June 24, 2013

## How do I secure fabric to wood?

I am making a headboard, and I sewed fabric around 1′x1‘ squares, and I need to attach them now to the 1/4″ plywood that I have. I have tried spray adhesive, all purpose glue and even tried super glue. The squares keep coming off! Anyone have any other ideas? I really need to limit using nails and screws since I live in an apartment, and cant have too many holes in the walls.

Im not sure how much tension will be on this fabric, whether or not your wrapping it around the wood really tight. If it is not super tight i would try hot glue… That is one of the few things hot glue is good for in my opinion. The problem is that all glues and adhesives take time to set, and your material will move. If possible i would use a limited amount of staples in combination with the hot glue.
Make sure you get the smallest staples(for a t50 staple gun) you can find because your would is only 1/4″ thick.

Let me know if i can help in any other way.

## How to find an estimate value of three variables?

I have an equation with six variables, let’s say x1, x2, x3, x4, x5 and x6. I also have a 170 samples for each of x1, x2, and x3. how can I find an optimal estimate of the other three variables x4, x5, and x6. could it be done with iteration in matlab?? and how can I do that!

If you have more variables than equations, the possible number of solutions is infinite.

You need to have a different approach.

## How to determine x and y using the regular falsi and secant method?

The sum of two numbers is 20. If each number is added to its square root, the
product of the two sums equals 155.55. We want to determine the two numbers to within 10^-4.
How can the two numbers say x and y can be determined by
•applying the regular falsi method with the starting x0 = 0 and x1 = 7?
•applying the secant method with the starting x0 = 8 and x1 = 10?

Right, we have two numbers x and y such that

x + y = 20 and (x + ?x).(y + ?y) = 155.55

We note that x and y are interchangeable, that is, if x = a and y = b is a solution, then so is x = b and y = a. Assuming that x ? Y (which is easily checked since then x = y = 10) and that x < y, then x 10.

We may reduce the problem to a single variable by substituting the first equation above into the second to obtain

(x + ?x).[(20 - x) + ?(20 - x)] = 155.55

so that the problem reduces to finding the root of f(x) = 0, where

f(x) = (x + ?x).[(20 - x) + ?(20 - x)] – 155.55.

As a first step in the solution we substitute a few integer values in the range x = 0 to 10 (or graph it, if you have a graphing calculator), which shows that f(x) changes sign from negative to positive as x increases through 6 to 7.

The regular falsi (frequently known as the ‘false position’ method in the UK) and secant techniques are suggested for the solution of this problem. These are numerical methods which estimate the linear slope of the function from two values of f(x) to obtain an improved estimate of the root. The regular falsi method selects two points either side of the root ie of opposite signs, which guarantees convergence, while the secant method simply extrapolates the straight line through two values of f(x) to its point of intersection with the x-axis.

It should be noted that both methods in their simple form have significant limitations, and are only employed for educational purposes in this form. In particular, starting values must give reasonably accurate values for the slope of f(x), otherwise the approximation may converge very slowly or not at all. As a result, I have changed the suggested starting values for both methods to avoid this problem.

(a) the regular falsi method

This employs initial estimates x(0) and x(1) lying either side of the root to calculate improved estimates x(n+1) for the position of the root from

x(n+1) = [x(0).f{x(n)} - x(n).f{x(0)}]/[f{x(n)} - f{x(0)}]

Note that the position and value of x(0) remains unchanged, while the improved value x(n) of x(1) moves towards (but remains on the positive side of) the root. As a result the rate of convergence falls off as the approximation progresses, and may become very slow if the initial estimates x(0) and x(1) are too far from the root.

As a result, I adopted x(0) = 6 and x(1) = 7 for the starting values in the table below. The precision of 10^(-4) requires calculations to be performed to 5 decimal places.

N . . . . X(n) . . . . . .f{x(n)}
0 . . . 6.0 . . . . . . -5.64205
1 . . . 7.0 . . . . . . .4.62302
2 . . . 6.54964 . . . 0.37401
3 . . . 6.51547 . . . 0.02678
4 . . . 6.51303 . . . 0.00190
5 . . . 6.51286 . . . 0.00009
6 . . . 6.51285 . . . 0.00003

(b) the secant method

The numerical approximation in this case is similar to that used in the Newton-Raphson method, and given by the equation

x(n+1) = x(n) – f{x(n)}.[x(n) - x(n-1)]/[f{x(n)} - f{x(n-1)}]

Again, I employed starting values closer to the root than those suggested, in order to obtain well-behaved convergence. The results obtained were

n . . . . .x(n) . . . . . . F{x(n)}
0 . . . 8.0 . . . . . . 11.90190
1 . . . 7.0 . . . . . . . 4.62302
2 . . . 6.36487 . . . -1.54583
3 . . . 6.52402 . . . . 0.11406
4 . . . 6.51308 . . . . 0.00240
5 . . . 6.51284 . . . .-0.00004

Hence the final solution was x = 6.51284 and y = 13.48716 with a precision around 1 in the last digit. As a check these give

(x + ?x).(y + ?y) = 155.5499