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

${\begin{aligned}&=\left|0.5^{\circ }\times (60H+M)-6^{\circ }\times M\right|\\&=\left|0.5^{\circ }\times (60H-11M)\right|\end{aligned}},$

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)