Talk:Clock angle problem

Mathematics C‑class Low‑priority

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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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 this:

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

The first 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 is wrong.

The fourth is ok.

The other example was

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

and should become

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

so the example should like

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

Tricky, and awkward indeed

.

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)[reply]