graphing
https://scienceblogs.com/taxonomy/term/18483/feed
enExperimental Rope Logarithms
https://scienceblogs.com/dotphysics/2010/06/01/experimental-rope-logarithms
<span>Experimental Rope Logarithms </span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>Sue from <a href="http://mathmamawrites.blogspot.com/2010/05/logarithms-and-ropes-as-found-in.html">Math Mama Writes...</a> sent me an email about wrapping a rope around a pole. In that post, Sue thinks about rope looped around a post. When you wrap a rope around a post, the friction between the rope and the post can help you hold something (like a horse) that is much stronger than you.</p>
<p>The first case she thinks about is using several posts. What if you wrap a rope around one post and pull? What if you use 2, or three posts? The idea is that if one posts 'multiplies' the force by 10, two posts would have an effect of multiplying by 100 and so forth. That seems reasonable.</p>
<p>The next thing, what about one post with just half a turn? What about multiple turns around the same post? This is what I looked at first since it is the easiest (you don't need multiples of the same post).</p>
<h2>Setup</h2>
<p>Here is a basic diagram of my experimental setup.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-9b4757d92a7f9457afa0e8d57ca313c3-2010-06-01_untitled_1.jpg" alt="i-9b4757d92a7f9457afa0e8d57ca313c3-2010-06-01_untitled_1.jpg" /></p>
<p>Where these force probes are <a href="http://vernier.com">Vernier</a> probes using Logger Pro. One of the probes is clamped to the table and the other held by me. First, let me test one probe connected to another by a string (no wrapping). Here is a plot of force 1 vs. force 2:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-9e6a0b0fd36bd3ebc3cc088188bdd77d-2010-06-01_0turnscmbl.jpg" alt="i-9e6a0b0fd36bd3ebc3cc088188bdd77d-2010-06-01_0turnscmbl.jpg" /></p>
<p>Fitting a linear function, I should get a slope of 1, but I get 1.152. This means that the two probes are not perfectly calibrated (I guess I should have checked this before hand). Now what about half of a turn around the post? (I calibrated the probes - they match a little better now)</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-38f1b8fe8b4508309d1fe801be1897e0-2010-06-01_untitled_2.jpg" alt="i-38f1b8fe8b4508309d1fe801be1897e0-2010-06-01_untitled_2.jpg" /></p>
<p>So, for 1/2 turn, there is a force difference with the relationship:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-554f1b4dbcfd552f806f8a2bb6e7b0b3-2010-06-01_la_te_xi_t_1_1.jpg" alt="i-554f1b4dbcfd552f806f8a2bb6e7b0b3-2010-06-01_la_te_xi_t_1_1.jpg" /></p>
<p>Now, what about one complete turn?</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-7ee84eb1834ed07bbd7cdc81a1132c35-2010-06-01_untitled_3.jpg" alt="i-7ee84eb1834ed07bbd7cdc81a1132c35-2010-06-01_untitled_3.jpg" /></p>
<p>This gives a relationship of:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-5d54beb4b3366594e80ab2450a4e324d-2010-06-01_la_te_xi_t_1_2.jpg" alt="i-5d54beb4b3366594e80ab2450a4e324d-2010-06-01_la_te_xi_t_1_2.jpg" /></p>
<p>I tried two loops, but I just could get good data. How about multiple posts? I already have data for one loop around one post. What about two posts?</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-19326291661bfddf37b8e315dfca5e6a-2010-06-01_untitled_4.jpg" alt="i-19326291661bfddf37b8e315dfca5e6a-2010-06-01_untitled_4.jpg" /></p>
<p>Can I get three posts? Yes, I can.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-5293cbd09034938fcf6aa44b2953b309-2010-06-01_untitled_5.jpg" alt="i-5293cbd09034938fcf6aa44b2953b309-2010-06-01_untitled_5.jpg" /></p>
<p>I am not going to try 4 posts. I know my force probes would be unhappy.</p>
<h2>Back to the theory</h2>
<p>For multiple posts, the idea is that if 1 posts multiplies the force by a factor of <em>a</em>, then two posts should have a total multiplicative factor of <em>a</em><sup>2</sup> and three would have a factor of <em>a</em><sup>3</sup>. In general, if one posts produces a force multiplication of <em>a</em> and the number of posts is <em>n</em>, then:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-d70ad6272ca8978d4643c82b74115573-2010-06-01_la_te_xi_t_1_3.jpg" alt="i-d70ad6272ca8978d4643c82b74115573-2010-06-01_la_te_xi_t_1_3.jpg" /></p>
<p>Let me take my three whole turn data points. I already have the ratio of F<sub>1</sub> to F<sub>2</sub> (the slope of the graphs). So, re-writing the above expression, I get:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-92a8f0e518b37bfb08d66587af99e82f-2010-06-01_la_te_xi_t_1_4.jpg" alt="i-92a8f0e518b37bfb08d66587af99e82f-2010-06-01_la_te_xi_t_1_4.jpg" /></p>
<p>If I plot the natural log of the ratio of forces vs. the number of wraps, I should get a constant. Here is that plot.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-d3098d6fd60fc3129e36c13cd1050cb3-2010-06-01_untitled_6.jpg" alt="i-d3098d6fd60fc3129e36c13cd1050cb3-2010-06-01_untitled_6.jpg" /></p>
<p>So, the natural log of a is the slope of this line - .8825. This would give a value for <em>a</em> as 2.41. This should be the same as the slope for one turn, but it is not. Oh well, it isn't too far off.</p>
<p>What about n = 1/2? Does that work? The natural log of the slope for half a turn is 0.385 which is fairly close to half the value of ln(a).</p>
<h2>The physics</h2>
