tag:blogger.com,1999:blog-41640600369522092382017-05-24T19:42:01.046-07:00Badger MechanicsIntuitive descriptions of fundamental engineering conceptsJoe Schonemannoreply@blogger.comBlogger6125tag:blogger.com,1999:blog-4164060036952209238.post-11689734346095046622012-08-21T16:00:00.001-07:002012-08-22T16:41:35.904-07:00Conservative Fields: Examples<br /><div style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em; text-align: left;"></div><br />In my <a href="http://badgermech.blogspot.com/2012/08/conservative-fields-theory.html" target="_blank">previous post</a> on the basic theory behind conservative vector fields, I described some of the mathematical properties these fields will have:<br /><br />- A conservative field has no curl: <img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{\bigtriangledown}%20\times%20\overrightarrow{F}%20=%20\overrightarrow{0}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /><br /><br />- By extension, a vector of 3 components (M<b>i </b>+ N<b>j + </b>P<b>k) </b>can be checked for conservancy with a component test:<br /><br /><div style="text-align: center;"><a href="http://latex.codecogs.com/gif.latex?\frac{\partial%20M}{\partial%20z}%20=%20\frac{\partial%20P}{\partial%20x}" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\frac{\partial%20M}{\partial%20z}%20=%20\frac{\partial%20P}{\partial%20x}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a> <img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\frac{\partial%20N}{\partial%20z}%20=%20\frac{\partial%20P}{\partial%20y}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /> <img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\frac{\partial%20M}{\partial%20y}%20=%20\frac{\partial%20N}{\partial%20x}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: center;"><br /></div><div style="text-align: left;">- A conservative field can be 'integrated' to find a potential function. Conversely, any gradient of a function is conservative: </div><div style="text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{F}%20=%20\bigtriangledown%20f" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: left;"><br /></div><div style="text-align: left;">This is all good stuff, but I like to see visual examples whenever possible. When we're talking graphs of vector fields and functions in space, this is entirely possible! I'm going to go through some examples now, using the graphs to emphasize the math.<br /><br /><a name='more'></a></div><div style="text-align: left;"><br /></div><div style="text-align: left;">For this post, I'll be staying with 2-dimensional vectors in the x-y plane. We can still write a 2-d vector as (x<b>i </b>+ y<b>j </b>+ 0<b>k</b>) when necessary--if the cross product has to be evaluated, for example. The importance of using 2-d vectors, however, is that it allows me to graph the potential function, rather than simply graphing the level curves of that function. Scalar functions of 3 variables are tough to visualize. Anyway, on with the show!</div><div style="text-align: left;"><br /></div><div style="text-align: left;"><b>Gravity: A Central Force</b></div><div style="text-align: left;"><b><br /></b></div><div style="text-align: left;">I'll start with a toss-up, the force field of gravity. This is probably the most-used example of a conservative field ever, but it still has a cool-looking potential function. <a href="http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation" target="_blank">Newton's Law of Universal Gravitation</a> is given in vector form as:</div><div class="separator" style="clear: both; text-align: center;"></div><div style="text-align: center;"></div><div style="text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{F}%20=%20\frac{-Gm_1m_2}{\left%20|%20\overrightarrow{r}%20\right%20|^2}\hat{r}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: center;"><br /></div><div style="text-align: left;">Here, <b>r</b> is the position vector between point 1 and point 2. (We are dealing with point masses here, a useful approximation). G, m<span style="font-size: xx-small;">1</span> and m<span style="font-size: xx-small;">2 </span>are all scalar parameters that I don't really care about now, so I'm just going to ignore them--they don't change the way the functions will look. By changing the unit vector into its constituent parts, the vector function looks like this:</div><div style="text-align: center;"></div><div style="text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{F}%20=%20\frac{\overrightarrow{-r}}{\left%20|%20\overrightarrow{r}%20\right%20|^3}%20=%20\frac{(-x\hat{i}%20-%20y\hat{j})}{(\sqrt{x^2+y^2})^3}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: center;"></div><div style="text-align: center;"><br /></div><div style="text-align: left;">Graphically, the nature of the force is easy to see. It's also pretty easy to imagine that this force is conservative. Since it's such a common example of a conservative force, I'm not going to bother with any component tests.</div><div style="text-align: center;"><br /></div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><img height="265" src="http://www.euclideanspace.com/maths/geometry/space/fields/gravity.png" style="-webkit-user-select: none; margin-left: auto; margin-right: auto;" width="400" /></td></tr><tr><td class="tr-caption" style="text-align: center;">Image from www.euclideanspace.com; Maple failed me on this one.</td></tr></tbody></table><div style="text-align: left;">An important note: Gravity is what's called a <i>central force</i>. Central forces have 2 main characteristics:</div><div style="text-align: left;"><br /></div><div style="text-align: left;">- They are always oriented towards a central point.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">- Their magnitude varies <i>only </i>as a function of distance from this point. </div><div style="text-align: left;"><br /></div><div style="text-align: left;">Springs are another common central force. Anyway, the main idea is that central forces are <i>always</i> conservative. This fact is extremely important in the study of dynamics.</div><div style="text-align: left;"><br /></div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-OkyPMGY_CXg/UDQEweFCS_I/AAAAAAAAABc/m_UK6UTmTg0/s1600/grav+potential.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://2.bp.blogspot.com/-OkyPMGY_CXg/UDQEweFCS_I/AAAAAAAAABc/m_UK6UTmTg0/s320/grav+potential.jpg" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Gravitational potential. This one I made in Maple.