Talk:Clock angle problem

Reverting to last sensible version

Lots of content seems to have been deleted. I've reverted back to an older edit. Apologies if I've accidently deleted some content. Nick Connolly 03:25, 12 October 2007 (UTC)

Note:

    Why is + M is needed after the $(60H)$ in the above equation?
    Say the time is 5:24.  We can't just calculate the 5.  We need to immediately 
    and NOT separately include the minutes i.e include the 24 minutes.  Because 5:24
    means the hour handle is NOT exactly at 5 now.  Since it's 24 min past then the hour
    handle would have moved a little bit away or a little bit past from the 5 O'Clock 
    mark.

I removed this from the article and posted it here. It should be in the discussion not formally placed in an article. --Pr. Ultracrepidarian (talk) 02:03, 26 October 2010 (UTC)

Degrees or no degrees

User WillemienH first made this edit, which I undid, and then made this edit which I undid too. Both edits left the article in a inconsistent state. Adding the degree units seems a bit awkward, so I propose to leave all units out altogether. - DVdm (talk) 17:20, 30 October 2015 (UTC)

I guess we were busy editing the page at the same time (sorry), i only later realised it needed more editing. My idea is to add degrees everywhere, also I wanted to make the formula's to be more readable by non-mathicatically proficient readers (add \times where there are multiplications, remove unneeded detail, but do add ^{\circ} where it is about degree. and that ilk. WillemienH (talk) 18:12, 30 October 2015 (UTC)

Ah, that explains. Your last edit had these:

$\theta _{\text{hr}}={\frac {1}{2}}M_{\Sigma }={\frac {1}{2}}0.5^{\circ }\times (60H+M)$

$\theta _{\text{min.}}=6M6^{\circ }\times M$

$\theta _{\text{hr}}={\frac {1}{2}}0.5^{\circ }\times (60\times 5+24)=162^{\circ }$

$\theta _{\text{min.}}=6^{\circ }\times 24=144^{\circ }$

and the previous edit had this:

${\begin{aligned}\Delta \theta &=\vert {\frac {1}{2}}(60\times 2-11\times 20)\vert ^{\circ }\\&=\vert {\frac {1}{2}}(120-220)\vert ^{\circ }\\&=50^{\circ }\end{aligned}}.$

The first line was

$\theta _{\text{hr}}={\frac {1}{2}}M_{\Sigma }={\frac {1}{2}}(60H+M)$

and, with degrees, should become

$\theta _{\text{hr}}=0.5^{\circ }\times M_{\Sigma }=0.5^{\circ }\times (60H+M).$

The second line was

$\theta _{\text{min.}}=6M$

and should become

$\theta _{\text{min.}}=6^{\circ }\times M.$

The third line is wrong.

The fourth line is ok.

The other example was

${\begin{aligned}\Delta \theta &=\vert \theta _{\text{hr}}-\theta _{\text{min.}}\vert \\&=\vert {\frac {1}{2}}(60H+M)-6M\vert \\&=\vert {\frac {1}{2}}(60H-11M)\vert \end{aligned}}$

and should become

${\begin{aligned}\Delta \theta &=\vert \theta _{\text{hr}}-\theta _{\text{min.}}\vert \\&=\vert 0.5^{\circ }\times (60H+M)-6^{\circ }\times M\vert \\&=\vert 0.5^{\circ }\times (60H-11M)\vert .\end{aligned}}$

So the example should be

${\begin{aligned}\Delta \theta &=\vert 0.5^{\circ }\times (60\times 2-11\times 20)\vert \\&=\vert 0.5^{\circ }\times (120-220)\vert \\&=50^{\circ }.\end{aligned}}$

Tricky, and awkward indeed . I.m.o. the most tricky and confusing part for these non-mathicatically proficient readers is this:

${\begin{aligned}&=\vert 0.5^{\circ }\times (60H+M)-6^{\circ }\times M\vert \\&=\vert 0.5^{\circ }\times (60H-11M)\vert .\end{aligned}}$

Right?

Does anyone have access to the original source, so we can verify and, in the spirit of good wikipedians, stick with the source? - DVdm (talk) 18:55, 30 October 2015 (UTC)

No :) I will rewrite the article tomorow to how I wanted it:

• Replace 1/2 everywhere with 0.5 (because 0.5 is used in the text just above it)
• add \times everywhere there is a multiplication
• add degrees where appropriate (so where it is inside the formula it just stays where it was before.

I don't think we need to worry about the original source, it just needs to be consistent and understandable for The man on the Clapham omnibus (or his daughter, who does not know mathematics) . WillemienH (talk) 20:31, 30 October 2015 (UTC)

I do think we need to worry about the original source . Anyone can come here and file for deletion of the entire article if the source is not available or if it doesn't directly support the content. Anyway, the new version okay. You forgot to adjust the fnal example, so I took care of that. Cheers. - DVdm (talk) 11:16, 31 October 2015 (UTC)

requested pending changes protection

I noticed this page is quite often corrected to what the editor thinks are the right equatrions.

To prevent this a bit I requested for pending changes: level 1 protection see Wikipedia:Protection policy#Pending changes protection and

Wikipedia:Requests for page protection#Clock angle problem

Hope you agree. WillemienH (talk) 10:33, 23 January 2016 (UTC)

The protection has been approved:

• Clock angle problem (edit | talk | history | links | watch | logs)

pending changes: level 1 : quite often editors want to edit this page to what they think are the correct formulas while it is correct as it is now.WillemienH (talk) 10:22, 23 January 2016 (UTC)

Pending-changes protected for a period of 1 year, after which the page will be automatically unprotected.--Ymblanter (talk) 11:04, 23 January 2016 (UTC)

Wikipedia:Requests for page protection/Rolling archive#23 January 2016

WillemienH (talk) 19:16, 23 January 2016 (UTC)

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