<p>Physically, how does this work? The normal model for friction says that the frictional force is proportional only to the force the two surfaces are pushing against each other. Not sure if that works here.</p>
<p><strong>Energy?</strong> This seems like it is cheating. Do you get more energy than you put in? No, because you don't even put in any energy if the rope doesn't move. This is not a simple machine. What about forces? This seems like it makes more force and thus violates Newton's third law. Well, it doesn't. If I pull with 1 Newton and a horse on the other end pulls with 10 Newtons, these are not "equal and opposite forces". You could think of it this way. The horse pulls on the rope with a magnitude of 10 Newtons. So, looking at the post, the horse pulls one way with 10 N, me the other way with 1 N. This leaves 9 Newtons. The ground exerts 9 Newton force on the post. If the horse were strong enough, it could pull the post out of the ground.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Tue, 06/01/2010 - 12:28</span>
Tue, 01 Jun 2010 16:28:17 +0000rallain108135 at https://scienceblogs.comActivities for Pi Day
https://scienceblogs.com/dotphysics/2010/03/14/activities-for-pi-day
<span>Activities for Pi Day</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>Oh, I know you missed it. Really, it wasn't your fault. Pi day fell on a Sunday, so how are you supposed to have pi-day activities in class? Don't let it stop you. You are better than that. Do the activity anyway. What to do? Here are some suggestions. (Suggestions aimed mostly at the high school level)</p>
<h3>Plot Diameter vs. Circumference</h3>
<p>This is a great one. Let your students find as many round things as they can (cylinders work the best - or flat stuff). Measure the circumference (you can use a string or a tape measure) and the diameter. Since the relationship between these two is:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-9c6732b0d80d71692533150ad7a01450-2010-03-14_la_te_xi_t_1_18.jpg" alt="i-9c6732b0d80d71692533150ad7a01450-2010-03-14_la_te_xi_t_1_18.jpg" /></p>
<p>A plot with circumference on the vertical axis and diameter on the horizontal axis should be a straight line with a slope of Pi. The great thing about this activity is that students can get a feel for where pi comes from. There are two many students that just think it is a number invented by mathematicians to make things more complicated and cool sounding.</p>
<p>Oh - as a bonus, students get to practice making graphs and finding the slope. I would recommend doing this one on real graph paper (and not in a spreadsheet).</p>
<h3>Rolling vs. Distance</h3>
<p>Really, this is the same thing as the activity above - but it looks different. Take a cylinder and roll it. Count the number of revolutions and measure the distance it rolled. Plot distance on the vertical axis and the number of revolutions on the horizontal axis. Here is the relationship between the two.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-0ca12caf928ac17559d0db7f94bcb7d6-2010-03-14_la_te_xi_t_1_19.jpg" alt="i-0ca12caf928ac17559d0db7f94bcb7d6-2010-03-14_la_te_xi_t_1_19.jpg" /></p>
<h3>LEGO Estimation of Pi</h3>
<p>I just thought of this one and have not actually tried it. Archimedes estimated Pi by drawing two 96-sided polygons inside and outside of a circle. He could then determine an upper and lower bounds for the value of Pi. You could try to reproduce this Lego pieces. Make an n-sided Lego polygon both inside and outside of a circle. Compare the perimeter of the polygons to the radius of the circle.</p>
<p>I might make this a future post, but if you try it out, let me know how it works.</p>
<h3>Monte Carlo Estimation of Pi</h3>
<p><a href="http://scienceblogs.com/dotphysics/2010/03/on_the_8th_day_god_made_pi.php">I had a more detailed post about this method for estimating Pi.</a> But maybe you don't want to look back - so here is the short version. If you randomly put points in a 1 x 1 square, some will be more than 1 units from the corner and some will be less. Here are some random dots.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-4398cc88e0d9b555593ecca0939d5920-2010-03-14_scratch.jpg" alt="i-4398cc88e0d9b555593ecca0939d5920-2010-03-14_scratch.jpg" /></p>
<p>Since the dots that are less than 1 from the lower left corner make up 1/4th of a circle, the ratio of red dots to total dots should be:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-a38ebc2616a47a71f57168ebe4c053b2-2010-03-14_la_te_xi_t_1_20.jpg" alt="i-a38ebc2616a47a71f57168ebe4c053b2-2010-03-14_la_te_xi_t_1_20.jpg" /></p>
<p>So, this is pretty straight forward. But how could you do this? I made a program in <a href="http://scratch.mit.edu/projects/rhettallain/925746">Scratch</a> as well as python. You could use anything that has a random number generator. Here is a version in google docs:</p>
<iframe width="500" height="300" frameborder="0" src="http://spreadsheets.google.com/pub?key=timmmSdmlR7BLc_Bx8LS1iw&output=html&widget=true"></iframe><p>I didn't finish it, well mostly I did - but you would need to do some more work on it to finish. If you wanted, you could have groups in the class calculate the average for 100 points and then take the average average for all the groups.</p>
<h3>Non-Computer Monte Carlo</h3>
