Home PageMindat NewsThe Mindat ManualHistory of MindatCopyright StatusManagement TeamContact UsAdvertise on Mindat

Donate to MindatSponsor a PageSponsored PagesTop Available PagesMindat AdvertisersAdvertise on MindatThe Mindat Store

Minerals by PropertiesMinerals by ChemistryRandom MineralSearch by minIDLocalities Near MeSearch ArticlesSearch GlossaryMore Search Options

The Mindat ManualAdd a New PhotoRate PhotosLocality Edit ReportCoordinate Completion ReportAdd Glossary Item

StatisticsThe ElementsMember ListBooks & MagazinesMineral Shows & EventsThe Mindat DirectoryHow to Link to MindatDevice Settings

Photo SearchPhoto GalleriesNew Photos TodayNew Photos YesterdayMembers' Photo GalleriesPast Photo of the Day Gallery

# How to properly measure a sphere?

Posted by Kristi Hugs

Kristi Hugs
May 21, 2012 09:20PM

Hey everyone :)Ok, there seems to be some issues here on how to most accurately measure and list a sphere. I was always told to find the accurate diameter I should measure the circumference and divide by 3.14.

So that is what I have been doing. However, a lady I know, told me that my measurement was off and that when she used her calipers on the same sphere, she got a different number. She told me I was wrong. I do not like being wrong :) but I would like to know for sure :)

So, do calipers give a more accurate measurement? or does measuring around the sphere and dividing by 3.14 give a more accurate measurement?

Thanks all!!

Mira

Amir C. Akhavan
May 21, 2012 11:57PM

There's nothing wrong with the way you do it. And there are no general rules for determining the best way of measuring some value. You can do all sort of things, and as long as you "do it right" they will all give the same result.

Except maybe one "rule": Because complex methods come with pitfalls and the errors of the steps involved in the method add up, one should choose a simple method, and whenever possible, measure the value "directly" with a reliable instrument.

So I'd use a caliper until the diameter gets too large or small for the calipers I got.

One thing is important, though:

Every scientist knows that

*who*does the experiment/measurement is very important.

A.k.a as "wishful thinking".

So instead of aguing with the lady about the methological problems in your measurement, you should take a caliper and compare what you find to what the lady finds ;-)

Edited 1 time(s). Last edit at 05/22/2012 12:04AM by Amir C. Akhavan.

Don Saathoff
May 22, 2012 12:15AM

I feel that there so many variables here that two measurements that agree will be difficult!! Is it a perfect sphere.....are the calipers at the widest dimension.....ia the circumference measurement made with a thin or wide tape or string??!? Take three measurements at different times, average them, and add the abbreviation "Approx." to the measurement. Her measurements and your measurements can't be more thana couple of mm's different.....are they?

Don

Dean Allum
May 22, 2012 12:19AM

Mira,Your questions bring our the Geekiness in many of us mindaters (Amir has really shown restraint here).

While most "quartz spheres" are only approximate, There was a NASA project a decade ago to create perfect quartz spheres. While it met the goal to confirm part of the General Relativity theory (principle?), creating the spheres is pretty cool also.

Here is a link that tells the story:

http://science.nasa.gov/science-news/science-at-nasa/2004/26apr_gpbtech/

It took a team of scientists several months to "measure" these spheres to the required accuracy.

Regards,

Dean Allum

Owen Lewis (2)
May 22, 2012 02:01AM

As suggested by others, I think the very first questiion you need to ask is, 'How accurate do I need my result to be?'. The second question is, 'Is my measuring instrument sufficiently accurate to allow measurement to that level of accuracy?' . When you have the answer to those, you'd be a long way on to answering your own question.If your method of measurement of circumference is a tape measure wrapped around a sphere (assumed perfect (which it won't be) just to simplify the basic discussion), you are unlikely to be able to measure with greater accuracy than about +/- 0.5mm (worse for small spheres) So, at best, a suitably flexible but non-elastic tape should let you determine diameter to an accuracy of about +/- 0.17mm. OTOH a decent (not expensive) gauge will give you a direct read out of diameter to +/- 0.01mm for all diameters of <1mm up to the capacity of the gauge (150mm?). Accordingly, I suggest that a direct measurement of diameter with a good gauge should be at least one order of magnitude more accurate that is the method of circumferential measurement with a good tape and then dividing by pi.

