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. 

{img src= width= height= align= desc="Model 1" link= }Model 1  shows a chain of 40 jumbo gold paper clip links (from Staples), with three others with three different pairs of tangent points and  tangents .

{img src= width= height= align= desc="Model 1A" link= }Model 1A. shows the original chain compared to one with 20 paper clip links replaced by two black equal tangents  with 20 links concentrated at their intersection. The fact that the 20 remaining hanging links (10 on each side) maintain the same shape as the original chain, shows that the forces acting at the tangent points (at the bottom of the 10th link on either side) are the same as those acting at the corresponding points on the original  chain. Or, you could say, that the rest of the catenary links can’t tell the difference!

{img src= width= height= align= desc="Model 1B" link= }Model 1B shows the original chain compared to one with 16 links replaced by 2 unequal blue tangents , at the bottom of the 10th link on the left, and at the bottom of the 14th link on the right, with the 16 removed links concentrated at the intersection of the tangents. Again, the 10 remaining hanging links on the left and the 14 on the right maintain the same shape as their cousins on the original catenary, so the forces must be the same.

{img src= width= height= align= desc="Model 1C" link= }Model 1C shows the original chain next to one where 10 links on the right side, from the bottom of the catenary up, are replaced by two red tangent strings , with 10 links concentrated at the intersection point. Note that the tangent line beginning at the 20th link, (at the bottom, the Singular point of the catenary), runs horizontally, so the force acting this point acts horizontally. Again the remaining chain holds the same shape as the original whole catenary, so the forces must be the same.
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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.

{img src= width= height= align= desc="Model 2" link= }Model 2 shows, from the top left to the bottom center, 20 links of a 40-link symmetrically hanging chain. Instead  of the second  half  of the catenary, there is a string attached to the 20th bottom link which runs horizontally to a pulley (from Home Depot) , with 9 links hanging vertically, which counterbalances the half –catenary so that its bottom point is exactly where it would be if the whole catenary were there. This shows that the force at the bottom of the chain is equal to the weight of 9 paper clip links.

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

{img src= width= height= align= desc="Model 3" link= }  Bernoulli states  in his lecture  “…5. The weight P, which is held by any two arbitrarily situated strings AB and CB, exerts its forces on the points A and C in such a relation, that  the necessary force at A is to the necessary force at  C (after drawing vertical line BG), as the sine of  angle CBG is to the sine of angle ABG, and the force of the weight P is to the force at C as the sine of the whole angle ABC is to the sine of the opposite angle ABG. This is proven in every theory of   Statics.”  Right.  Footnote 2 of the translation in 21Century  gives a blackboard demonstration of  Bernoulli’s statement.  Model 3. is intended to help explain the explanation. 

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.

{img src= width= height= align= desc="Model 3A" link= }
{img src= width= height= align= desc="Model 3B" link= }

{img src= width= height= align= desc="Model 4" link= }Model 4  shows Figure 5 from the translation of Bernoulli’s lecture in 21Century, with an actual chain of 20 paperclips for AB and an actual weight at point E consisting of  20 paperclips. Using the relations of forces and angles discovered in Model 3, the relationship of the Force of the whole weight and the force at the bottom can be compared to the Sine of the whole angle AEB and the Sine of the angle opposite the point B, angle AEL. It is easily seen that the Sine of angle AEB is equal to Cos AEL.
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

{img src= width= height= align= desc="Model 4A" link= }
{img src= width= height= align= desc="Model 4B" link= }

{img src= width= height= align= desc="Model 5" link= }Model 5  shows that for this catenary curve of 20 paper clips, the segment  C=a  from Bernoulli’s text has a length of  8 paperclips: it is the constant  length of chain which has a weight equal to the constant force at B. First making a mark where every paper linked with the other, our researchers precisely drew and measured with a plane leveler the differentials dx and dy of  each  “differential paperclip”.  We then calculated the differential ratio dx/dy for each and compared it to the ratio s/a for each paperclip along AB(we first did the measurements in the original pencil before making them in permanent marker).  We have experimentally verified Bernoulli’s differential equation dx/dy=s/a! All of this brings one to ask, what unseen principle of change is expressing itself in this data, which, although expressed differently at every point of the chain is maintaining a relationship between the physics and the rate of change of the curve.

{img src= width= height= align= desc="Model 5A" link= }
{img src= width= height= align= desc="Model 5B" link= }

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!