</td></tr></tbody></table><div style="text-align: left;"><br /></div><div style="text-align: left;">The potential function for the field I used above is given by:</div><div style="text-align: left;"><a href="http://latex.codecogs.com/gif.latex?\frac{-1}{\sqrt{x^2%20+%20y^2}}" imageanchor="1" style="clear: left; display: inline !important; margin-bottom: 1em; margin-right: 1em;"><span style="clear: left; color: white; display: inline !important; margin-bottom: 1em; margin-right: 1em;"> </span><span style="clear: left; display: inline !important; margin-bottom: 1em; margin-right: 1em;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\frac{1}{\sqrt{x^2%20+%20y^2}}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></span></a></div><div style="text-align: left;"><br /></div><div style="text-align: left;">However, the actual gravitational function is defined to be negative. (I'm not exactly why, hopefully I can figure it out later). Anyway, the graph of the gravitational potential function is shown at right. Since we're modeling point masses, the function goes to negative infinity and the mesh breaks down at the origin. </div><div style="text-align: left;"><br /></div><div style="text-align: left;">An important feature of potential functions are 'areas of equal potential:' spots on the graph that share the same potential. These are called <i>equipotential,</i> or <i>level,</i> curves and surfaces. This is a 2-dimensional example, so this function's level curves are circular contours. In three dimensions, the level surfaces of equipotential would be spherical. </div><div style="text-align: left;"><br /></div><div style="text-align: left;"><b>Something A Bit More Contrived</b></div><div style="text-align: left;"><b><br /></b></div><div style="text-align: left;">It is not always so simple to observe whether a field is conservative or not, particularly in the cases of fluid flows or complicated electrical fields. Here's a vector function I made up:</div><div style="text-align: left;"><br /></div><div style="text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{F}%20=%20(2y%20+%202xy)\hat{i}%20+%20(2x%20+%20x^2%20-%203cos(y))\hat{j}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-jNopbRYBLYY/UDQKVr_UuPI/AAAAAAAAABo/N9M4P0gRF94/s1600/ex2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="http://3.bp.blogspot.com/-jNopbRYBLYY/UDQKVr_UuPI/AAAAAAAAABo/N9M4P0gRF94/s400/ex2.jpg" width="400" /></a></div><div style="text-align: center;"><br /></div><div style="text-align: left;">Here is its field plot. In this case, it seems that there might be some 'whirlpool' action happening on the left and right sides. Tough to be positive, though. This is where the component test comes in.</div><div style="text-align: left;"><br /></div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\frac{\partial}{\partial{y}}(2y%20+%202xy)%20=%20\frac{\partial}{\partial{x}}(2x%20+%20x^2%20-%203cos(y))%20\rightarrow%20(2%20+%202x)%20=%20(2%20+%202x)" style="margin-left: auto; margin-right: auto;" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></td></tr><tr><td class="tr-caption" style="text-align: center;"><span style="text-align: left;"><span style="font-size: x-small;">Since this vector field does not have a z-component, the other two component tests are trivial.</span></span></td></tr></tbody></table><div class="separator" style="clear: both; text-align: center;"></div><div style="text-align: left;"><a href="http://1.bp.blogspot.com/-OzjiBDwtgxk/UDQOGLR6dVI/AAAAAAAAAB4/DafRFlH4onA/s1600/ex2pot.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="400" src="http://1.bp.blogspot.com/-OzjiBDwtgxk/UDQOGLR6dVI/AAAAAAAAAB4/DafRFlH4onA/s400/ex2pot.jpg" width="400" /></a>Pass! We have a conservative field on our hands, appearances to the contrary. The potential function is then: </div><div style="text-align: left;"><br /></div><div style="text-align: left;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?p(x,y)%20=%202xy%20+%20x^2y%20-%203sin(y)" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: left;"><br /></div><div style="text-align: left;">With <i>p</i> for <i>potential</i>, of course. (Seriously, how hard is that, you textbook writers?) Take a moment to try and visualize the gradient that results from this function--it does indeed match up to the vector field plotted above.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">(Remember, the <a href="http://betterexplained.com/articles/vector-calculus-understanding-the-gradient/" target="_blank">gradient</a> points in the direction of a function's greatest increase).</div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><b>Time for a Counter-example</b></div><div style="text-align: left;"><b><br /></b></div><div style="text-align: left;">Any bets on whether this one will be conservative or not? </div><div style="text-align: left;"><br /></div><div style="text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{F}%20=%20(-x^2y%20+%20150)\hat{i}%20+%20(-xy%20-%20200)\hat{j}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-XGL2j2ah6lk/UDQQM5kATeI/AAAAAAAAACA/lBsEO00QcIc/s1600/ex3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="http://2.bp.blogspot.com/-XGL2j2ah6lk/UDQQM5kATeI/AAAAAAAAACA/lBsEO00QcIc/s400/ex3.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">The graph doesn't really look <i>that</i> different from the previous one, although there are some heavier indicators of curl off to the right. In any case, we can quickly find out that this field does indeed fail the component test:</div><div style="text-align: left;"></div><div style="text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?-x^2%20\neq%20-y" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: center;"><br /></div><div style="text-align: left;">So there's not cool potential function to look at, and finding the work through this field will be a lot tougher than usual. One thing we can do here is find the actual value of the curl. Letting the z-component equal zero;</div><div style="text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow\bigtriangledown%20\times%20\overrightarrow{F}%20=%20(-x^2%20+%20y)\hat{k}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div style="text-align: center;"><br /></div><div style="text-align: left;">With the curl entirely in the <b>k</b> (z) direction, out of the screen. This lines up with what we see above: There is no curl at the origin, but quite a lot at the corners of the graph. It always pays to check for meaning in the math you're doing.