Gord Howe
May 22, 2012 02:39PM

Mira;I agree with Owen. An inexpensive pair of digital calipers taking diameter measurements across centerline would be more than adequate to define the diameter. I would also measure it across several different axis and average it out. 150mm calipers won't measure a 150mm diameter but with a little inginuity a simple jig can be made up that will allow it to. If you have access to micrometers, even better. What you are trying to do with circumferences is tricky and as soon as you start multiplying or dividing you are multiplying any errors accumulated in your procedure.

Gord

Kristi Hugs
May 22, 2012 07:22PM

Thank you all! I love you :) I have invested in a nice digital caliper that measures from 0-150mm. None of my spheres are that big, so this will work just fine. It seems to me that there was such a discrepancy between my measurement and hers, I really had to wonder if I was doing something wrong.Thank you all again. You are awesome!!

Douglas Merson
May 22, 2012 10:57PM

If you are going to use a tape to measure a sphere, you should find a pi tape. When measured around the true diameter of the sphere, it will give you the diameter without having to do the math. The best method is the digital calipers that Mira got unless one has access to a CMM.Doug

Keith Wood
September 28, 2012 04:52AM

Only slightly related. On the topic of roundness, I read some time back that physicists had spent several years studying the roundness of the electron. I never quite understood how, but the result of the study was that the electron was declared to be the most perfectly round thing known, so much so that if its diameter were the size of the orbit of Neptune, its roundness would vary by no more than the width of a human hair. I think they planned to keep the study going longer to refine their measurements. Maybe with time they will get to Pluto with the human hair!Amazing what they can figure out!

Mira, maybe those guys could give you some tips on measuring spheres! They seem to be doing a pretty good job!

Edited 1 time(s). Last edit at 09/28/2012 04:53AM by Keith Wood.

Jolyon & Katya Ralph
September 28, 2012 09:28AM

> I was always told to find the accurate diameter I should measure the circumference and divide by 3.14. Unless in Alabama, when you have to divide it by 3 ;-)

Owen Lewis (2)
September 28, 2012 11:03AM

Jolyon,You mis-remember or had a bad teacher :-) Measuring a circumference (a little harder to do accurately by any other method) is best done by measuring the diameter (easy to do accurately) and then multiplying that value by π (pi) (3.146.......etc).

If you can't measure the diameter accurately, it can be calculated as follows:

1. For objects of large diameter (a planet even), measure the maximum angle subtended between accurate observation of the points furthest apart on the opposite sided of the sphere. Call that angle θ. Call the diameter d, (d=2r where r is the radius of the sphere). Call the distance from the observer to the centre of the sphere D; measure or calculate D.

tan θ = 2r/D from which, by substituting the determined values for tanθ and D, out pops the value for 2r (=d).

2. If the object is of a more handy size and you have no calipers suitable to measure d, but you have an accurate balance, then determine first the SG of the sphere by the hydrostatic method and hence determine its volume. The formula for the volume (V) of a sphere is stated as:

V=(4/3)*π(r^3)

Since we know π and have determined V, getting the value of 2r (=d) becomes a matter of simple maths.

P.S. Alabama. I had often wondered why the holes in the doughnuts and bagels one buys there are never round :-D However, I've noticed the same phenomenon in some other places too!

Edited 1 time(s). Last edit at 09/28/2012 11:28AM by Owen Lewis (2).

Dennis Tryon
September 28, 2012 03:40PM

pi is not 3.146---, it 1s 3.14159---.Dennis

Edited 1 time(s). Last edit at 09/28/2012 03:44PM by Dennis Tryon.

Mineralogical Research Company
September 28, 2012 05:08PM

For those interested in more explicit mathematical details, please see this.
Adam Kelly
April 14, 2014 05:22PM

As usual I see the humor in everything, and the joke starts like this.How many rockhounds does it take to measure a sphere?

Sorry, only registered users may post in this forum.

Mindat.org is an outreach project of the Hudson Institute of Mineralogy, a 501(c)(3) not-for-profit organization.

Copyright © mindat.org and the Hudson Institute of Mineralogy 1993-2017, except where stated. Mindat.org relies on the contributions of thousands of members and supporters.

Privacy Policy - Terms & Conditions - Contact Us Current server date and time: May 22, 2017 16:26:50

Copyright © mindat.org and the Hudson Institute of Mineralogy 1993-2017, except where stated. Mindat.org relies on the contributions of thousands of members and supporters.

Privacy Policy - Terms & Conditions - Contact Us Current server date and time: May 22, 2017 16:26:50