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## Integral calculus and Laplace transforms

Hi,

I am trying to solve the equation below to compute the dynamic pressure on a vertical sluice gate subject to rotational oscillation (q(t)) which is a sinusoidal function.

I am stumped evaluating the last integral that contains the cos (fo(s)) term for the phase function. It is not obvious to me how the integral is evaluated given the variable "s" is not the variable that is the subject of the integral.

When I type the equation below into Mathcad it gives an error saying the "s" variable is undefined.

Any clues to assist in solving this equaition would be appreciated.

Cheers

ross

and g = acceleration due to gravity (g = 9.81 m/sec2) and m is a "small positive constant".

y = vertical displacement, x = lateral displacement, q,r,do are constants.

15 REPLIES 15

## Re: Integral calculus and Laplace transforms

Your problem might be units. g is gravity, s has no defined units, mu is unitless. What would the radical evaluate to? If phi is phase, then it's radians (unitless.)

What units should s have to make the phase function work?

## Re: Integral calculus and Laplace transforms

As I interpret that function with all those missing constants (d0, tau) and functions (theta.t, theta) it depends on just the two parameters y and t and Mathcad is right, the s in Phi.o(s) in the last integral IS undefined.

So either you would have to check the formula or make s and additional parameter of the function.

What can you tell about Phi.s(x,y,t) - especially the meaning of the index s?

Proper units will be the next problem, I guess.

## Re: Integral calculus and Laplace transforms

Fred and Werner,

Thanks for taking the time to review my problem.

Equation 21 I have posted is derived using Laplace transforms and Cosine transforms, hence the variable "s" comes from the Laplace operator. The first 2 integrals in equation 21 are integrated over "s" so there is no problem with these. The third integal is integrated over tau which is a time based function so the integral can be evaluated successfully if the cos(phio(s)) term is omitted. What I do not understand is how the phase function phio(s) can be evaluated within this integral with the Laplace variable "s" undefined. For the phase function phio(s) to be unitless (i.e. radians) this implies that "s" must be in units of the inverse of the gravitational constant g or units of sec2/m. This phase function appears in more than one tech paper by the author so it doesn't seem to be an error.

Cheers Ross

## Re: Integral calculus and Laplace transforms

So should it be a formula in time-domain or in s-domain? Looks like things got mixed up or there is a double meaning of s.

## Re: Integral calculus and Laplace transforms

From the phase fct definition, there is clearly a minimum threshold value for s, which is mu^2/(4g). Presumably, this has some physical meaning. If s is the Laplace variable, is it possible that the cos phi(s) term should be inside the s-integral, and that it may have slipped position when the double integral term was separated?

Lou

## Re: Integral calculus and Laplace transforms

Lou, Werner and friends,

One of my work collegues has I believe pointed me in the right direction. The final integral needs to be evaluated assuming the Phase Function fo(s) is a constant. The resulting expression which contains the "s" variable is then inserted in the middle integral which is integrated over "s".

The way I achieved this was to evaluate the final integral symbolically in MathCad (goodness knows how you write software to do this!!) which gave a nasty looking expression in "s". This was then substituited in the middle integral of Equation 21 and evaluated over "ds". The results look plausible in terms of expected magnitude and phase of the pressure p(x,y,t).

Thanks for your interest and assistance.

Ross

## Re: Integral calculus and Laplace transforms

So you treat the formula as if the ds is not before the last integral but at the end, after the dtau?

Hmm!?

## Re: Integral calculus and Laplace transforms

Werner,

I have attached the mathcad file where I solve the third integral in equation 21 symbolically which gives a result containing the variable "s" and then insert the result into the second integral and integrate over "ds". Let me know what you think.

Cheers

ross

## Re: Integral calculus and Laplace transforms

As I see it you have definitely changed the formula given in the way I suspected above. I can't judge if this would make any sense for your application.