<p>Maybe you think computers will one day rule the world and you would rather not use them to calculate Pi. I can understand that. You could drop something so that it has a random distribution on a 1 x 1 square and then count the number in and outside of a circle. Maybe find some way to drop sand on a square paper? Make sure that sand is falling outside of the paper also, or it likely will not be randomly distributed.</p>
<h3>How Accurate can you get Pi by Measuring?</h3>
<p>What if you used the plot of circumference vs. diameter from above? How accurate of a value of Pi could you get?</p>
<h3>Other resources</h3>
<p>There are tons of great Pi sites out there. Here are just a few:</p>
<ul><li><a href="http://www.joyofpi.com/">The Joy of Pi</a> - tons of good stuff here.</li>
<li><a href="http://en.wikipedia.org/wiki/Pi#History">Wikipedia's Pi page</a></li>
<li><a href="http://ualr.edu/lasmoller/pi.html">The history of Pi</a></li>
</ul></div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Sun, 03/14/2010 - 17:21</span>
Sun, 14 Mar 2010 21:21:23 +0000rallain108066 at https://scienceblogs.comDIY Force probe
https://scienceblogs.com/dotphysics/2010/02/14/diy-force-probe
<span>DIY Force probe</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>I was thinking about some experiments that deal with friction and I wanted to show something with a force probe. The problem is that most people don't have one of these. So, I decided to try and make one out of simple things. In this case, I am using some straws, a rubber band and some paper clips. Let me draw a little sketch of how this thing works.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-f30652ef559b3615ef2dd2d88356cddc-2010-02-14_untitled.jpg" alt="i-f30652ef559b3615ef2dd2d88356cddc-2010-02-14_untitled.jpg" /></p>
<p>The basic idea is to use the rubber band to measure the force (by measuring the amount the rubber band stretches). The two paper clips do two things. First, it allows you to hook up the device to something (like hanging some Lego bricks on it) and it gives the straws a place to connect. The straws (one of those smaller coffee straws) slide and let me measure the stretch. I made a video of how I put one of these together. Not sure how useful the video is - I am not actually a professional hand model.</p>
<object width="480" height="295"><param name="movie" value="http://www.youtube.com/v/U3thDj-B6gM&hl=en_US&fs=1&" /><param name="allowFullScreen" value="true" /><param name="allowscriptaccess" value="always" /><embed src="http://www.youtube.com/v/U3thDj-B6gM&hl=en_US&fs=1&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="295"></embed></object><p>The only tricks: I used a string to pull the rubber band through the straws. Also, I bent one of the paperclips so the straw would fit snugly over it. The only thing holding the straws to their corresponding paper clips is friction. If that is not working for you, you could tape it or something.</p>
<p>Great, but what next? It needs to be calibrated somehow. In my quest to use everyday things, I am going to use Lego bricks - the big ones. I hung various numbers of bricks from the force scale and used a marker to mark on the small straw the location that the two straws meet. Here is a picture of one set of hanging bricks (I made a video, but you couldn't see everything).</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-a377133646d8569432b7d390583db476-2010-02-14_flip_share.jpg" alt="i-a377133646d8569432b7d390583db476-2010-02-14_flip_share.jpg" /></p>
<p>After hanging several different Lego masses, this is what my force scale looked like (I have decided to call it a 'scale' instead of a 'probe' - you know, because of the car from Ford).</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-ed04ed4b423faf402fcafd4d38c0d10b-2010-02-14_scale.jpg" alt="i-ed04ed4b423faf402fcafd4d38c0d10b-2010-02-14_scale.jpg" /></p>
<p>Wow. I didn't realize just how poorly that picture came out. Note to self: stop using your phone to take these kinds of pictures. Maybe you can't read the ruler, but hopefully you can see the marks on the inner straw. Here is a quick plot of the distance the rubber band is stretched vs. the number o Lego bricks.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-6af5d6c7135fb8ca0fe879e51c569df9-2010-02-14_untitled_1.jpg" alt="i-6af5d6c7135fb8ca0fe879e51c569df9-2010-02-14_untitled_1.jpg" /></p>
<p>Since the amount of stretch is linear with respect to the force, I can just use a ruler and measure amount of stretch to determine the force. Note: I am assuming that each Lego brick as the same mass. Of course, if you don't have those big Lego bricks, you could use washers or nuts or anything that has a fairly consistent mass.</p>
<p>If I did want to use the distance to determine the force, I would have to know the slope of that line. From the fit above, I get a slope of 0.847 cm/Lego. Thus after measuring the stretch, my force would be:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-5931800a393c842048812d7e023f45f6-2010-02-14_la_te_xi_t_1.jpg" alt="i-5931800a393c842048812d7e023f45f6-2010-02-14_la_te_xi_t_1.jpg" /></p>
<p>This would give the force in units of Lego bricks. Then if you know the weight of one Lego brick, you could convert this to Newtons.</p>
<h3>Things to think about</h3>
<ul><li>Does the rubber band stay consistent? As you use it, does it need to be re-calibrated?</li>
<li>Does the stretch per cm change with temperature (hint).</li>
<li>Can you use different rubber bands to measure different forces? (thick vs. thin rubber band)</li>