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">I currently don't know <i>why </i>exactly a non-conservative field won't yield a potential function. This is a topic I will come back to later, whenever I decide to do a more complete discussion on vector calculus. For now, it'll be back to particle dynamics, which is much more fun.</div><a href="http://latex.codecogs.com/gif.latex?\frac{-1}{\sqrt{x^2%20+%20y^2}}" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><br /></a><a href="http://latex.codecogs.com/gif.latex?\frac{-1}{\sqrt{x^2%20+%20y^2}}" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><br /></a><span style="clear: left; color: white; display: inline !important; margin-bottom: 1em; margin-right: 1em;"><a href="http://latex.codecogs.com/gif.latex?\frac{-1}{\sqrt{x^2%20+%20y^2}}" imageanchor="1" style="clear: left; display: inline !important; margin-bottom: 1em; margin-right: 1em;"></a></span>Joe Schonemanhttps://plus.google.com/118177381588067007820noreply@blogger.com0tag:blogger.com,1999:blog-4164060036952209238.post-36804266808915700532012-08-19T19:00:00.000-07:002012-08-21T16:02:29.604-07:00Conservative Fields: Theory<div class="separator" style="clear: both; text-align: left;"></div><div class="p1"><span class="s1">Next stop: Vector Calculus!</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">To continue in the present direction--a discussion of Energy methods in Particle Dynamics--it’s necessary to take a quick stop over in mathematician territory. An unfriendly place, it is. Look alive.</span></div><div class="p2"><br /><span class="s1"></span></div><div class="p1"><span class="s1">The basic idea that will recur in this discussion is that, in certain types of vector fields, the <i>net work</i> done <i>on an </i><i style="font-weight: bold;">object </i>moving from one point to another<i> </i>is independent of the path taken. This is called <i>path-independence</i>, and is a very important property of <i>conservative fields</i>. </span><br /><br /><a name='more'></a><br /><span class="s1">Conceptually, I like to think of a river: </span></div><div class="p1"><span class="s1"><br /></span></div><div class="p1"><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-J25y7_IlJ98/UDGOWtI7t1I/AAAAAAAAABE/qC_Xlk5wz_E/s1600/river.jpeg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" height="267" src="http://2.bp.blogspot.com/-J25y7_IlJ98/UDGOWtI7t1I/AAAAAAAAABE/qC_Xlk5wz_E/s400/river.jpeg" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Wisconsin is great, but the morning commute can get tiresome.</td></tr></tbody></table>Yeah, there's one. The path you take in swimming across this river has no effect on how much work you do: Maybe I swim straight across, fighting the current the whole way. Maybe you swim far upstream, float downstream, and step out at the same point I did. The river has done the same amount of <i>net work</i> on both of us, regardless of our paths. In this context, it's easy to see what's going on. Things can get trickier, though.<br /><br />Conservative fields are both physically and mathematically important, and there are a lot of ways to answer the question, "What is a conservative field?" These competing definitions are all equivalent, but can get confusing. I'll start with my personal favorite: A conservative field is <i>irrotational,</i> or contains no vortices. I like this because it makes physical sense. The whole idea of conservative fields and forces is that the direction you travel through them does not matter. Imagine a whirlpool: Swimming in circles <i>with</i> the current will give you a free ride. Swimming in circles <i>against</i> the current will cost you. A whirlpool cannot be conservative. </div><div class="p1"><br /></div><div class="p1">Mathematically, this is equivalent to saying "If <b>F </b>is conservative, the <i>curl</i><b> </b>of <b>F </b>is <b>zero</b>:"</div><br /><div class="separator" style="clear: both; text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{\bigtriangledown}%20\times%20\overrightarrow{F}%20=%20\overrightarrow{0}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="" style="clear: both; text-align: left;">Carrying out the cross product leads to a 3-part component test for a vector (M<b>i </b>+ N<b>j </b>+ P<b>k</b>):</div><div class="" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><a href="http://latex.codecogs.com/gif.latex?\frac{\partial%20M}{\partial%20z}%20=%20\frac{\partial%20P}{\partial%20x}" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><span style="color: white;"> </span><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\frac{\partial%20M}{\partial%20z}%20=%20\frac{\partial%20P}{\partial%20x}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a> <a href="http://latex.codecogs.com/gif.latex?\frac{\partial%20N}{\partial%20z}%20=%20\frac{\partial%20P}{\partial%20y}" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\frac{\partial%20N}{\partial%20z}%20=%20\frac{\partial%20P}{\partial%20y}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a> <img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\frac{\partial%20M}{\partial%20y}%20=%20\frac{\partial%20N}{\partial%20x}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div class="separator" style="clear: both; text-align: left;">For 2-dimensional vectors in the <i>x-y</i> plane, only the third equality need be tested.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">So what now? If the vector field passes this component test, we know it is conservative, and we know that the work through that field is path-independent. Specifically, the <i>Fundamental Theorem of Line Integrals, (</i>much like the <i>Fundamental Theorem of Calculus)</i>, states that:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://latex.codecogs.com/gif.latex?\int_{a}^{b}\overrightarrow{F}%20\cdot%20d\overrightarrow{r}=%20f(b)-f(a)" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\int_{a}^{b}\overrightarrow{F}%20\cdot%20d\overrightarrow{r}=%20f(b)-f(a)" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">For conservative <b>F</b>. What about little <i>f</i>, though? Well, it's obviously a scalar function. In fact, the vector field <b>F</b> is actually the gradient of scalar <i>f</i>, and so we call little <i>f</i> the <i>scalar potential</i> of a vector field <b>F</b>.<i style="font-weight: bold;"> </i></div><div class="separator" style="clear: both; text-align: left;"><i style="font-weight: bold;"><br /></i></div><div class="separator" style="clear: both; text-align: left;">I don't know who was the genius who decided to relate capital and lowercase F like this. It's difficult enough to read, it's even worse trying to explain verbally. I often refer to <i>f </i>as the "integral" of <b>F</b>; even though that's not precise, it's the same idea.