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Alltrade 940759 Powerbuilt Digital Torque Adaptor for 1/2-Inch Driver
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Hitachi C8FSE 8-1/2-Inch Sliding Compound Miter Saw
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BK Precision 9110 100W Multi Range 60V/5A DC Power Supply
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The BK Precision 9110 is a new type of power supply. Unlike conventional power supplies with fixed output ratings, the 9110 automatically recalculates voltage/current limits for each setting. The 9110 provides 100W output power in any Volt/Amp combination within the rated voltage (60V) and current (5A) limits...
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TRIGGER FOR BRAKES (CIVILIAN FLIGHTS) OR RAPID FIRE (MILITARY FLIGHTS) CONTROLFULLY AMBIDEXTROUS JOYSTICK3 REMOVABLE COMPONENTS ALLOW THE JOYSTICK TO BE PERFECTLY TAILORED FOR LEFT-HANDED OR RIGHT-HANDED USEH...
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Olympus Linear Recorder (LS-11)
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Precision Linear Expansion Apparatus
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This precision apparatus will produce accurate determinations of the coefficient of expansion. Readings are repeatable to 0.005 mm. -- This rigidly constructed apparatus requires no electrical accessories...
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Kicker 07DS400 4-Inch 100mm Coax Speakers (Pair)
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Here are some more information for Linear Precision:

Interpolation
Piecewise Linear interpolation is one of the simplest methods used for interpolation. As the name implies it fits a straight line to each consecutive pair of data points and uses the appropriate straight line for interpolation. For linear interpolation, a polynomial of degree one is constructed between each pair of known data points. For instance, given two data points (xa, ya) and (xb, yb), the interpolant is given by :
y = ya + (x - xa)(yb - ya) at the point (x, y)
(xb - xa)
Although linear interpolation is quick and easy, it is not very precise and the interpolant is not differentiable at the point xk. Let us assume that we wish to interpolate a function g, and suppose that x lies between xa and xb, and that g is twice continuously differentiable. Then linear interpolation error is given by
|¦(x) - g(x)| £ C(xb - xa)² where C = 1 max |g¢¢(y)|
8 ye[xa, xb]
Essentially the error is proportional to the square of the distance between the data points. Linear interpolation results in straight lines between each pair of points and all derivatives are
discontinuous at the data points. As it never overshoots or oscillates, it is frequently used in chemical engineering despite the fact that the curves are not smooth. To obtain a smoother curve, cubic splines are frequently recommended.
Spline interpolation uses low degree polynomials in each of the intervals, and selects the polynomial pieces such that they fit together smoothly. The natural cubic spline is piecewise cubic and twice continuously differentiable, furthermore its second derivative is zero at the end points. For cubic spline interpolation, a polynomial of degree three is constructed between each pair of known data points. Given n + 1 observation points x0
j=0
The cubic spline interpolant S(x), through these points is defined by
Piecewise Cubic: S(x) = Sj(x) for xj £ x £ xj+1, j = 0, 1, 2, …, n-1 where each Sj(x) is a cubic polynomial.
Interpolation Conditions: S(xj) = ¦(xj) for j = 0, 1, 2, …, n.
Continuity Conditions: Sj(xj+1) = Sj+1(xj+1) for j = 0, 1, 2, …, n - 2,
S¢j(xj+1) = S¢j+1(xj+1) for j = 0, 1, 2, …, n - 2,
S¢¢j(xj+1) = S¢¢j+1(xj+1) for j = 0, 1, 2, …, n - 2,
Two End Conditions are also required:
Free or Natural Boundary: S¢¢(x0) = S¢¢(xn) = 0
Clamped Boundary: S¢(x0) = ¦¢(x0) and S¢(xn) = ¦¢(xn)
Not a Knot Condition: S¢¢¢(x) is continuous at the knots x1 and xn-1
Cubic splines have error of order 4
Barycentric lagrange interpolation
Given a set of k + 1 data points ((x0), (y0)), …, ((xk), (yk)) where no two xj are the same, the interpolation polynomial in the Lagrange form is a linear combination
k
L(x) = å yjij(x)
j=0
Of Lagrange basis polynomials
k
ij(x) = Õ (x - xi) = (x - x0) … (x - xj-1) (x - xj+1) … (x - xk)
i=0,i¹j (xj - xi) (xj - x0) (xj - xj-1) (xj - xj+1) (xj - xk)
Using the quantity
i(x) = (x - x0) (x - x1) … (x - xk)
We can rewrite the Lagrange basis polynomial as
ij(x) = i(x) 1____
(x - xk) Õk (xj - xi)
i=0,i¹j
Or, by defining the barycentric weights
wj = 1___
Õk (xj - xi)
i=0,i¹j
We can simply write
ij(x) = i(x) _wj__
(x - xj)
This is known as the first form of the barycentric interpolation formula. This allows the interpolation polynomial to be evaluated as
k
L(x) = i(x) å _ wj__ yj
j=0 (x - xj)
The barycentric interpolation formula can also be easily restructured to incorporate an additional node xk+1 by dividing each of the wj, j = 0, …, k by xj - xk+1 and constructing the new wk+1 as above. We can further simplify the first form by considering the barycentric interpolation of the constant function g(x) º 1:
k
g(x) = i(x) å _ wj__
j=0 (x - xj)
Dividing L(x) by g(x) does not adjust the interpolation, yet produces
k
å _ wj__ yj
j=0 (x - xj)
L(x) = _______________
k
å _ wj__
j=0 (x - xj)
This is referred to as the second or true form of the barycentric formula. The advantage of being in this form is that i(x) does not need to be evaluated for each evaluation of L(x). The barycentric formula is forward stable for any set of interpolating points with a small Lebesgue constant. Therefore the barycentric formula can be significantly more accurate than the modified Lagrange formula unless there is a poor choice of interpolating points.
About the Author
The author graduated with a BSc Hons in Mathematics from the University Of St. Andrews before moving to Heriot Watt University in Edinburgh where he attained an MSc in Applied Mathematics. The author is also an associate member of the Institute of Mathematics and its Applications. During his study and since leaving university the author has held a number of customer services and marketing roles within major retail organisations.For a free marketing course visit http://www.scottemcclelland.info
Scott E McClelland MSc Applied Mathematics AMIMA.
Authors Homepage: http://www.whoisscottemcclelland.info/
Need to cut thin sheet metal -- Minimal cut width -- precision -- cheap tool (shear, tin snips, nibbler)?
I need to cut linear and non linear cuts through sheet metal (mild stuff, not more than 1, or 2 mm thick)
the width of the cut should be smallish, and the tool, precise since the piece i'm working on isnt very large
my question: should i use shears, tin snips, or a nibbler
I think the snips would provide precision, and a small cut width, but I dont know if they'll be strong enough to easily cut through the electrical outlet cover
shears and nibblers are more expensive, and probably will have wider cuts, but may be easier.
anyone who has experience cutting thin sheet metal, please guide me on which tool to use.
(btw, i'm looking for something cheapish, or rentable)
Tin snips will work fine from what you have described. They are not very expensive and can be handy to have around for other things later.
NASA Hubble Space Telescope Daily Report #5048
NASA Hubble Space Telescope Daily Report #5048
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