I took a historical interest in a 21 mile 11,000ft rise cableway in Chilecito, Argentina. It was built to reach the Mina La Mejicana in the Famatina region. In 1905 it set a world record for length and it still is for rise. I worked out the catenary for all 275 towers on a spreadsheet and put it into a KML file so I could take a ride down the line in Google Earth.
In the course of this activity I felt that some boiled down observations could bring catenaries down from higher math to levels that could be easily entered in Excel.
The horizontal force along a catenary cable is equal everywhere, pulling to left on the left of the low point and to the right on the right of it. The vertical force at a support tower is equal to the weight of the cable from it down to the low point. The vertical force at a point on the way down to the low point is equal to the weight of the remaining cable to the low point. The horizontal and vertical vectors at a given point give the angle of the cable at that point and the tension in it. The point tension increases from the low point to a maximum at the tower.
If a tower mount is not clamped then the horizontal tension in any adjacent catenary will be the same. This leads to using a spreadsheet to connect and adjust the sags in a series of catenaries over uneven terrain.
The span and the tower height are not the only givens, the ratio of the tension to the unit cable weight is also a user choice. The heavier the cable the deeper the low point will be pulled, varying the sharpness of the sag. If the cable had zero weight it would be a horizontal line with zero sag. If the total weight per half length equaled the tension it would drop straight down and rise straight back up to the next tower with a span of zero feet. The key point for all the curved shapes in between is that there is only one shape of hyperbolic curve, we are just using the tension/weight ratio and span dimensions to scale where our span fits across that curve.
To calculate a catenary by hand or in Excel, choose:
-a trial horizontal tension N
-a sample cable weight per unit length Q
-the horizontal span distance L
Working in feet & lb or in meters & kg, we find:
A = N/Q the tension to unit weight ratio.
If there is a difference in the elevation of the tower tops, delta He, we find:
M = A*asinh(((delta He/A)/2)/sinh((L/2)/A)) the low point's offset from middle of the span.
Otherwise we use the mid point of the span as the low point.
Either way, using Lp as the distance of a point of interest from the low point we find:
Hp = A(cosh(Lp/A)-1) the height of the point above the low point.
This is also described as the sag if it is figured for an end point.
It is valid for symmetrical spans, for unequal tower top elevations with the low point between them (the OP's question, also called an inclined catenary), for unequal tower top elevations with the low point to one side of them and for any intermediate point.
Sp = A*sinh(Lp/A) is the running length along the cable from the low point.
2x the Sp for the end point of a symmetrical catenary finds the whole span.
Hp is the y where our Lp fits the x of our hyperbolic curve and Sp is the arc length along the curve.
For graphs just generate an incremented table of Lp points as + or - offsets from the low point, copy down the Hp height calculation and graph Lp & Hp as x & y. Start and finish with the Lp for the actual top points to make the top data calcs exact. For the inclined catenary I plotted a second calc moving the lower tower out to where it would be the same height as the taller one. Made this a dotted line and then it was easier to see how the subject span was just part of a longer hyperbolic curve.
The calcs can be done directly in Excel. Goal seek would be needed if you want to go from a desired sag re ground clearance back to the allowable tension or weight. But the one goal seek can solve an entire span or string of spans. If your terrain is extreme you can get locations where the upper tower is actually trying to lift the lower tower (the mid point calc gives an answer outside your span). Then you have to reduce N until the sag droops away from the low tower. Or have a receiving tower design that can resist the uplift, this complicates the cable attachment if it is a cableway. Overall there must have been a lot of trial and error in manual calculation days. I recollect once seeing Plexiglas curve templates being used by transmission line designers to solve fitting to route profiles.
For a point load at the low point, figure out an amount of L about that point that has a cable weight equal to your load. Cut that length out of your graph and slide the two pieces back together. The vertical tension left at the cable ends will equal your load. Could probably just goal seek an Lp offset back up the sag until the point's vertical tension/Sp cable weight matches half your cargo. Oops, actually you have to get back to having the original span. Major distributed point loads such as suspension bridge suspenders would be more complicated. Loads distributed uniformly along the run of the cable are just a heavier cable.
Okay, that's enough recapping of what many of you already know better than I do. Let's go for a ride on the Chilecito cableway, it's in the attached KML below. Set Google Earth's Tools/Options/Touring line settings to Camera tilt 75deg, Camera range 15 meters and Speed 175. Then select the "Top to Bottom" line in Places and click the Play Tour button at the bottom of the Places box.
Sagen.AT has a paper the builder's (Adolf Bleichert & Co) Senior Engineer wrote on the project. Note that the plan views are flipped top to bottom. Google translate will fix the language.
[link "Sagen.At"]
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Wikipedia has articles on the cableway
[link "Mina La Mejicana"]
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[link "Materialseilbahn Chilecito-La Mejicana"]
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The site is now a tourist attraction, search Google Image for Chilecito Cable Carril for site pictures.
Bill
