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	<title>Comments on: Teaching the gentle art of estimations</title>
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	<link>http://blog.sciencegeekgirl.com/2009/07/07/teaching-the-gentle-art-of-estimations/</link>
	<description>explorations and inspirations... in how we learn science</description>
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		<title>By: Anonymous Coward</title>
		<link>http://blog.sciencegeekgirl.com/2009/07/07/teaching-the-gentle-art-of-estimations/comment-page-1/#comment-1676</link>
		<dc:creator>Anonymous Coward</dc:creator>
		<pubDate>Fri, 10 Jul 2009 20:01:10 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sciencegeekgirl.com/?p=692#comment-1676</guid>
		<description>&gt;The calculation of the exact force of impact is meant to relate to the ball, not the hand.

I suspect the forces may be related in some sort of equal-and-opposite fashion.  Or am I misunderstanding the problem?

&gt;The ultimate point that the force on the hand (and thus the force on the ball at moment of impact) is larger than mg is unchanged.

I certainly agree with that point.

&gt;I’m not so sure that his reasoning of the time that it takes for the signal to pass from one side of the ball to the other has “no physical validity”! That seems to be a pretty strong statement for a standard estimation of signal propagation.

I apologize for my provocative language, but I stand by the basic point.  (As much as an Anonymous Coward can stand by anything.)

I think everyone agrees that we need to calculate some F=ma equation, and to estimate the acceleration we need an estimate of its duration.  It seems to me that what Dr. Mahajan has done to estimate this is to pick a number out of a hat (the speed of sound), but it&#039;s not the relevant number for the problem.  The important one is the spring constant of the materials involved, as Tom/swansont pointed out above.*

The spring constant is why if I drop a wine glass on a concrete floor it is more likely to break than if I drop it on a trampoline.  

If you&#039;re interested in examples of how to estimate accelerations in a variety of situations, I&#039;d point out that Rhett Allain&#039;s &quot;http://blog.dotphys.net/&quot; blog has a variety of nice examples of calculations of this sort.


*Caveat: accurately working out the problem is more complicated than just the static spring constant due to the fact that we&#039;re looking at dynamics and non-point-objects: at high velocities the mass density of the various objects involved will start to become significant.  Working out the crossover between the inertia-dominated regime and the spring-constant-dominated regime is left as an exercise for the reader.</description>
		<content:encoded><![CDATA[<p>&gt;The calculation of the exact force of impact is meant to relate to the ball, not the hand.</p>
<p>I suspect the forces may be related in some sort of equal-and-opposite fashion.  Or am I misunderstanding the problem?</p>
<p>&gt;The ultimate point that the force on the hand (and thus the force on the ball at moment of impact) is larger than mg is unchanged.</p>
<p>I certainly agree with that point.</p>
<p>&gt;I’m not so sure that his reasoning of the time that it takes for the signal to pass from one side of the ball to the other has “no physical validity”! That seems to be a pretty strong statement for a standard estimation of signal propagation.</p>
<p>I apologize for my provocative language, but I stand by the basic point.  (As much as an Anonymous Coward can stand by anything.)</p>
<p>I think everyone agrees that we need to calculate some F=ma equation, and to estimate the acceleration we need an estimate of its duration.  It seems to me that what Dr. Mahajan has done to estimate this is to pick a number out of a hat (the speed of sound), but it&#8217;s not the relevant number for the problem.  The important one is the spring constant of the materials involved, as Tom/swansont pointed out above.*</p>
<p>The spring constant is why if I drop a wine glass on a concrete floor it is more likely to break than if I drop it on a trampoline.  </p>
<p>If you&#8217;re interested in examples of how to estimate accelerations in a variety of situations, I&#8217;d point out that Rhett Allain&#8217;s &#8220;http://blog.dotphys.net/&#8221; blog has a variety of nice examples of calculations of this sort.</p>
<p>*Caveat: accurately working out the problem is more complicated than just the static spring constant due to the fact that we&#8217;re looking at dynamics and non-point-objects: at high velocities the mass density of the various objects involved will start to become significant.  Working out the crossover between the inertia-dominated regime and the spring-constant-dominated regime is left as an exercise for the reader.</p>
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		<title>By: sciencegeekgirl</title>
		<link>http://blog.sciencegeekgirl.com/2009/07/07/teaching-the-gentle-art-of-estimations/comment-page-1/#comment-1673</link>
		<dc:creator>sciencegeekgirl</dc:creator>
		<pubDate>Fri, 10 Jul 2009 06:11:19 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sciencegeekgirl.com/?p=692#comment-1673</guid>
		<description>Thanks for your thoughtful comments, Tom and Anonymous.  To clarify, Anonymous, the calculation for the ball is not, I believe, intended to relate directly to the hand.  The reasoning for the force on the hand being larger than mg (and larger than 2mg) is the heuristic reasoning earlier in the post.  The calculation of the exact force of impact is meant to relate to the ball, not the hand.  One aspect I didn&#039;t mention was that the hand is on the table.  The ultimate point that the force on the hand (and thus the force on the ball at moment of impact) is larger than mg is unchanged.

So what is the exact force on the ball?  You disagree that the time of impact is on the order of microseconds.  I took this calculation directly from Dr. Mahajan, and it was rather an afterthought since the point isn&#039;t the symbolic manipulation.  I will do some more thinking on the matter.  I&#039;m not so sure that his reasoning of the time that it takes for the signal to pass from one side of the ball to the other has &quot;no physical validity&quot;!  That seems to be a pretty strong statement for a standard estimation of signal propagation.

