ThreeDimensionalConfigurations/ChallengesPt6 - Revision history

Jet53man: /* For Specific Tip Angle */

2022-02-19T22:31:20Z

For Specific Tip Angle

← Older revision		Revision as of 18:31, 19 February 2022
Line 1,663:		Line 1,663:


	Then, for this specific tip angle, the Riemann flow velocity is,		<span id="SpecificTipAngle">Then, for this specific tip angle</span>, the Riemann flow velocity is,

	<table border="0" cellpadding="5" align="center">		<table border="0" cellpadding="5" align="center">

2022-02-19T19:39:06Z

For Specific Tip Angle

← Older revision		Revision as of 15:39, 19 February 2022
Line 1,797:		Line 1,797:
	</table>		</table>

	Hence,		<span id="Orthogonal">Hence,</span>

	<table border="0" cellpadding="5" align="center">		<table border="0" cellpadding="5" align="center">

2022-02-19T19:37:35Z

Plot Off-Center, Slightly Flattened Ellipse

← Older revision		Revision as of 15:37, 19 February 2022
Line 3,663:		Line 3,663:
	</table>		</table>

	As we have already stated — but setting <math>z' = 0</math> and ignoring the <math>\mathbf{\hat{k}'}</math> component because there is no motion in that direction —		As we have [[#gradP\|already stated]] — but setting <math>z' = 0</math> and ignoring the <math>\mathbf{\hat{k}'}</math> component because there is no motion in that direction —
	<table border="0" cellpadding="5" align="center">		<table border="0" cellpadding="5" align="center">

Line 3,696:		Line 3,696:
	\boldsymbol{\hat\imath'} \biggl(\frac{x'}{a^2}\biggr)		\boldsymbol{\hat\imath'} \biggl(\frac{x'}{a^2}\biggr)
	+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{( y'\cos\theta )\cos\theta}{b^2}\biggr]		+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{( y'\cos\theta )\cos\theta}{b^2}\biggr]
	+ \biggl[\frac{( z_0 + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}		+ \biggl[\frac{( z_0 + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\} \, .
	</math>		</math>
	</td>		</td>
	</tr>		</tr>
	</table>		</table>

			Since, properly normalized, <math>\nabla G'</math> is identical to <math>\nabla P'</math>, and since we have [[#Orthogonal\|already shown]] that <math>\mathbf{u'}_\mathrm{EFE}</math> is everywhere orthogonal to <math>\nabla P'</math>, it must be true that <math>\nabla G'</math> is everywhere orthogonal to <math>\mathbf{u'}_\mathrm{EFE}</math>.

			<b><font color="red">Hooray!!</font></b>

	=See Also=		=See Also=

2022-02-19T19:30:50Z

For Specific Tip Angle

← Older revision		Revision as of 15:30, 19 February 2022
Line 1,774:		Line 1,774:
	</table>		</table>

	As we have already stated,		As we have [[#gradP\|already stated]],
	<table border="0" cellpadding="5" align="center">		<table border="0" cellpadding="5" align="center">

2022-02-19T19:28:45Z

For Arbitrary Tip Angles

← Older revision		Revision as of 15:28, 19 February 2022
Line 1,268:		Line 1,268:
	</table>		</table>

	Hence,		<span id="gradP">Hence,</span>
	<table border="0" cellpadding="5" align="center">		<table border="0" cellpadding="5" align="center">

2022-02-19T19:06:04Z

Plot Off-Center, Slightly Flattened Ellipse

@@ Line 3,560: / Line 3,560: @@
++
 \boldsymbol{\hat\jmath'} \biggl[ \frac{y' - y'_0}{(y')^2_\mathrm{radius}} \biggr]
 </math>
    </td>

2022-02-19T18:38:11Z

Plot Off-Center, Slightly Flattened Ellipse

@@ Line 3,570: / Line 3,570: @@
 <tr>
    <td align="right">
-<math>\frac{1}{2}\nabla P'</math>
+<math>\frac{1}{2} \nabla P'(x', y', z')</math>
    </td>
    <td align="center">
@@ Line 3,578: / Line 3,597: @@
 <math>
 \boldsymbol{\hat\imath'} \biggl(\frac{x'}{a^2}\biggr)
-+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{( y'\cos\theta - z'\sin\theta )\cos\theta}{b^2}\biggr]
++ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{( y'\cos\theta  )\cos\theta}{b^2}\biggr]
-+ \biggl[\frac{( z_0 + z'\cos\theta + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}
++ \biggl[\frac{( z_0 + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}
 </math>
    </td>

2022-02-19T18:32:53Z

Plot Off-Center, Slightly Flattened Ellipse

@@ Line 3,557: / Line 3,557: @@
    <td align="left">
 <math>
-\boldsymbol{\hat\imath'} \biggl[ \frac{y' - y'_0}{(y')^2_\mathrm{radius}} \biggr]
+\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{(x')^2_\mathrm{radius}} \biggr]
++
-\boldsymbol{\hat\jmath'} \biggl[ \frac{x'}{(x')^2_\mathrm{radius}} \biggr]
+\boldsymbol{\hat\jmath'} \biggl[ \frac{y' - y'_0}{(y')^2_\mathrm{radius}} \biggr]
 </math>
    </td>

2022-02-19T18:23:43Z

Plot Off-Center, Slightly Flattened Ellipse

@@ Line 3,522: / Line 3,522: @@
 </tr>
 </table>
-This seems to work perfectly!
+This seems to work perfectly!  We used Excel to generate trajectories using this expression and the results matched earlier determinations of these trajectories to machine precision.
 =See Also=

2022-02-18T21:30:37Z

Tipped Frame