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Anyway, finding a vector field's potential function is fairly straightforward, I'm not going into it here. Probably later. The important thing is that, much like standard one-dimensional differentials from elementary calculus, the value of the conservative work function depends only on two endpoints--this leads to the <i>loop property</i>, which state: "Travelling on a closed loop through a conservative field yields no net work."</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\oint\overrightarrow{F}%20\cdot%20d\overrightarrow{r}=%200" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">So, to review:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">- Work in a conservative field is path-independent.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">- We can check for the conservancy of a field with a 3-part component test. (Single-component in 2 dimensions). </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">- If a field is conservative, then it must also have a potential function, which is a scalar function. The field in question will be that function's gradient.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">- The net work can then be evaluated by taking values in that potential function between the two endpoints we're interested in.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Coming up: <a href="http://badgermech.blogspot.com/2012/08/conservative-fields-examples.html" target="_blank">Visual Examples for Visual Learners</a>.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"></div>Joe Schonemanhttps://plus.google.com/118177381588067007820noreply@blogger.com0tag:blogger.com,1999:blog-4164060036952209238.post-68190060624438959002012-08-15T18:48:00.001-07:002012-08-15T18:53:47.394-07:00Derivation of the Work/Energy Principle<br /><div class="p1"><span class="s1"></span> </div><div class="p1"><span class="s1"></span> </div><div class="p1"><span class="s1">In my previous post on <a href="http://www.badgermech.blogspot.com/2012/08/acceleration-energy-and-momentum.html" target="_blank">methods for particle dynamics</a>, I stated that, by integrating Newton’s Second Law over a distance, we could make use of ‘Work/Energy’ methods. This is true as far as it goes, but is certainly not the whole truth. It’s important to realize what’s actually happening, lest we fall into a trap of some sort.</span></div><br /><div class="p2"><span class="s1"><br /></span></div><div class="p2"><span class="s1">Newton’s Second Law, is, of course, a vector equation:</span></div><div class="separator" style="clear: both; text-align: center;"><a href="http://latex.codecogs.com/gif.latex?\overrightarrow{F}%20=%20m\overrightarrow{a}" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{F}%20=%20m\overrightarrow{a}" style="font-family: Arial, Helvetica, sans-serif; font-size: 12px; margin-top: 10px; text-align: center;" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="p1"> So I have to be a little more specific when I say 'Integrate over a distance.' Start tossing around vectors willy-nilly, and you'll end up with results that don't make any sense. Anyway, there are two main questions regarding what needs to happen here:</div><div class="p1"><br /><a name='more'></a><br /></div><div class="p1">1) In order to integrate a function properly, there should be some 'infinitesimal' (very small) quantity involved. (i.e., 'dx' would be the tiny change in x that we always see in textbooks). What quantity should we multiply each side of the equation by in order to integrate?</div><div class="p1"><br /></div><div class="p1">2) Which vector operation should be used here, dot or cross product?</div><div class="p1"><br /></div><div class="p1">Answering the first leads to the second. We want to evaluate the total <b>force</b> exerted over some <b>distance</b>, right? For a particle with position vector <b>r</b>, its change of position from moment to moment will be d<b>r</b>. (d<b>r</b>/dt is the velocity of that particle, without the 'dt' d<b>r</b> is just a very small change of position). If we add up the force exerted over every infinitesimal distance, we'll get what we want.</div><div class="p1"><br /></div><div class="p1">Now, to properly extract a result from the vectors <b>F </b>and d<b>r</b>, we have to choose a proper operation. The cross product would produce...a perpendicular vector. Moreover, if <b>F </b>was exerted entirely in the direction of d<b>r</b>, then our vector would have no magnitude--in other words, this is not the proper operation to use here. The dot product it is!</div><div class="p1"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://latex.codecogs.com/gif.latex?\int_{1}^{2}\overrightarrow{F}\cdot%20d\overrightarrow{r}%20=%20\int_{1}^{2}m\overrightarrow{a}\cdot%20d\overrightarrow{r}" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\int_{1}^{2}\overrightarrow{F}\cdot%20d\overrightarrow{r}%20=%20\int_{1}^{2}m\overrightarrow{a}\cdot%20d\overrightarrow{r}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">So we're travelling along a path from point 1 to point 2, integrating along the way. Note that, since the dot product has been invoked, there will be <b>no direction information </b>left here. Energy methods are purely scalar-based, and this is both a blessing and a curse.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">At this point, we're halfway there: The left-hand integral is defined as <b>work</b>. It has units of <b>[N*m]</b> and is better thought of as 'useful work.' More on that later.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">To proceed on the right-hand side, though, requires some trickery. How can we possibly integrate <b>a </b>in terms of d<b>r</b>? By finding an element they share in common: <b>v</b>, the velocity vector.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="" style="clear: both; text-align: left;">Acceleration, of course, is the time derivative of velocity: </div><div class="" style="clear: both; text-align: left;"> </div><div class="separator" style="clear: both; text-align: center;"><a href="http://latex.codecogs.com/gif.latex?\overrightarrow{a}%20=%20\frac{d\overrightarrow{v}}{dt}" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{a}%20=%20\frac{d\overrightarrow{v}}{dt}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a></div><div class="" style="clear: both; text-align: left;"><br /></div><div class="" style="clear: both; text-align: left;">So we can perform that substitution into the integral above. What about d<b>r</b>? Since velocity is the time integral of position:</div><div class="separator" style="clear: both; text-align: center;"><a href="http://latex.codecogs.com/gif.latex?