Will think more about this tomorrow, gentle readers.  

And check out the video posted by Anonymous, it&#039;s pretty neat.</description>
		<content:encoded><![CDATA[<p>Thanks for your thoughtful comments, Tom and Anonymous.  To clarify, Anonymous, the calculation for the ball is not, I believe, intended to relate directly to the hand.  The reasoning for the force on the hand being larger than mg (and larger than 2mg) is the heuristic reasoning earlier in the post.  The calculation of the exact force of impact is meant to relate to the ball, not the hand.  One aspect I didn&#8217;t mention was that the hand is on the table.  The ultimate point that the force on the hand (and thus the force on the ball at moment of impact) is larger than mg is unchanged.</p>
<p>So what is the exact force on the ball?  You disagree that the time of impact is on the order of microseconds.  I took this calculation directly from Dr. Mahajan, and it was rather an afterthought since the point isn&#8217;t the symbolic manipulation.  I will do some more thinking on the matter.  I&#8217;m not so sure that his reasoning of the time that it takes for the signal to pass from one side of the ball to the other has &#8220;no physical validity&#8221;!  That seems to be a pretty strong statement for a standard estimation of signal propagation.</p>
<p>Will think more about this tomorrow, gentle readers.  </p>
<p>And check out the video posted by Anonymous, it&#8217;s pretty neat.</p>
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		<title>By: Anonymous Coward</title>
		<link>http://blog.sciencegeekgirl.com/2009/07/07/teaching-the-gentle-art-of-estimations/comment-page-1/#comment-1672</link>
		<dc:creator>Anonymous Coward</dc:creator>
		<pubDate>Thu, 09 Jul 2009 23:02:32 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sciencegeekgirl.com/?p=692#comment-1672</guid>
		<description>I just want to echo Tom&#039;s comment above regarding the miscalculation of the forces involved in the rock-on-hand stuff above.

Even if the rock is infinitely stiff, the hand beneath it isn&#039;t, and will deflect by a few millimeters, not a few microns.  This is why one can drop stuff on one&#039;s hand without it shattering.

It&#039;s ironic that an article lamenting the inability of students to use physical reasoning or do order-of-magnitude estimates would use a method of no physical validity to arrive at a numerical answer which is off by orders of magnitude.

P.S.  For a fun video of hand vs rock (with the hand deflecting by more than a few millimeters):
http://www.youtube.com/watch?v=mOLp4doE51Q</description>
		<content:encoded><![CDATA[<p>I just want to echo Tom&#8217;s comment above regarding the miscalculation of the forces involved in the rock-on-hand stuff above.</p>
<p>Even if the rock is infinitely stiff, the hand beneath it isn&#8217;t, and will deflect by a few millimeters, not a few microns.  This is why one can drop stuff on one&#8217;s hand without it shattering.</p>
<p>It&#8217;s ironic that an article lamenting the inability of students to use physical reasoning or do order-of-magnitude estimates would use a method of no physical validity to arrive at a numerical answer which is off by orders of magnitude.</p>
<p>P.S.  For a fun video of hand vs rock (with the hand deflecting by more than a few millimeters):<br />
<span style="text-align:center; display: block;"><a href="http://blog.sciencegeekgirl.com/2009/07/07/teaching-the-gentle-art-of-estimations/"><img src="http://img.youtube.com/vi/mOLp4doE51Q/2.jpg" alt="" /></a></span></p>
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		<title>By: Aaron F.</title>
		<link>http://blog.sciencegeekgirl.com/2009/07/07/teaching-the-gentle-art-of-estimations/comment-page-1/#comment-1671</link>
		<dc:creator>Aaron F.</dc:creator>
		<pubDate>Thu, 09 Jul 2009 19:44:32 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sciencegeekgirl.com/?p=692#comment-1671</guid>
		<description>Jean-Michel---

Ha! Awesome! ^_^</description>
		<content:encoded><![CDATA[<p>Jean-Michel&#8212;</p>
<p>Ha! Awesome! ^_^</p>
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		<title>By: Jean-Michel</title>
		<link>http://blog.sciencegeekgirl.com/2009/07/07/teaching-the-gentle-art-of-estimations/comment-page-1/#comment-1669</link>
		<dc:creator>Jean-Michel</dc:creator>
		<pubDate>Thu, 09 Jul 2009 08:46:27 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sciencegeekgirl.com/?p=692#comment-1669</guid>
		<description>The correct answer to the  &quot;Age of the capitaine&quot; question exists : it is 39 years old. This is an old french riddle that plays on the double meaning of the word quarantaine which means &quot;quarantine&quot; and &quot;forty years old&quot;. In french the answer is : &quot;Il a 39 ans car il va vers la quarantaine&quot;.  Which has two translations :  &quot;He will soon be forty&quot; and &quot; He goes to the quarantine&quot;. ( The correct wording of the question is : A ship transporting  26 sheep and 10 goats is arriving at the harbor.  How old is the captain? )</description>
		<content:encoded><![CDATA[<p>The correct answer to the  &#8220;Age of the capitaine&#8221; question exists : it is 39 years old. This is an old french riddle that plays on the double meaning of the word quarantaine which means &#8220;quarantine&#8221; and &#8220;forty years old&#8221;. In french the answer is : &#8220;Il a 39 ans car il va vers la quarantaine&#8221;.  Which has two translations :  &#8220;He will soon be forty&#8221; and &#8221; He goes to the quarantine&#8221;. ( The correct wording of the question is : A ship transporting  26 sheep and 10 goats is arriving at the harbor.  How old is the captain? )</p>
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