pro e can plot curvs with equations we give. I am in urgent need of the following equation.Any mathematician kindly help me..:

i need equation for a twisted pair wire(2 wires only wound in form of a helix along a circle.Such as a two

TWISTED CABLE PICTURE

The figure below shows a twisted cable .i need equation for such a twisted curve made along a helix. as such a twisted wire in form of a telephone cable.RIGHT!!

TELEPHONE CABLE

Sameerdubey.sbp (talk)

If you need parametric equations, try (r*cos(t), r*sin(t), a*t) and (-r*cos(t), -r*sin(t), a*t) for the two curves, where r and a are constants you can choose to decide the proportions. Rckrone (talk) 04:06, 9 July 2012 (UTC)[reply]

BTW, you shouldn't double post on two Ref Desks like this, it causes confusion. At most, you should post a note on one desk (either here or the Science Desk), directing people to your question on the other desk. StuRat (talk) 04:52, 9 July 2012 (UTC)[reply]

ok pardon me.note:this is posted in science desk also.now please help me. 203.197.246.3 (talk)

Do you have a question about this ? If not, try reading interpolation. StuRat (talk) 22:03, 9 July 2012 (UTC)[reply]

I do, yes it just took me a bit longer to compose and here it is.

In general, if I need to do some interpolation, is it better to interpolate the input a and then evaluate f(a) or should I evaluate f(a) and then interpolate f(a). f is a highly nonlinear function and a little expensive to compute as well. In addition, between any two adjacent values, I have to interpolate six values or so. Because f is a little expensive, I was thinking of just plugging in a, get f(a), and then use cubic splines to interpolate to the higher resolution. But then someone just told me to always interpolate the input, never the output. I guess that makes sense because f is non-linear but I thought cubic splines will take care of that. Typical length of a is about 120 and after interpolation the length will be on average about 1200. Any comments? What do the people around here think? 192.12.184.6 (talk) 22:13, 9 July 2012 (UTC)[reply]

I'd think it would be OK to interpolate the output. You might get slightly different results, but I don't see how it can be argued that either one is more correct than the other. I suspect that the person you talked to started with the assumption that the interpolated input version is correct, so therefore judge the slightly different interpolated output version to be incorrect. But, just to be sure, I suggest you try both methods, and see which is better. StuRat (talk) 22:20, 9 July 2012 (UTC)[reply]

I can think of one argument for interpolating the output, in that it may allow you to more evenly distribute your data points (say if you add more or fewer points, depending on the distance between existing points). So, you might want to determine if this is a significant factor, in your case. StuRat (talk) 22:23, 9 July 2012 (UTC)[reply]

I don't know what is meant here by interpolating the inputs. Wouldn't you then have to evaluate f for all the intermediate points whereas the point of the interpolation was to evaluate f at only a few points? Dmcq (talk) 13:18, 10 July 2012 (UTC)[reply]

They only have a few data points, and both want to apply some formula to them, and interpolate between them. For example, the data points could be 15% of a group of people's gross incomes, and they want to plot all their net incomes as the output. StuRat (talk) 19:23, 10 July 2012 (UTC)[reply]

I'm missing something. Suppose sin(x) is very difficult to calculate, what would one do differently with it when setting up some scheme to approximate values between x=0 and pi to some desired accuracy or using a limited number of evaluations of sin(x)? Dmcq (talk) 14:11, 11 July 2012 (UTC)[reply]

Well, just using 0 degrees and 90 degrees as the inputs, if we did a linear interpolation to get one input variable, or one output variable, this would be the diff (X in degrees):

INTERPOLATED    INTERPOLATED
    INPUT          OUTPUT
============    ============
 X    SIN(X)     X    SIN(X)      
--    ------    --    ------
 0         0     0         0
45    0.7071     -       0.5   
90         1    90         1

Of course, we could do a lot better than a linear interpolation, and, with more actual data points, the difference would be far less, even with linear interpolation. StuRat (talk) 18:35, 11 July 2012 (UTC)[reply]

But that input interpolation has calculated three values of the function rather than 2. That doesn't sound like interpolating to me, it sounds like just calculating the value of the function. Dmcq (talk) 00:04, 12 July 2012 (UTC)[reply]

The 45 was obtained by linear interpolation midway between 0 and 90. StuRat (talk) 00:06, 12 July 2012 (UTC)[reply]

This doesn't make any sense. You're not "interpolating" the input, you're just figuring out to which input you want to compute the output. We've established that the desired input is 45°. Now the question is whether to compute the function value at 45° directly (which is expensive), or by interpolation (which is inaccurate). The answer is of course "depends on how expensive is the function, and how much accuracy you need". -- Meni Rosenfeld (talk) 08:40, 12 July 2012 (UTC)[reply]

The 0 and 90 are supposed to represent actual data determined by surveys, etc., while the 45 does not. Thus, it was interpolated between the actual data points. This sine wave example isn't that good, so perhaps we should return to my example, where various people were polled to find their gross income, and say we got adjacent data points of $25,362 and $26,119. We now want to interpolate data points between those (not necessarily linearly), and put them through our calculation process to find the estimated net income for each case, and we have the choice of doing the interpolation either before or after. StuRat (talk)

Please give more details, I'm struggling to fill in the blanks in your last example (better the OP but it seems you understand what he's after). -- Meni Rosenfeld (talk) 18:36, 14 July 2012 (UTC)[reply]