\overrightarrow{v}%20=%20\frac{d\overrightarrow{r}}{dt}%20\therefore%20d\overrightarrow{r}%20=%20\overrightarrow{v}dt" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\overrightarrow{v}%20=%20\frac{d\overrightarrow{r}}{dt}%20\therefore%20d\overrightarrow{r}%20=%20\overrightarrow{v}dt" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="" style="clear: both; text-align: left;">Putting everything together, this is what it looks like:</div><div class="separator" style="clear: both; text-align: left;"><br /></div> <a href="http://latex.codecogs.com/gif.latex?\int_{1}^{2}\overrightarrow{F}\cdot%20d\overrightarrow{r}%20=%20\int_{1}^{2}m\left%20(\frac{%20d\overrightarrow{v}}{dt}%20\right%20)\cdot%20\overrightarrow{v}dt" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\int_{1}^{2}\overrightarrow{F}\cdot%20d\overrightarrow{r}%20=%20\int_{1}^{2}m\left%20(\frac{%20d\overrightarrow{v}}{dt}%20\right%20)\cdot%20\overrightarrow{v}dt" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></a><br /><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">The 'dt' terms cancel, and the <b>a </b>vector is gone! Work is independent of both <i>time</i> and <i>acceleration</i>! At this point, the right-side integral is easy to evaluate:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><img id="equationview" name="equationview" src="http://latex.codecogs.com/gif.latex?\int_{1}^{2}\overrightarrow{F}\cdot%20d\overrightarrow{r}%20=%20\int_{1}^{2}m\left\overrightarrow{v}%20\right%20\cdot%20\overrightarrow{dv}%20=%20\frac{mv_{2}^2}{2}%20-%20\frac{mv_1^2}{2}" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">(I sort of skipped a step, but it's trivial: <b>v </b>dot d<b>v </b>is simply v*dv, since they are in the same direction. Good to remember that we are no longer dealing with vector quantities, though).</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Those final terms are, of course, <b>kinetic energy</b>. (We know this thanks to good friend <a href="http://en.wikipedia.org/wiki/Willem_%27s_Gravesande" target="_blank">Willem Jacob Gravesande</a>, among others). </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">So that wasn't too bad, as derivations go. A few closing thoughts on work and energy:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">- These are line integrals we're talking about here. (I'm new with LaTex, so I couldn't deduce how to properly indicate a line integral). Technically, this is important; practically, it's not. <b>v </b>dot d<b>v </b>is an <i>exact differential</i>, and the line integral of an exact differential behaves like a normal integral--it's path independent</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">- Speaking of path independence, this is just one more quantity that Work and Energy are independent of. The others: time and acceleration.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">- Direction of travel <b>does</b> matter, but only in a broad sense: If the force <b>F </b>is applied in the direction of travel, it does positive work, increasing your speed. If <b>F</b> is applied against your direction of travel, you slow down. Makes sense. </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">- Beware of traps such as: 'How much work does it take to hold a 45 lb weight above your head for 30 seconds?' You are obviously applying 45 lbs of force to hold the object up, which is probably sapping a lot of energy from your muscles, but no work is done from a mechanics standpoint. Furthermore, <b>constraint forces</b>--those applied <i>not</i> in the direction of travel--do no work. Unless the constraint moves. But that's another story.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><span style="font-size: x-small;">The online LaTex editor I'm using can be found <a href="http://www.codecogs.com/latex/eqneditor.php" target="_blank">here</a>. Also, reference for this section came, once more, from Engineering Mechanics: Dynamics; Plesha/Costanzo/Grey, 1st edition.</span></div><div class="p1"></div><div class="p1"></div><br />Joe Schonemanhttps://plus.google.com/118177381588067007820noreply@blogger.com0tag:blogger.com,1999:blog-4164060036952209238.post-39105280275800353832012-08-07T23:11:00.003-07:002012-08-07T23:11:31.264-07:00Kerbal Space Program: Day 1<br /><div class="p1"><span style="font-family: Times, Times New Roman, serif;">So, from time to time, certain gaming blogs write 'Game Diaries' documenting their experiences playing, say, Minecraft or Civilization. They're a lot of fun to read, and usually get me in the mood for playing the game in question. I recently discovered an interesting little gem called <a href="http://kerbalspaceprogram.com/">Kerbal Space Program</a>, which seems like a cartoonish rocket-building game but is actually an incredibly detailed simulation of orbital mechanics. Its misadventures and smiling green spacemen lend themselves to storytelling, so here we go:</span></div><div class="p1"><span style="font-family: Times, Times New Roman, serif;"><br /></span></div><div class="p1"><span style="font-family: Times, Times New Roman, serif;"></span></div><a name='more'></a><br /><div class="p1"><br /></div><div class="p1"><span class="s1"><span style="font-family: Helvetica Neue, Arial, Helvetica, sans-serif;"><br /></span></span></div><div class="p1"><span class="s1"><span style="font-family: Helvetica Neue, Arial, Helvetica, sans-serif;">“So it’s Jeb, then?</span></span></div><div class="p1"><span class="s1"><span style="font-family: Helvetica Neue, Arial, Helvetica, sans-serif;">“Kerman’s our best shot. I’d be lying if I told you his scores were off the charts, but you’ve seen the others...”</span></span></div><div class="p1"><span class="s1"><span style="font-family: Helvetica Neue, Arial, Helvetica, sans-serif;">“The man can hold his breath for over 5 minutes without so much as a whimper.”</span></span></div><div class="p1"><span class="s1"><span style="font-family: Helvetica Neue, Arial, Helvetica, sans-serif;">“If that’s what you think we need, sir.”</span></span></div><div class="p1"><span class="s1"><span style="font-family: Helvetica Neue, Arial, Helvetica, sans-serif;">“It’s not what we need, Lieutenant. It’s what <i>he</i> needs.”</span></span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">A bright morning greeted the <i>Gary I </i>as it trundled out to Kerban Space Center. Rising above the eastern sea, the sun was a flaming ball of heat, chasing off cool morning breezes.