The common geometry of the Rocking-Toilet Seat remains the same configuration throughout. Methods of Manufacture match those of current toilet seats. The coil springs when inserted and provide the strength and flexibility needed.
It is well known that dogs when relieving themselves, go through a series of pumping motions while walking, slowly, which enables the dog to gradually eject the stool from his body. In a similar manner, a person, will find relief by rocking front-to-back, or side to side, continuously, on a spring supported, stool seat, until the bowels are discharged. The “Rock-Away,” will be particularly useful to those adults, and children, subject to constipation.
a.) The device is clipped into place atop the toilet seat by sliding the clips into place. The Rock-Away can be readily lifted, and removed from the toilet seat by removing the 4 clips.
b.) The Rock-Away is designed to be light, compact, easy to attach.
c.) The four spring units and the stool seat will not be exposed since they are inside (2) layers of abrasive resistant nylon.
d.) The Rock-Away will have the same shape as the toilet seat.
e.) The Rock-Away may be used by both children and adults.
f.) The Rock-Away does not become soiled with feces or urine because it is located above and at a sufficient distance from the anus and the uretha.
g.) Both backward forward and sideward angular motion of the Rock-Away may be used alone or with assistance.
h.) For new water closet installations, wherein there is no existing toilet seat already in place, the Rock-Away's bottom seat will be extended 2 inches with bolts supplied, for a normal toilet seat connection.
Assume torsional modules of elasticity.
Load=80 lb., Mean Spring Diameter=1.5″, C=11, 500,000 psi.
Assume allowable working stress's=60,000 psi
Try No. 6 wire=0.192 inches Find Wahl Factor
Trial wire size
stress less than 60,000 psi
Use #7 Wire
Assume clearance between loaded coils={fraction (1/16)}″
Then Pitch=L=B+f+d={fraction (1/16)}″+0.0191+0.192″=0.444″
Clearance=L−f−d+0.444″−0.191−0.177″=0.076″
Assume solid Length is less than 1{fraction (7/16)}″
Therefore Solid Length=h=8×0.177=1.4″
Free Length=H=8×0.444″=3.5″
Pitch L Per Coil=of Loaded Spring=1
Pitch 1 per coil=L−f=0.444−0.191=0.253″/coil
Assume Hw=2.5″ and end coils are squared.
When N=8
H=8×0.444″+3×0.177=4.1″
Total Deflection=N=(L−d)=8(0.444″−0.177″)=2.136″
Working load: Working Deflection: Maximum Load: Maximum Deflection
Maximum Working Load=2.136″×80 lb/1.53″=112 pounds
Solid height load
Working Stress: Maximum Stress: Working Deflection: Maximum Deflection
The Ratio of Mean Spring Diameter to Wire Diameter, i.e., the “Spring Index” should be between 6 and 9, wherein 9 is ideal.
U.S. Patent Documents 5311617May, 1994Ammatelli4/237; 4/2545666673Sep., 1997Ammatelli4/234; 4/237; 297/3355412815May, 1995Ellis4/239; 4/2375996133Dec., 1999Fletcher4/239; 4/237