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Any geometry experts out there?

  • Thread starter Thread starter ForceFed70
  • Start date Start date Feb 9, 2007
F

ForceFed70

That's why they call it "dope"
Founding Member
Dec 6, 1999
4,818
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69
BC Canada
Feb 9, 2007
#1
  • Feb 9, 2007
  • #1
If so, maybe you can solve this problem for me?

Take a look at this diagram:

Assume that point A is fixed and the diagram is being rotated.
If I want point C to move 0.5" how far would I need to move point B?

Hope it makes sense. Thanks!
 

zookeeper

Founding Member
Aug 25, 2001
3,415
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109
Rogue River, Oregon
Feb 9, 2007
#2
  • Feb 9, 2007
  • #2
I'm not seeing the diagram
 

NasaGT

Founding Member
Sep 19, 2002
1,993
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49
Virginia
Feb 9, 2007
#3
  • Feb 9, 2007
  • #3
No pic, just Red X
 

zookeeper

Founding Member
Aug 25, 2001
3,415
63
109
Rogue River, Oregon
Feb 9, 2007
#4
  • Feb 9, 2007
  • #4
You don't suppose the red x is the diagram do you? Hmmmm.....
 
3

351carlo

New Member
Mar 9, 2004
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Philadelphia
Feb 10, 2007
#5
  • Feb 10, 2007
  • #5
Yes, I'm an expert. No I don't know how far point B moves....

e-mail me the picture, I'd be glad to help. If you're on AIM my sn is see-dee-jay-arghh-three-two-zero. (All the phonetic spellings are letters and the numbers are just the numerical digits). Email is fio-rella-@udel.-edu except take out all the (-'s)
 
F

ForceFed70

That's why they call it "dope"
Founding Member
Dec 6, 1999
4,818
1
69
BC Canada
Feb 10, 2007
#6
  • Feb 10, 2007
  • #6
ARGH!! This was working.

Let's try again...



OK, can you see it now?
 

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D.Hearne

New Member
Sep 29, 2000
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south louisiana
Feb 10, 2007
#7
  • Feb 10, 2007
  • #7
Take a compass (these are available in your local drug store's school supplies dept.) and use it. Place one point on "A", then adjust it to where the other point (there's a pencil on this end) on "B", draw a curved line with it 1/2" away from "B". Then do the same with "A" and "C"
 

red65

Member
Apr 12, 2003
411
0
17
Ann Arbor, MI
Feb 10, 2007
#8
  • Feb 10, 2007
  • #8
Your sketch is a little ambiguous. Is the distance from A to B 13.75" (half of 27.5")?

Making that assumption, you just set the rotation angles for points B and C about point A equal. Using basic geometry,

s = (theta) x r

where s is the distance traveled - aka 'arc length' (inches)
theta is the rotation angle (radians)
r is the radius (inches)

Set theta equal for points B and C and do a little algebra to get:

s(B) = s(C) x r(B)/r(C)

If my assumption about r(B) being 13.75" is correct, then:

s(B) = 0.5" x 13.75"/40"
s(B) = 0.172"

If it's not, plug in the correct value for r(B) and off you go.

You can do the same math using similar triangles, and you'll get the same answer, just a different approach. Hope that helps.
 
3

351carlo

New Member
Mar 9, 2004
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Philadelphia
Feb 10, 2007
#9
  • Feb 10, 2007
  • #9
Well I got your email just now and got your answer but it looks like Red65 got to ya first. I'd go with similar triangles, but it's the same idea as arc length. When the sides are in proportion to each other, you use that ratio to find out how much a change in one would affect a change in the other, assuming they will remain in the exact proportions to each other.

With this example you would want to sent your sides AC and AB as the equal sides in the triangle. The ratio of these sides is 13.75/40 as already used in the arc length formula. Just like the arc length formula you know that if it is moved .5 for the whole triangle, then it is moved .5 x ratio for the smaller triangle.

This yields the same answer (obviously since the simplified arc length formula is the same as this) of 0.171875 or 0.172 or if you're using a hacksaw simply .17
 
F

ForceFed70

That's why they call it "dope"
Founding Member
Dec 6, 1999
4,818
1
69
BC Canada
Feb 10, 2007
#10
  • Feb 10, 2007
  • #10
That rocks! Thanks guys!

Red65, your assumption is correct the distance from A to B is 1/2 of 27.5 = 13.75

Been way too long since I had to do much more than simple math.
 
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