# Question about RF wavelengths and steel buildings **URL:** \ **Category:** Factual Questions\ **Created:** [January 21, 2003, 10:41pm UTC](https://boards.straightdope.com/t/question-about-rf-wavelengths-and-steel-buildings/150078 "2003-01-21T22:41:50Z")\ **Posts on this page:** 5\ **Page:** 1
**Author:** ![friedo](https://avatars.discourse-cdn.com/v4/letter/f/8edcca/32.png) [@friedo](https://boards.straightdope.com/u/friedo)\ **Post date:** [January 21, 2003, 10:41pm UTC](https://boards.straightdope.com/t/question-about-rf-wavelengths-and-steel-buildings/150078/1 "2003-01-21T22:41:50Z")
A friend of mine is interested in the answer to the following question: > [@](#): > > Assume a building made of (grounded) steel studs. Assume the openings > between the studs is slightly less than one wavelength of the RF you’re > interested in. Will the RF get through, or will the studs act like a faraday > cage? If the RF gets through, will it be attenuated? If so, is the > attenuation related to the size of the space between the studs? If yes, how? I told him that I’m pretty sure the RF can’t get through the holes in the building if they are smaller than the wavelength, which I’m pretty sure is correct. I don’t know anything about the attenuation issues, though. Help from Smart Physics People is appreciated. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [January 21, 2003, 10:52pm UTC](https://boards.straightdope.com/t/question-about-rf-wavelengths-and-steel-buildings/150078/2 "2003-01-21T22:52:33Z")
In a word: [yes.](http://www.physlink.com/Education/AskExperts/ae176.cfm) ---
**Author:** ![Napier](https://avatars.discourse-cdn.com/v4/letter/n/ce73a5/32.png) [@Napier](https://boards.straightdope.com/u/Napier)\ **Post date:** [January 21, 2003, 10:59pm UTC](https://boards.straightdope.com/t/question-about-rf-wavelengths-and-steel-buildings/150078/3 "2003-01-21T22:59:30Z")
“RF can’t get through the holes in the building if they are smaller than the wavelength” This is how I understand it, from the point of view of building enclosures to keep electromagnetic interference in or out. Note that the dimension of the hole for this purpose is its longest dimension. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [January 21, 2003, 11:22pm UTC](https://boards.straightdope.com/t/question-about-rf-wavelengths-and-steel-buildings/150078/4 "2003-01-21T23:22:47Z")
Have your friend get a copy of the book _Noise Reduction Techniques in Electronic Systems_ by Henry W. Ott, John Wiley and Sons. He has an excellent dissertation on shielding. To quote from the book: _As a practical matter, the intrinsic shielding is of less concern that the leakage through seams, joints and holes. … The amount of leakage from a shield discontinuity depends mainly upon 3 things; 1) **The maximum linear dimension** , 2) The wave impedance, 3) the frequency._ He goes on to say _A rectangular [opening] forms a slot antenna. Such and entenna even if very narrow can cause considerable leakage it is longer than 1/100 wavelength._ He then gives a formula for attenuation of a shield having holes in it of maximum dimension l at wavelength L. S in db = 20log(L/2l) So it looks like the length of the metal studs is more critical than the spacing. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [January 21, 2003, 11:27pm UTC](https://boards.straightdope.com/t/question-about-rf-wavelengths-and-steel-buildings/150078/5 "2003-01-21T23:27:39Z")
By the way. That formula was for one hole and the shielding effectiveness (attenuation) is reduced by approximately the square root of the number of holes. I don’t believe metal studs will attenuate reasonably high frequencies very much.