<li>Could you use two rubber bands in parallel? What would happen then? Could you use two rubber bands in series?</li>
</ul></div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Sun, 02/14/2010 - 02:06</span>
Sun, 14 Feb 2010 07:06:20 +0000rallain108046 at https://scienceblogs.comRP 8: The price of a piece of LEGO
https://scienceblogs.com/dotphysics/2009/12/29/rp-8-the-price-of-a-piece-of-l
<span>RP 8: The price of a piece of LEGO</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>This idea comes from my friend Thomas. His son is like mine in that they both think LEGO are awesome, and they are correct. For some reason, Thomas decided to calculate the price per piece of LEGO in each set. To promote repeatability, I decided to do this also. Looking at the <a href="http://www.lego.com">catalog at LEGO.com</a>, I can get both the price of each set and how many pieces it has. Just a note, I looked at almost all of the Star Wars LEGO series and some other select themes. I didn't include any sets that had been marked down in price. I will put the first plot on down below, maybe this would be a good time for you to guess the average price per piece.</p>
<p>This is a plot of all the different themes mixed together.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-b13cd576c24297c5af0c705caee89cb9-2009-12-26_legoprice_1.jpg" alt="i-b13cd576c24297c5af0c705caee89cb9-2009-12-26_legoprice_1.jpg" /></p>
<p>From that plot, it seems that price per piece is fairly consistent. The slope of the linear function fitting the data gives $0.097 per piece. The one data point highlighted that seems of a little bit is the Republic Dropship with AT-OT. It is listed on LEGO as "exclusive" and is $0.14 per piece. A couple of those real expensive sets make it difficult to see the lower stuff. Let me zoom in on so that those sets are not included.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-38a901558b8d2e2b9d591aff3a2a24dc-2009-12-26_lego_2.jpg" alt="i-38a901558b8d2e2b9d591aff3a2a24dc-2009-12-26_lego_2.jpg" /></p>
<p>Here you can see I labeled a couple of the stray points. Another interesting thing is that the function that fits the data has a non-zero y-intercept. I guess this would mean that if you bought a LEGO box with zero pieces, it would still cost $6.18 (I guess that is for the packaging, instructions and stuff)</p>
<p>Do the different themes have different prices per piece? Here are the average prices per piece for different themes.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-18d57a7cf21b630a24db9a20632b7f80-2009-12-26_lego_3.jpg" alt="i-18d57a7cf21b630a24db9a20632b7f80-2009-12-26_lego_3.jpg" /></p>
<p>The cheapest per piece is the technic. This may be because the technic sets have lots of those really, really tiny pieces which are likely cheaper. Also, the bionicle sets are interesting. Most of these are for these big guys that all cost $12.99 and have "around" 50 pieces.</p>
<p>Finally, LEGO store has for sale individual lego pieces. I guess you could order all the pieces you need for a particular set instead of buying the set itself. I looked at about the first 100 pieces that were listed (not sure what order they were listed in) and I made a histogram of the prices.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-fd00b0ded9526c97a9cddb403f384df5-2009-12-26_lego_4.jpg" alt="i-fd00b0ded9526c97a9cddb403f384df5-2009-12-26_lego_4.jpg" /></p>
<p>I left off two points on this histogram. In the first 100 items, there was a piece that was $4.25 and there was a piece that was $0.54 (I left them off because they made the chart look odd). Including those two points, the average price per piece for the first 100 was $0.1795 with a standard deviation of $0.4238.</p>
<p>So, what is the point? I am really not sure. I have seen a lego program that lets you virtually build stuff, but it is really loud here right now and I can't find it.</p>
<h3>UPDATE</h3>
<p>I was thinking about this some more, one thing in my mind was that my friend said the average price per piece was something different than mine. I realized that I did not do the same thing he did. I fit a linear function to the set price vs. number of pieces data. In this fit, there is a non-zero price-intercept. What my slope says is: "If I increase the number of pieces by 1, what will the increase in the price be?". For just the star wars Lego sets, this value is $0.09951. In the bar graph above, I report the average price per piece as $0.11. This number is the price of the set divided by the number of pieces in the set and averaged for all sets.</p>
<p>If the linear function had a zero intercept, these two numbers would be the same. The first method is better because it takes into account the idea that there is some base cost to a lego set. If you had a lego set with just one piece, would it cost $0.11? (well, it would if you ordered it from the lego parts store - but it wouldn't be a 'set') I think this is a great example of the difference between slope and 'y/x' - which I find students often confuse.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Tue, 12/29/2009 - 02:00</span>
Tue, 29 Dec 2009 07:00:00 +0000rallain108005 at https://scienceblogs.comMore projectile motion than you can shake a stick at
https://scienceblogs.com/dotphysics/2009/10/20/more-projectile-motion-than-you-can-shake-a-stick-at
