 

[zekeman]

Closed form solution
Let
D=C+Jsinb
B=Rsina
A=H-Rcosa
F=sqrt(A^2+B^2)

Then

x=inverse sin(D/F)-inverse tangent(A/B)

Checked against Baker's solution

D=25+160*sin(70)=175.35
B=224*sin(110)=211.75
A=100-224*cos70=23.38
F=213
x=inverse sin(D/F)=55.41 minus inverse tangent(A/B)=6.3
x=55.4-6.3=49.1
compared with Baker's 49.55

 

[dgallup]

Pro/E WF4 gets 49.55 as well.

However, plugging zekeman's equations into excel gives

D 175.3508193
B 210.4911471
A 23.3874879
F 213

asin(D/F) 55.41078923
atan(A/B) 6.340079533
X 49.0707097


----------------------------------------

The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.

 

[zekeman]

Couple of errors in evaluating
should be
D=25+160*sin(70)=175.35
B=224*sin(110)=210.49
A=100-224*cos70=23.38
F=211.785

asin(D/F) 55.889
atan(A/B) 6.338
X 49.55

which is now agreed to be the answer

Dgallup,
Thanks for your input
Your F was my old F.

 

[Walterke]

x=inverse sin(D/F)-inverse tangent(A/B)

I'd like to know how you got to that solution. That's the part I'm having trouble with

Everyone else: thanks for the help!

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

 

[Walterke]

I took me a bit more time but I managed to get a result. Thanks everyone for the help!

@Zeke, I'm still curious as to how you did it.

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

 

[rb1957]

we seem to be using lenths from John's sketch.

zeke, i'm curious; what's the relationship betwwn the 160 side length, and the angle of the opposite corner ?
similarly the 224 length and the opposite 110deg angle ?

i guessing it's some math relationship for closed polygons (similar to sine rule for triangles) ?

 

[rb1957]

interstingly, the angle "x" is the complement (i think) to the unknown angle at the lower RH corner.

"x" appears as the angle of the RH side to the vertical.

 

[zekeman]

I tossed the solution once I did it and will resurrect it later when I get some time.

However, for those interested the trick was to rotate the vector diagram so that the unknown link is horizontal.

Then you write a single equation in x by equating the y ooordinate of the J link end , Jsinb,by adding the y coordinates (from the left of the datum link), thru all the links, C,H and R, fairly straightforward.

 

[zekeman]

Ok, the unknown link is horizontal and the links and their angles to the horizontal are

C 90
H -90-x
R -90-x+180-a=90-a-x
J b-180
so equating the y coordinate of J and setting equal to the sum of the remaining links from the left

-Jsinb=C-Hcosx+Rcos(a+x)
Hcosx-Rcosacosx+Rsinasinx=Jsinb+C
(H-Rcosa)cosx+Rsinasinx=Jsinb+C
Using my notation above
Acosx+Bsinx=D
You can now solve the quadratic equation for sinx,after substituting cosx=sqrt(1-sin^2x) and you are done or
do it using vectors which I find easier to use as follows :
Construct the vector Bi+Aj =F*
where the angle of F* to the x direction is alpha=atan(A/B)
Now rotate F* an angle x and the vector becomes
(Bcosx-Asinx)i +(Acosx+Bsinx)j
Now F maks an angle alha +x to the x axis
And the projection of F* on the y axis is
F*.j=Acosx+Bsinx which is equal to D
From the trigonometry,that angle alpha+x=asin(D/F)
whence x=asinD/F-atan(A/B)

 

[zekeman]

I am trying to upload a sketch of the above. Hope it works

 

[Walterke]

Thanks again zeke!

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000