</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1"><i>My fate? </i>Wondered Jebediah Kerman, <i>or my destiny?</i> Hand-picked after years of training and months of testing, Jeb was to be the first Kerb lofted into space. No stranger to the skies, he was still not acclimated to the concept of leaving behind his precious jewel of a planet, not even for a few minutes. Staring at his craft as it reached the gurney, he felt precious little confidence in the beaming engineers who looked on from afar. Still, though...<i>if anyone can fly that thing, I can.</i></span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">That was just the problem, though. More than anything else, Kerbals were a people composed of rank amateurs. The rocket assembly area was a veritable circus tent of clownish bumbling--wrenches dropped on toes, pants lit on fire, a true study in the art of slapstick. The Kerbonaut Selection Committee, unsure quite which characteristics to select for, had made a good show of improvisational technique: An obstacle-course one day, swimming the next, a chess tournament, and a surprise pie eating contest were all good examples of the curriculum. So, on reflection, it was quite possible that <i>nobody </i>on Kerbin could fly that rocket, precisely because it was Kerbals who’d built the thing.</span></div><div class="p2"><span class="s1"></span><br /></div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-qP2CtCwCCj4/UCIAVoyT_cI/AAAAAAAAAAk/xwSJpoLjP5w/s1600/1.png" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://4.bp.blogspot.com/-qP2CtCwCCj4/UCIAVoyT_cI/AAAAAAAAAAk/xwSJpoLjP5w/s320/1.png" width="312" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Gary I in the factory</td></tr></tbody></table><div class="p1"><span class="s1"><i>Gary I </i>was a single-stage, sub-orbital design, boasting 3 single-stack fuel tanks and a fixed nacelle. 2 stubby boosters had been comically affixed to the body, and 3 winglets were to be Jeb’s sole means of control once airborne. ‘Keep it simple, succotash,’ was chief engineer R. Glabowski’s response to any design criticisms--Jeb, for one, was not quite sure how well those wings would function above 20,000 meters. ‘Yer headed up there to find out!’ reminded Glabowski.</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">So he was. The mission profile was simple and not particularly ambitious: Break free of the atmosphere by reaching an altitude of at least 70 km, and splash down in the ocean, east of Kerban Space Center. Everyone was certain that 3 tanks of fuel would be enough to finish the job, especially with 2 boosters attached. In fact, some were worried that Jeb might accidentally achieve orbit and be left with no way to return home: Doomed to live his remaining days in a cramped, one-man cockpit.</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">‘I’m a good jumper,’ he would reassure fawning ladies at the KSC bar. ‘I’ll pole-vault my way back down if I have to.’ </span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">In fact, Jeb was terrified of heights, and kept his eyes averted as he was strapped into <i>Gary I.</i> Soon, though, the hatch slammed shut and the gurney retracted. Jeb was alone.</span></div><div class="p1"><span class="s1">“Gary I, mission control, radio check.”</span></div><div class="p1"><span class="s1">“Five Five mission control, awaiting instruction.”</span></div><div class="p1"><span class="s1">“Gary I, Gary I, be advised, refer to this station as Dragon, over.”</span></div><div class="p1"><span class="s1">“Roger, uh...Dragon.” Jeb sat. His nails needed clipping, and he tried to bite at them; his fingers ran into the glass faceplate. </span></div><div class="p1"><span class="s1">“Gary I, do you require a flight tutorial at this time?”</span></div><div class="p1"><span class="s1">“Negative Dragon, not necessary.”</span></div><div class="p1"><span class="s1">“Are you sure you don’t require a tutorial?”</span></div><div class="p1"><span class="s1">“That’s--yes, that’s affirmative.”</span></div><div class="p1"><span class="s1">A droplet of sweat trickled between Jeb’s eyes. </span></div><div class="p1"><span class="s1">“Gary I, would you like us to ask you about the flight tutorial on your next launch?”</span></div><div class="p1"><span class="s1">“What? No! Can we just launch the rocket!”</span></div><div class="p1"><span class="s1">“That’s affirmative, Gary I, you are go for ignition.”</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1"><i>Game time.</i> <i>3...2...1...</i> Jeb slammed the throttle forward.</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">“I say again, Gary I, you are go for ignition.”</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1"><i>Maybe a little tutorial wasn’t the worst idea...</i>Slamming the ‘engage’ button in front of him, Jeb was thrown backwards as <i>Gary I</i> leapt away from Kerbin. 100...200...the rocket hit 400 m/s within moments! Never in his life had he felt such speed, so light, so powerful! The boosters died after 30 seconds, and Jeb regained his composure: He had stability to maintain.</span></div><div style="text-align: left;"><a href="http://2.bp.blogspot.com/-SHUtkPk4DZA/UCIAXlhR27I/AAAAAAAAAAs/8strnpC8RIQ/s1600/2.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="240" src="http://2.bp.blogspot.com/-SHUtkPk4DZA/UCIAXlhR27I/AAAAAAAAAAs/8strnpC8RIQ/s320/2.png" width="320" /></a></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">60 seconds in, this became a problem. The rocket began oscillating back and forth, Jeb’s efforts at control seemed to feed into the wobble. Nothing seemed to be working properly: With a glance at the altimeter, he realized he was well over 30 km into the sky! There was no hope of regaining control! As this realization sunk in, the rocket truly broke loose; the engines still at full burn, it tumbled end-over-end in a drunken dance through the sky.</span></div><div class="p2"><span class="s1"></span><br /></div><br /><div class="p1"><span class="s1">Resisting the nausea, Jeb reached forward to blow the stack charges, hoping to evade the death carnival he’d been dragged into. He was kicked hard in the back, but no joy...a sickening mosaic of sky, sea, and land still filled the windscreen. </span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1"><i>Remain calm,</i> thought Jeb, panicking. He repeatedly slammed the console, hoping against hope. For long seconds, nothing changed. Then, as ‘28,000’ whizzed past on the altimeter, he was free. Peacefully floating in the air, Jeb breathed a sigh of relief.