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Topic: TO DIVIDE AN ANGLE IN ANY NUMBER OF EQUAL PARTS
Replies: 23   Last Post: Mar 4, 2013 4:05 AM

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 Peter Scales Posts: 192 From: Australia Registered: 4/3/05
Re: TO DIVIDE AN ANGLE IN ANY NUMBER OF EQUAL PARTS
Posted: Oct 9, 2010 11:27 AM

> To: Peter Scales
> From: Shyamal Kumar Das
> Sir,
> Long time, no see ???.i mean no response. Hope ,
> this e-mail finds you in good health and spirit.
> Now, I am writing you this letter because of the
> following reasons:
> 1) I ?ve realized that again I ?ve made mistake
> applying unitary method which is not correct . Again
> sorry sir.
> 2) Any way, what is very much annoying me is the
> drawback of my method. The difference between arc
> length and chord length and how to overcome that have
> spoilt my sleep at night.
> 3) I ?ve calculated out something which ,according to
> me, is very much interesting and
> relevant.
> CALCULATION :
> A) When , the angle subtended at the center= 180
> deg.,
> The difference between arc length and chord length=
> pi*R-2R=R( pi-2 ) =R ( 3.14-2) = 1.14R
> B) When , the angle subtended at the center= 90
> deg.,
> The difference between arc length and chord length=
> pi/2*R- v2R= R( 1.571-1.414)= 0.16R
> C) When , the angle subtended at the center= 60
> deg.,
> The difference between arc length and chord length=
> pi/3*R- R = R( 1.047-1 ) =0.047R
> D) When , the angle subtended at the center= 30
> deg.,
> The difference between arc length and chord length=
> pi/6*R-2Rsin15 deg
> = R(0.5233- 2*0.2588) =
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> =0.0057R
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> E) When , the angle subtended at the center=
> 22.5 deg.,
> The difference between arc length and chord length=
> pi/8*R- 2Rsin11.25deg
> =R(0.3925-
> =R(0.3925-
> =R(0.3925- 2*0.1951)=
> =R(0.3925- 2*0.1951)= =0.0023R
> F) When , the angle subtended at the
> center= 1/5 of 22.5deg = 4.5deg
> The difference between arc length and chord length=
> pi/40- 2Rstn2.25deg
> =R( 0.0785-2*0.0393 ) ???
> OBSERVATION & INFERENCE :
> Hence, it is observed that, when angle subtended at
> the center tends to zero, the difference between arc
> length and chord length also tends to zero. Thus, we
> can infer that when angle subtended at the center is
> 4.5deg, the arc and chord lengths are almost same.
> In my previous letter where I took the example of
> 180deg., to be divided into 5 equal parts, the
> cut-off arc / chord length has been taken for
> graduation, against subtended angle = 4.5deg. at
> the center.
> Please correct me ,if I am wrong. I am eagerly
> waiting for your reply on this. If you are O.K with
> my calculation, observation and inference, then I
> shall only hope that you will approve my method which
> is almost correct. Thanks and regards. Shyamal Kumar
> Das 09.10.2010

Sir,

Sorry to be tardy in replying. It is not due to lack of interest, but pressure of other duties!

The accuracy you need to achieve depends on the use to which you intend the method to be applied.

I had thought that the following two benchmarks may be useful in your assessment:

1. For student or artisan use, with a school-type compass and reasonably sharp pencil, to obtain a result within 1/4 of a degree, and

2. For precise engineering drawing on good quality paper with a professional compass and a hard, very sharp pencil to obtain a result within 1/10 of a degree.

Give these numbers some thought. You may think they are a bit too precise. The matter is open to debate.

You might like to experiment and see what accuracy you can achieve. Each time you place the compass point, or scribe an arc, you introduce some small error. Random errors tend to cancel out. Systematic errors tend to accumulate.

Eventually you should be able to refine your rules of application to achieve some stated accuracy.

Then it will be up to users to decide whether your method is more or less useful in practice compared to other methods that are available.

Regards, Peter Scales.