Fluid Power Journal

Problem: Find the Cylinder Rod

By Ernie Parker, CFPAI, CFPSD, CFPS, CFPMM, CFPMT, CFPMIP, CFPMMH, CFPMIH, CFPE

During the school year of 1975-1976 and with the help of my students, we built the first fully functional hybrid car using two ten-gallon accumulators with a Brady GT body on a VW chassis. One of our challenges was to find how much actual hp we needed to overcome the various resistances while moving the vehicle at 60 mph. One of our options was to use an oscilloscope with a load cell, but that would require a generator in the back of our tow vehicle. Remember this was 1975. A second option was to use a cylinder by placing a little oil in the rod end and putting it in series with a tow rope. Then we could use a pressure gauge with a long hose and read the pressure as we were driving. The oil would prevent the cylinder from bottoming out and not giving a true pressure reading.

The challenge in this Air Teaser is to solve for the rod diameter using a 3-inch bore cylinder, where the resistance would produce a pressure on the gauge that would be equal to the hp at 60 mph.

Helpful formula: hp = (pounds of pull x feet/sec.) / 550

See The Solution

We need to find a solution where the resistance would produce a pressure of 1 psi for each hp.

60 mph = 88 feet / sec.

1 hp = ?  lbs. of pull x 88 / 550

1 hp = 6.25 lbs. of pull

3” diameter cylinder = 7.068 sq. in.

Since 1 hp = 1 psi, then 6.25 lbs. requires 6.25 sq. in (effective rod-end area)

7.068 – 6.25 = .818 sq. in. of rod

A = D² x .7854

Working backward, rod diameter = 1. inches.

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