</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">‘Now whatever ya do, don’t pop chutes above 5,000 meters! Deceleration’ll pull the thing right off of ya!’ intoned R. Glabowski, from some distant memory. Looking skyward, Jeb confirmed that the parachute had, in fact, pulled him off the rocket stack, and then realized that his speed was in excess of 450 m/s! Only the pilot was extended, but could his chute possibly survive the entire fall? </span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">“Gary I, this is Dragon. Do you wish to report an equipment malfunction?”</span></div><div class="p1"><span class="s1">“Affirmative Dragon, I lost control at about 30 km up, I had to jettison the rocket. On the way down now...hope my chute holds!”</span></div><div class="p1"><span class="s1">“Gary I, please report any pilot errors during the debrief. Do you have equipment malfunction at this time?”</span></div><div class="p1"><span class="s1">“How about you just kiss my little green ass! I’ve gotta land this thing, Gary I out!”</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">Jeb turned off the radio and prepared to land, popping off his helmet and blowing up the inflatable life raft he’d been strapped into. The altimeter and airspeed indicator ticked down; one quickly, one slowly. </span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">A thundering explosion sounded from below--for a moment, Jeb thought he’d been killed. But he realized it was just his rocket stage crashing into the sea. Then, without further ado, the chute fully deployed and yanked Jeb’s capsule to a stately 6 m/s. Fluttering in a stately breeze, he could see KSC to the west, disappearing over the horizon. <i>Short flight, long swim</i>.</span></div><div class="p2"><span class="s1"></span><br /></div><div class="p1"><span class="s1">He clicked the radio back on.</span></div><div class="p1"><span class="s1"><br /></span></div><div class="p1"><span class="s1">"</span>Dragon, Dragon be advised: Gary I has landed." The hatch popped, shorting the radio as Jeb clambered into the ocean.</div><div class="p2"><span class="s1"></span><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-FJ9vNinflmE/UCIBM24MdPI/AAAAAAAAAA0/d6dKn2EgEWU/s1600/screenshot5.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="300" src="http://1.bp.blogspot.com/-FJ9vNinflmE/UCIBM24MdPI/AAAAAAAAAA0/d6dKn2EgEWU/s400/screenshot5.png" width="400" /></a></div><div class="p1"><br /></div>Joe Schonemanhttps://plus.google.com/118177381588067007820noreply@blogger.com0tag:blogger.com,1999:blog-4164060036952209238.post-46509799495351937872012-08-06T13:40:00.001-07:002012-08-15T18:49:24.077-07:00Acceleration, Energy, and Momentum<div class="p1"><span style="font-family: Georgia, 'Times New Roman', serif;">In any Dynamics class, 3 main methods of analysis will be taught -- Acceleration, Energy Methods, and Impulse/Momentum. Within the class, it’s pretty easy to figure out which method the professor wants: If we’re in Chapter 4, it must be Energy, right?</span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;"><br /></span></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">What about later on? While I’m still only through the first-level Dynamics course in my curriculum, I do assist others with the class occasionally (I’m a walk-in tutor for some of the 200/300 level engineering courses here), and haven’t yet developed a good set of rules to delineate between the methods. Here’s a first effort, hopefully I can add more in the future.</span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span></span></div><a name='more'></a><span style="font-family: Georgia, Times New Roman, serif;"><br /></span><br /><div class="p1"><b><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Acceleration Methods: Newton’s Second Law</span></span></b></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Newton’s Second Law, <i>F = ma</i>, is the foundation for the entire discipline of Statics, and as such is a good transition from a Statics course into Dynamics. As such, the problem-solving steps here are very similar to what a statics student will be accustomed to:</span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><ul><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Draw a Free Body Diagram</span></span></li><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Sum the forces in each orthogonal component</span></span></li><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">?</span></span></li><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Profit!</span></span></li></ul><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Of course, we now have moving bodies to worry about, so the missing step is to draw a Kinetic Diagram of the object in question. This simply involves drawing the applicable velocity and acceleration vectors. Then, we can equate our acceleration components to our force components and solve the problem. </span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">This gets a bit tricky in different coordinate systems, but it’s the process that counts.</span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">So when are Force/Acceleration methods the right tool for the job? Well, since we’re relating force to acceleration, take a look at the information given: If we have several known forces acting throughout a time interval on an object, or have an easily described acceleration, then <i>F = ma</i> will be easy to solve. Or at least possible to solve.</span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><b><span class="s1"></span><br /></b></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;"><b>Energy Methods: Newton’s Second Law, Over a Distance</b></span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Perhaps ‘easy’ isn’t the correct word, or perhaps we’re more concerned with efficiency of the solution. Take, for example, a bungee jumper. </span></span><br /><br /><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">In this case, the force applied is a function of the distance travelled, which yields a 2nd order-ode. Since <i>ads = vdv</i>, we can still solve by direct integration. But what’s the broader principe here? What have we actually done? Essentially, we’ve integrated both sides of <i>F = ma</i></span><span style="font-family: Georgia, Times New Roman, serif;">: </span></span></div><ul><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">On the left, Force over a distance, yielding Work.