<span>More projectile motion than you can shake a stick at</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>I don't really know what that title actually means. So, I have been having problems with my PASCO projectile launcher devices. I will just call them launchers (they are really cannons). In my previous post, I <a href="http://scienceblogs.com/dotphysics/2009/10/more-measurements-of-the-projectile-velocity/">looked at the launch speed from a launcher shot horizontally and vertically</a>. The problem was that I was getting different launch speeds for the vertical and horizontal shot. So, here is my plan: shoot the ball and a variety of angles from 0 to 90 degrees and see how the launch speed changes. I will only use the data from video analysis (of course using <a href="http://www.cabrillo.edu/~dbrown/tracker/">Tracker Video Analysis</a>)</p>
<p>PASCO makes two launchers. A normal sized one and a mini-me-launcher. </p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/uploads/2009/10/pascolarge.jpg" alt="Pascolarge" height="197" width="156" /><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-19a3e3dc7eba6c61ee1fd9762c91ba4e-minilauncher.jpg" alt="i-19a3e3dc7eba6c61ee1fd9762c91ba4e-minilauncher.jpg" /></p>
<p>So, what data do I actually get from the video?</p>
<ul><li>time</li>
<li>horizontal (x) position</li>
<li>vertical (y) position</li>
<li>I measured the angle from the launcher</li>
</ul><p>Also, after getting the data, Tracker Video fit a function for me so that I have:</p>
<ul><li>The slope of the x-t graph (this is the x-velocity)</li>
<li>The parameters of the 2<sup>nd</sup> order polynomial that fits the y-t data</li>
<li>The time of the first data point. This is not zero for two reasons. First, the video frame was likely taken some moment after the ball left. Second, I didn't cut the video up for each shot. So, the 30 degree shot may have taken place at 15 seconds into the video.</li>
</ul><p>So, I have two ways to get the launch velocity.</p>
<h3>Launch Velocity Method 1</h3>
<p>The x-velocity is pretty straight forward to get. Since the x-velocity is constant, I can just find it from the slope of the x-position time graph. Actually, this is a good measure to see if things are working correctly (a straight x-t graph).</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-35127b9f5037625f4a46744b3731ec25-zero_deg_x_pos.jpg" alt="i-35127b9f5037625f4a46744b3731ec25-zero_deg_x_pos.jpg" /></p>
<p>With the angle and the x-velocity, I can find the total velocity when the ball was launched.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-c14e617ca3747f46c707423605228c7b-untitled2.jpg" alt="i-c14e617ca3747f46c707423605228c7b-untitled2.jpg" /><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-1e210cc1f64f80858cceb227378b4c95-la_te_xi_t_12.jpg" alt="i-1e210cc1f64f80858cceb227378b4c95-la_te_xi_t_12.jpg" /></p>
<p>A couple of problems. What happens when I shoot the ball straight up? Then the x-component of the velocity is zero. So, it doesn't work for this. Also, the higher the ball is shot, the less accurate this will be. Also, this assumes I have the launcher at the correct angle. I used the Tracker Video angle measuring tool and it essentially agreed with my measurement.</p>
<h3>Launch Velocity Method 2</h3>
<p>This next method finds the total launch speed from both the x- and y-velocites using:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-5a98d071715a87d461f01ac9dce5826a-la_te_xi_t_1_14.jpg" alt="i-5a98d071715a87d461f01ac9dce5826a-la_te_xi_t_1_14.jpg" /></p>
<p>How do I find the initial y-velocity? I suspect this is where my error from my previous post came. If I wait until the next video frame, I will not be finding the initial velocity. Well, when I fit a polynomial to the y-time data, I get:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-a8a32674346a22ef127156a0fd76fa2b-la_te_xi_t_1_22.jpg" alt="i-a8a32674346a22ef127156a0fd76fa2b-la_te_xi_t_1_22.jpg" /></p>
<p>NOTE: That a is a parameter, it is not the acceleration. If I take the derivate with respect to time, I get:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-230bee5bd0272b44d046b0ee7b00a07a-la_te_xi_t_1_32.jpg" alt="i-230bee5bd0272b44d046b0ee7b00a07a-la_te_xi_t_1_32.jpg" /></p>
<p>That would be great if I knew the time that I want to call "initial" velocity, but I don't. I can find the position of the ball initially though. In fact, when I took data I chose the launch position as the origin. In this way, when the ball is at x = 0, y = 0, that is the velocity I want. Someone please tell me why I only wrote down the polynomial fit for the y-direction and not the x? That was dumb. Anyway, I have the function for y(t). I can find the time that corresponds to y = 0 meters and then use this time for the velocity as a function of time.</p>
<p>This doesn't really work well when theta is 90 degrees. For this case, I plotted y vs. v<sub>y</sub> and I get something like:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-02c236a7d05038530fe35750890dea37-largegun_y_vs_vy.jpg" alt="i-02c236a7d05038530fe35750890dea37-largegun_y_vs_vy.jpg" /></p>
<p>This gives me y as a function of vy. I set the origin to y = 0 so that I can solve for the "roots" of this equation. The positive root will give me the y-velocity at y=0.</p>
<h3>Data</h3>
<p>You have been very patient. Here is what I found. First, for the small launcher.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-649fdd4874cbedf47ddb56b222531ed2-launchspeedsmalllaunche.jpg" alt="i-649fdd4874cbedf47ddb56b222531ed2-launchspeedsmalllaunche.jpg" /></p>
