Degrees of Freedom
Ileana Streinu
Christie Rice
Victoria Manfredi
Generally, degrees of freedom refer to the ability to move. A line segment that has neither vertex pinned down has three degrees of freedom because the position of the line segment is dependent on three pieces of information: the (x,y) coordinates for translation are two pieces of information and the angle that the line segment has been rotated from the original line segment is the third piece of information. (Click on the line segment below to move it and see the three different degrees of freedom.)
Applet of a Line Segment With Three Degrees of Freedom
degFree2.cdy
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The two pieces of information, x and y, needed for translation are seen from the following applet.
Applet Showing Translation of a Line Segment
degFree3.cdy
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By pinning down one of the line segment vertices, the two degrees of freedom needed for translation are lost. But since rotation is still possible, (click with the mouse on vertex B to rotate the line segment) the line segment still has one degree of freedom.
Applet of a Line Segment With One Degree of Freedom as Vertex A is Pinned Down
degFree4.cdy
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If both vertices of the line segment are pinned down, there are no degrees of freedom as the line segment can't move.
Applet of a Line Segment With Zero Degrees of Freedom as Both Vertices are Pinned Down
degFree.cdy
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A quadrilateral is a mechanism with one degree of freedom when one bar of the quadrilateral is pinned down. Why is this?
Once the two vertices on either end of a bar in the quadrilateral are pinned down, and the lengths of each bar in the quadrilateral are known, then the other two vertices are dependent on only one piece of information which we'll call x, hence one degree of freedom. The coordinates for the vertices pinned down must be known as they are fixed.
Where does x fit into this? Let's start by constructing a quadrilateral with one bar pinned down:
Applet of the Quadrilateral We Want to Construct (Note: see what happens when you mouse click on point3.) point1 and point2 are the vertices of the pinned bar.
degFree6.cdy
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What we know about this quadrilateral:
Pinned vertices:
point1 = {0,0}
point2 = {8,0}
Unpinned vertices:
point3 = ?
point4 = ?
Lengths of each bar of the quadrilateral:
point1 to point2 = 8
point2 to point3 = 7
point3 to point4 = 6
point4 to point1 = 5
We know the coordinates for point1 and point2. Together point1 and point2 give the fixed bar of the polygon. But how dow we get point3 and point4? Let's first find point3. Well, point3 has to be on the circle that has point2 as its center (if we go in a counterclockwise direction around the quadrilateral). So let's find the equation for the circle that has point2 as its center, and then we'll have an equation for all the coordinates that could be point3.
Applet of the Circle with point2 as its Center
degFree7.cdy
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equation for a circle:
(x - a)2 + (y - b)2 = r2
We have already chosen the radius of this circle to be 7 because we said the distance from point2 to point3 was 7. We know the center of the circle is (8,0) because those are the coordinates we set for point2. So plugging the numbers into the equation we get:
(x - 8)2 + (y - 0)2 = 72
(x - 8)2 + y2 = 49
x2 - 16x + 64 + y2 - 49 = 0
y2 = - x2 + 16x - 15
y = (- x2 + 16x - 15) 1/2
We will need to plug in a number for x to get point3. Generally, point3 of the quadrilateral is:
point3 = {x, (- x2 + 16x -15) 1/2}
Note the x in the coordinates for point3. It is this x that is the one unknown piece of information on which the two unpinned vertices depend. This x will show up again.
Let's now find point4.
point4 has to be on the circle that has point1 as its center but it also has to be on the circle that has point3 as its center. So we want the intersection of two circles: the circle wth point1 as its center and the circle with point3 as its center. Two circles will intersect in two places as we shall see. Let's start finding these intersections by first finding the equation of the circle with point1 as its center and then finding the equation of the circle with point3 as its center.
Finding the equation of the circle with point1 as its center:
Applet of the Circle with point1 as its Center
degFree9.cdy
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We have chosen the radius for the circle with point1 as its center to be 5 because we said the distance from point 4 to point1 was 5. We know the center of the circle is (0,0) because those are the coordinates we set for point1 So plugging the numbers into the equation we get:
(Note: to avoid confusion with our unknown x, the x in the equation for a circle has been switched to z.)
(z - 0)2 + (y - 0)2 = 52
z2 + y2 = 25 <-- equation of circle with point1 as its center
Finding the equation of the circle with point3 as its center:
Applet of the Circle with point3 as its Center
degFree8.cdy
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We have chosen the radius for the circle with point3 as its center to be 6 because we said the distance from point2 to point3 was 6. We know the center of the circle is {x, (- x2 + 16x -15) 1/2} because those are the coordinates we found for point3. So plugging the numbers into the equation we get:
(Note: to avoid confusion with our unknown x, the x in the equation for a circle has been switched to z.)
(z - x)2 + (y - (- x2 + 16x -15) 1/2)2 = 62
(z - x)2 + (y - (- x2 + 16x -15) 1/2)2 = 36 <-- equation of circle with point3 as its center
Finding where the circle with point1 as its center and the circle with point3 as its center intersect:
Applet of the intersections of the circle with point1 as its center and the circle with point3 as its center.
degFree5.cdy
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By solving the following two equations simultaneously we will get the coordinates for point 4. But in order to solve the two equations simultaneously we need some value for x. We have arbitrarily selected x to equal 5. Because x is the only unknown piece of information that we need to know and that we have not been given, a quadrilateral with one bar pinned down has one degree of freedom.
Our two equations:
z2 + y2 = 25 <-- equation of circle with point1 as its center
(z - x)2 + (y - (- x2 + 16x -15) 1/2)2 = 36 <-- equation of circle with point3 as its center
Substituting 5 in for x we get:
z2 + y2 = 25
(z - 5)2 + (y - (- 52 + 16(5) -15) 1/2)2 = 36
Solving these two equations simultaneously we find two sets of answers:
answer set one:
z = 1/65(135-32*351/2), y = 2/65(27*101/2-20*141/2)
answer set two:
z = 1/65(135-32*351/2), y = 2/65(27*101/2-20*141/2)
Therefore, (z,y) is point4. But we have two sets of (z,y) coordinates for point4, so does that mean that the quadrilateral actually has two degrees of freedom because point4 can be in two places? No, because although we found two possible sets of coordinates for point4, we needed to be told only one piece of information, what x was, in order to do so. Note: of course you also need to know the coordinates of the pinned bar and the lengths of each bar in the quadrilateral as otherwise it could be any quadrilateral. Thus, a quadrilateral with one bar pinned has only one degree of freedom.
Mouse click and drag point4 to make the following two quadrilaterals look the same. Each of the quadrilaterals uses one of the two sets of coordinates found for point4
Applet of the Quadrilateral Implementing Answer Set One for point4.
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Applet of the Quadrilateral Implementing Answer Set Two for point4.
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The Mathematica code to do the above calculations, and showing the
resulting quadrilaterals can be found here.