Paperclip Pedagogy for Calculus and Catenaries
Paperclip Pedagogy for Calculus and Catenaries by the Boston LYM
In his Lecture on the catenary, in assumption 2 Bernoulli states that if a section AC between points A and C on the chain is replaced by two weightless strings tangent to the curve at A and at C, with a weight equal to the missing chain hanging at their intersection point, then the force at A will be the same as it was in the original whole catenary, and the force at C will be the same as it was in the original catenary.
Bernoulli states in assumptions 3 and 4 of his lecture that if the fastening points are changed along the catenary curve, the shape of the remaining curve does not change, so therefore the forces acting at each point along the curve do not change, consequently the force at the bottom of the chain will be the same.
From the top right to the center we have the 15 bottom-right links of the same 40-link catenary, again balanced against the weight of 9 paper clips. From the left again, the 8 bottom-left links of the same catenary, balanced against 9 paper clips. At the top of the model, the two (!) bottom links of the original catenary, balanced in position against 9 paper clips. This shows that the force at the bottom of this 40-link chain is equal to the weight of 9 paper clips no matter how much of the original chain is left.1
Nothing is moving at point A. There you see line BA, meeting a horizontal line, labeled in red Fasinα and a vertical line, labeled in red Facosα. The horizontal line leads to a pulley which has giant paperclip weights hanging from it, which pulley redirects the force of the weight of the giant paperclips to act horizontally to the left at point A. The vertical line leads up to another pulley which has another weight hanging from it, which pulley redirects the force of that weight to act vertically upward at point A. Since nothing is moving at Point A is not moving, there must be a force Fa acting along line AB, down and to the right, which is equal and opposite Fasinα plus Facosα. Therefore Fa=Fasinα + Facosα. Or, Fasinα equals the horizontal component of Fa and Facosα equals the vertical component of Fa. Similarly, at Point C, in green , we have Fc=Fcsinβ+Fccosβ, Fcsinβ the horizontal component, FCcosβ the vertical component of Fc. Now look at point B. Nothing is moving. Therefore the forces acting there cancel out. Specifically the horizontal forces acting cancel out, and the vertical forces acting there cancel out. Horizontally , Fasinα pulls left, Fcsinβ pulls right, Fp does not pull either way horizontally. Therefore Fcsinα=Fcsinβ or Fa:Fc :: sinβ:sinα, as Bernoulli stated. Vertically, Facosα pulls up, Fccosβ pulls up, Fp pulls down, so Facosα+Fccosβ=Fp . From there some Algebra and the trigonometric law sin(α+β)=sinαcosβ+cosαsinβ see Model 3A. lead to Fp; Fc= sin(α+β):sinα as Bernoulli states and the footnote elaborates. In Model 3. , we have five sets of giant paper clips hanging from pulleys, as follows: Fp= 18 paperclips, Fasinα=5 paperclips, Fcsinβ=5 paperclips, Facosα=13.5 paperclips, Fccosβ=4.5 paperclips. Note that Fasinα=Fccosβ= 5 paperclips, and Fp=Facosα+Fccosβ= 18 paperclips. The 18 giant paperclips correspond to the 18 small paper clips of catenary segment AC in the model. We could not find adequate pulleys to use with the small clips, so we used large pulleys and giant clips for the weights. Important tip: do not confuse the forces Fa and Fc with the line segments AB and CB. The forces act along these lines but are not equal to them. In our model, α=angle ABG is about 21 degrees and β=angle CBG is about 49 degrees.
From this we derive that the Force of whole weight is to the force at bottom as EL is to AL and consequently from Model 1 that the length of whole chain S to the length of chain equalling the force at the bottom, (a) is equal to Force E/Force B. The topmost paperclip represents the infinitesimal hypotenuse aA of the infinitesimal right triangle aAH, with infinitesimal legs dx and dy, the differentials of vertical and horizontal change, the revolutionary, new thought object of Leibniz’s Differential Calculus. 2 Thus the unseen principle expresses itself in the differential triangle as the relationship of s/a = dx/dy
Stay tuned for more! Coming up next : Models demonstrating Bernoulli’s proof for Leibniz’s catenary construction from the logarithmic (exponential curve) and illustrating LaRouche’s January 1989 memo on the catenary. Notes: 1 If a mathematics professor asks you “ Voulez vous Cauchy avec moi?”, call the police! 2 How to make a catenoid: Fill a bucket with cold water. Add 1 cup of dishwashing liquid. Then drop the cat in! Created by: admin last modification: Friday 10 of June, 2005 [05:35:23 UTC] by admin |
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