<p>The red line is the data from calculating the launch speed only from the angle and the x-velocity. This one does not have a value for theta = 90 degrees, you know...because. Also, to get a sense of how the function fits varied, here is a plot of the acceleration of the ball for the different shots.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/uploads/2009/10/gforsmalllauncher.jpg" alt="Gforsmalllauncher" height="387" width="512" /></p>
<p>The acceleration for all of these seems off of the accepted -9.81 m/s<sup>2</sup> value. That could easily be due to a scaling problem. The acceleration for the theta = 0 degrees shot is probably off because there were very few data points in that case.</p>
<p>Here is the similar data for the large launcher:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-ee30cf92d41f8c2930f4a844c971f847-largelanucher.jpg" alt="i-ee30cf92d41f8c2930f4a844c971f847-largelanucher.jpg" /></p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-68434baf0a62b74fba2e3c5632abcc61-largecannong.jpg" alt="i-68434baf0a62b74fba2e3c5632abcc61-largecannong.jpg" /></p>
<h3>The Answer</h3>
<p>I don't think the speed of the launcher depends significantly on the angle. First, if it did, I would expect to see some type of trend for v<sub>0</sub> vs angle. I don't see a trend. Second, I don't suspect that the acceleration of the ball changed with each shot, yet it produced a similar looking plot.</p>
<p>I must have just made some type of mistake in my previous attempt. So, I shall certify the PASCO launcher as being an acceptable device. Oh - and I really should have error bars on those graphs, but I have not decided the best way to get the uncertainty from a video analysis. I am still thinking about that one.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Tue, 10/20/2009 - 16:22</span>
Tue, 20 Oct 2009 20:22:16 +0000rallain107962 at https://scienceblogs.comPhysics of Fantastic Contraption I
https://scienceblogs.com/dotphysics/2008/10/23/physics-of-fantastic-contraption-i
<span>Physics of Fantastic Contraption I</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>One of my students showed me this game, [Fantastic Contraption](<a href="http://fantasticcontraption.com/">http://fantasticcontraption.com/</a>). The basic idea is to use a couple of different "machine" parts to build something that will move an object into a target area. Not a bad game. But what do I do when I look at a game? I think - hey! I wonder what kind of physics this "world" uses. This is very similar to [my analysis of the game Line Rider](<a href="http://scienceblogs.com/dotphysics/2008/09/the-physics-of-linerider/">http://scienceblogs.com/dotphysics/2008/09/the-physics-of-linerider/</a>) except completely different. </p>
<p>Fantastic Contraption gives the unique opportunity to build whatever you want. This is great for creating "experiments" in this world. </p>
<p>The first step is to "measure" some stuff. The game includes three types of "balls" and two types of connectors. The balls are:</p>
<ul><li>Clockwise rotating</li>
<li>Counterclockwise rotating</li>
<li>Non-driven</li>
</ul><p>Connectors:</p>
<ul><li>wood lines - these can not pass through each other</li>
<li>water lines - these can pass through each other, but not the ground</li>
</ul><p>First question: Do the different balls have the same mass? This can be tested by creating a little "balance"</p>
<p>![Screenshot 05](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-052.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<!--more--><p>
Now, I can test this by adding two of the same balls on each side (well, one on each side). It is still balanced. Now for two different types of balls:</p>
<p>![Screenshot 06](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-062.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Note: the blue ball does not spin and the yellow is a clockwise spinner. They look balanced. What about a blue and a conterclockwise spinner? Still balanced. So, it appears all the balls have the same mass.</p>
<p>What is the linear mass density for the two types of sticks? To measure this, I created a device with a ball at one end and the pivot NOT in the center, but it still balances:</p>
<p>![Screenshot 10](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-101.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Here you can see three forces acting on the device: the gravitational force on the ball, the gravitational force on the stick, and the pivot point pushing up. Since the stick is clearly not a point object, I have to draw it's gravitational force at the center of the stick. (I am not going to derive that right now, you will just have to trust me).</p>
<p>Newton's laws says that the forces must add up to the zero vector if the object is staying at rest. This means (in the y-direction, where y is up):</p>
<p>![Screenshot 11](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-111.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Here m<sub>s</sub> is the mass of the stick and m<sub>b</sub> is the mass of the ball. This would make the the gravitational pull on the ball -m<sub>b</sub>g (notice it is the y-component, so I can have it negative). From all of this, I could solve for the force the pivot pushes on the balance, but what good is that? What I am really looking for is the mass of the stick. To do this, I need to consider torque. Here is the real definition of torque:</p>