</span></span></li><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">On the right, we do some chain rule work to end up integrating the velocity term. Acceleration times a distance is equivalent to velocity times a change in velocity, right? Anyway, this results in 1/2mv^2, a term we call kinetic energy.</span></span></li></ul><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Note that this is not at all rigorous, and skips some pretty important stuff which I will hopefully<a href="http://www.badgermech.blogspot.com/2012/08/derivation-of-workenergy-principle.html" target="_blank"> cover later</a>. For now, though, the gist of the idea: Work causes a change in kinetic energy! A simple idea, but one that we retrieved simply by manipulating Newton’s second law. The principles also extend to potential energy, such as the gravitational energy our jumper above benefits from.</span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">More importantly, this is a template we can use to talk about changes in energy as a result of work done on an object, or vice versa. Rather than drawing the traditional FBD and labeling forces (it’s still good practice to do this anyway), we can simply draw an Energy Diagram of the situation at large. So, in the case of the jumper, there are a few points of interest:</span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><ol class="ol1"><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Standing on the platform. Lots of potential energy, no kinetic energy.</span></span></li><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Cord begins to stretch. Still plenty of potential energy, and a good amount of kinetic energy.</span></span></li><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Wherever the jumper stops. (Hopefully above the ground--we haven’t reached impact analysis yet!) Kinetic energy is 0, potential energy is either neglected or small, and ALL of that energy has been transferred to the cord. At this point, our analysis stops: Simple work-energy principles aren’t sufficient to, say, determine the ‘snap-back’ of the cord. </span></span></li></ol><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">So that’s where the limitations of Work-Energy show up: We can only find out broad characteristics of an object at different points along its path. Since the conservative forces we model are ‘path-independent,’ we actually can’t find out anything about the path of the jumper--only make an estimate of how far he falls before stopping.</span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">But sometimes, a quick-and-dirty feasibility analysis is what’s needed. Work divided by time is power, so Work/Energy gives an easy way to put upper and lower bounds on power requirements for a motor, for example. Roller coaster problems are also classic examples of the method: ‘How tall a drop to get this coaster around this loop?’ </span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><b><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Momentum/Impulse: The 2nd Law Over Time</span></span></b></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Finally, we have a class of problem that occurs very often in real life, but can’t be adequately modeled with either of the above methods: Short, violent impacts. Car crashes, baseball bats, a sprinter’s foot...the examples are endless. The problem faced in these situations is that very complicated, volatile accelerations can occur over very short time intervals. Attempting to solve and integrate large, discontinuous acceleration will give unreliable results.</span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">So, we’ll again integrate Newton’s 2nd Law, this time with respect to time. </span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><ul><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">On the left side, a force acting for some amount of time is called an impulse. </span></span></li><li class="li1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">On the right side, integrating acceleration with time is velocity. We call mass times velocity momentum. When people say ‘inertia,’ they’re generally speaking of ‘momentum.’ (i.e. “The truck has a lot of inertia at 60 mph.’) </span></span></li></ul><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">So what this gives is a way to relate impulses directly to changes in velocity. Additionally, we can analyze collisions between objects, if we know their velocities before or after the collision. </span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Anytime you’re presented with quick, violent forces and want to look at how they change the velocity of something, use impulse/momentum. </span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1"><span style="font-family: Georgia, Times New Roman, serif;">Momentum also leads into more advanced mechanics concepts, such as angular momentum, and allows analysis of mass flows (fluids) and variable mass systems (rockets). </span></span></div><div class="p2"><span style="font-family: Georgia, Times New Roman, serif;"><span class="s1"></span><br /></span></div><div class="p1"><span class="s1" style="font-size: x-small;"><span style="font-family: Georgia, Times New Roman, serif;">My Dynamics textbook is Engineering Mechanics: Dynamics by Gray, Costanzo, Plesha. It hasn’t received good reviews on Amazon, although the 2nd edition is better-organized than the 1st edition I own. I think the ‘Gold Standard’ of Dynamics texts is anything by Hibbeller, and old editions can be had quite cheaply, if you’re so inclined.</span></span></div><br />Joe Schonemanhttps://plus.google.com/118177381588067007820noreply@blogger.com0tag:blogger.com,1999:blog-4164060036952209238.post-19353091742075416792012-08-02T12:39:00.002-07:002012-08-02T12:40:12.189-07:00IntroductionHello! My name is Joe, and I'm an Engineering Mechanics student at the University of Wisconsin - Madison. I'm here to discuss the physical and mathematical concepts present in my coursework, with the aim of gaining a better understanding for myself and those who make their way here. With any luck, I'll be able to carry this forward into my professional career, should such a thing ever manifest itself. Until then, look for helpful posts on such banalities as:<br /><br />- The history and origins of calculus<br />- Operations in vector fields<br />- The cross product, and why it's calculated as a determinant<br /><br />Interesting stuff, right? We'll see. Thanks for stopping by!Joe Schonemanhttps://plus.google.com/118177381588067007820noreply@blogger.com0