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6-Contributor

## Issue with calc

Hi

I have attached an mcdx file. I don't understand what the issue is when lambda*c is raised to the power of e. Please let me know a work around for this problem.

Sriprakash

7 REPLIES 7
21-Topaz II
(To:sshastry)

Hi,

you have to check the formula, t may be that is missing even a term.

6-Contributor
(To:sshastry)

F.M.

There is no term missing. I see that you have changed lambda to the power of 3 instead of 2. However, even then I don't get a result. There is something wrong. Kindly help. The problem seems to be with e to the power of lambda X c.

Sriprakash

21-Topaz II
(To:sshastry)

Hi S. S.,

You have, however, multiplied lambda for the unit meter, which is incorrect.

Correcting results:

Greetings

F.M.

6-Contributor
(To:-MFra-)

‌Hi F.M

LAmbda is a factor without units. However the program insists on giving it a unit of 1/m. So I multiplied the units by m to ensure that the result is without units. Anyway, the problem at hand seems to be to find a way out to compute e raised to lambda x c. Kindly let me know if there is a solution. Do you feel this is due to the units of lambda itself.

Sriprakash

21-Topaz II
(To:sshastry)

if lambda must be dimensionless, then c must also be such!

there is something else to be corrected:

23-Emerald III
(To:sshastry)

It's not "the program insists on giving it a unit of 1/m". That is YOUR doing.

YOU gave units to k, h and Ecm. With the formula provided for lambda, (without you multiplying it with m) the unit of 1/m comes out.

And, as FM states, lambda MUST have the unit 1/m or else c cannot have unit m. Note that you are using lambda time c as the argument to the sin as well as the exp functions. In both those functions the argument cannot have a unit; so the units of lambda and c must cancel out.

Luc

6-Contributor
(To:LucMeekes)

Luc

Many thanks for throwing light on the problem.

Sriprakash

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