<p>![Screenshot 12](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-121.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>This definition is a little more complex than I want to go into (but I had to say it). The torque is technically a vector resulting from the cross product of a force and a vector from the point of rotation to the point the force is applied. The scalar version of torque can be written as:</p>
<p>![Screenshot 13](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-13.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Here, r is the distance from the point that you want to calculate the torque about (I chose the pivot point) and the point where the force is applied. ? is the angle between the force and the distance to point about which to calculate the torque. In this case, the angle is 90 and sin(90) = 1. Another important consideration is the sign of the torque. I will arbitrarily call counterclockwise torques positive and clockwise torques negative.</p>
<p>So, how do I use torque? Well, I need to know the distance from the pivot point to the center of the ball and from the pivot point to the center of the stick. I can use [my favorite free video anlaysis program, tracker,](<a href="http://www.cabrillo.edu/~dbrown/tracker/">http://www.cabrillo.edu/~dbrown/tracker/</a>) to do this (even though it is just an image)</p>
<p>I will use the diameter of one of the balls as my unit (from the center of an attachment point circle to another one). Doing this, I get the distance to the ball and the center of the stick as:</p>
<p>![Screenshot 15](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-151.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>- Here I am using "U" as my distance unit - described above.<br />
- To find the distance from the pivot to the center of the stick required some trickeration. I measured the length of the stick. I then used half that distance and measured from the one end of the stick to find the center. Knowing that point, I could then measure to the pivot point. Using these measurements in the torque equation:</p>
<p>![Screenshot 16](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-162.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Note that the torque due to the pivot does not contribute at all. This is because I calculated the torques about the pivot point. The distance from the pivot point to the pivot point is zero (thus zero torque).</p>
<p>So, I have the mass of the stick in terms of the mass of the ball. I can also get the linear mass density of the stick:</p>
<p>![Screenshot 17](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-171.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Cool - I should stop here. No!!!! I am on a roll. I will now calculate the linear mass density for the "water" stick. I can't do quite the same thing because the water would fall through the pivot. Instead, I will do the following. First, I will make a stick with two ball (one on each end) balance. Then I will replace one of the balls with "hanging" water so that it is still balanced. At this point, the mass of the water stick will be the same as the ball (I could have done this with the wood stick if I had thought of it then).</p>
<p>![Screenshot 18](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-181.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>You may not be able to tell, but this is two overlapping full water sticks and one shorter one. I will have to combine the length of all of these. This gives a total length of water = 8.5 U. So, the linear mass density for water is:</p>
<p>![Screenshot 19](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-191.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Interesting. The linear density is half that of the sticks. Must be dense sticks. I tried putting a wood stick versus a water stick that was twice as long - they balanced.</p>
<p>**Acceleration of falling objects**</p>
<p>Do things accelerate? Is there air resistance? I created an engine that just kind of "flung" a ball up. I used [copernicus](<a href="http://www.danicsoft.com/projects/copernicus/">http://www.danicsoft.com/projects/copernicus/</a>) to capture the video from the screen. Then [tracker video](<a href="http://www.cabrillo.edu/~dbrown/tracker/">http://www.cabrillo.edu/~dbrown/tracker/</a>) to get position time data. Here is what I found:</p>
<p>![Screenshot 20](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-202.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>This shows that it does indeed acceleration. Using [the ideas from a previous post on graphing](<a href="http://scienceblogs.com/dotphysics/2008/09/basics-making-graphs-with-kinematics-stuff-part-ii/">http://scienceblogs.com/dotphysics/2008/09/basics-making-graphs-with-ki…</a>), the acceleration of the object is twice the coefficient in front of the squared term, this means that:</p>
<p>![Screenshot 21](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-212.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>If this is on Earth, then this acceleration should be 9.8 m/s<sup>2</sup>. With this assumption, I can find the conversion from U to m:</p>
<p>![Screenshot 22](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-223.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>**What is left?**</p>
<p>Questions to answer:</p>
<ul><li>Is there air resistance? From the above data, maybe not. To test this, I need to launch a ball with very high speed. If the horizontal velocity changes, then there is likely air resistance</li>
<li>Make a pendulum, does it oscillate at the expected rate (assuming the dimensions from here)? I already started to set this up, but there is CLEARLY some type of frictional force slowing it down.</li>
<li>Friction - what is the coefficient of friction? Does this game follow the model for friction where the frictional force is some coefficient times the normal force?</li>
<li>What kinds of torque are these rotating balls capable of</li>
<li>What is the moment of inertia of these balls? Are the cylinders or spheres?</li>
</ul><p>I will probably answer some of these questions - but if someone ones to answer them first, I will gladly link to your results OR post them here.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Thu, 10/23/2008 - 11:14</span>
Thu, 23 Oct 2008 15:14:32 +0000rallain107744 at https://scienceblogs.comCollaborating on a data analysis project: students do the math with the Google Docs spreadsheet program
https://scienceblogs.com/digitalbio/2007/11/13/classroom-collaborations-math
<span>Collaborating on a data analysis project: students do the math with the Google Docs spreadsheet program</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>For many years, I had my biotech students do projects where each group of students would analyze their own data, in addition to all of the data gathered by the class. I would draw a table on the white board and each group would enter their data. At the end of the class, all the groups would copy all the results into their notebooks, then analyze them in Microsoft Excel.</p>
<p>This worked pretty well, but it wasn't perfect. There were always cases where one group would be really slow, or someone had to leave early, or I needed to use the board and couldn't. </p>
<p>And, this method <em>certainly</em> wouldn't work for an on-line course. You don't want students to have to post lots of data to forums or e-mail spread sheets back and forth. </p>
<p>Google Docs spreadsheet program has solved this problem!</p>
<!--more--><p> Students can gather data and collaborate on-line on the <i>same</i> document at the <i>same</i> time.
</p>
<p>And I can control whether students are able to edit a document, or simply view the document and download the results.
</p>
<p>I can even share the document with the world, if I so choose.
</p>
<p>And when I tried this out in my class a couple of weeks ago, it worked really well!</p>
<p>Here's what we did:
</p>
<ol><li>Before class, I set up the table for data entry.</li>
<li>All my students went to <a href="http://www.google.com">www.google.com</a> and signed up for a gmail account (if they didn't already have one).</li>
<li>I clicked the Share tab and entered a list of my students' gmail addresses.</li>
<li>Then, I clicked Invite Collaborators to send an e-mail to the students.</li>
</ol><div class="inset center"><img src="http://scienceblogs.com/digitalbio/wp-content/blogs.dir/460/files/2012/04/i-d6c4fc71dda1c2099af1b10016553bb5-share.gif" alt="i-d6c4fc71dda1c2099af1b10016553bb5-share.gif" /></div>
<ol><li value="5">The students clicked the link in their e-mail to access the spread-sheet.</li>
<li>They entered their results in the spread sheet, simultaneously, as they worked on identifying their bacteria via blastn. This went on during and between class periods.</li>
<li>Then in the next class, I used the Sort function to sort data, show them what happened, and discuss some of the issues related to data analysis and bioinformatics, for example:</li>
</ol><blockquote><ul><li>being a good citizen and entering your data in community databases (<i>I had never realized before this fall that so many of the information-poor entries in GenBank are that way because of the people who entered the information so poorly!</i>)</li>
</ul><ul><li>consistent data entry (<i>it's hard to analyze your data when all the entries describe the same thing differently</i>)</li>
</ul><ul><li>the work that you have to do to clean up your data <i>before</i> you can analyze it, like fixing names, making all the entries consistent. </li>
</ul></blockquote>
<div class="inset center"><img src="http://scienceblogs.com/digitalbio/wp-content/blogs.dir/460/files/2012/04/i-eb5dd2e4b74f2e0d69c5c159d83919ed-spreadsheet.gif" alt="i-eb5dd2e4b74f2e0d69c5c159d83919ed-spreadsheet.gif" /></div>
<p>Then, all the students downloaded data to their own computers for analysis and graphing. The students could have made some graphs in the Google document, but I wanted <em>all</em> the students to get practice. Second, I wanted to them to use PivotTables to analyze the results for the two different biomes (creek vs. forest) and that feature isn't available in Google Docs.
</p>
<p>If you'd like to take a look at our results and see the table, I also published our data. You can find it at: <a target="_blank" href="http://spreadsheets.google.com/pub?key=pGTLOe_72ueskC9w3c3AaUw">JHU_bacteria_2004</a>
</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/sporte" lang="" about="https://scienceblogs.com/author/sporte" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">sporte</a></span>
<span>Tue, 11/13/2007 - 03:20</span>
Tue, 13 Nov 2007 08:20:16 +0000sporte69577 at https://scienceblogs.com