Presentation on theme: "Car Jack Mast Design Update"— Presentation transcript:
1 Car Jack Mast Design Update Presented by Doug Eddy and Dr. Sundar Krishnamurty at UMass Amherst for Hoppe Tool on 7/16/10
2 Control of Scissors’ Motion This mechanism is under constrained.
3 Potential RemedyTheoretically, the bottom pivots far apart shortens travel and close together affects stability.
4 Alignment without Binding Connecting links neededbetween center pivotsHow well will the 3sides move together andkeep top in line withbottom?
5 Consider an actual Car Jack Design The screw is horizontaland moves up anddown during lifting.
6 Consider a Conventional Scissor Lift The shape is rectangular.Parallel scissors arerigidly connected at outerpivots. Legs are made ofrectangular tubing forstrength and stability.
7 Engineering Analysis A simplified explanation is found at: The information is verified and correct.This reveals a design challenge with the multiscissor arrangement.The screw force required is very large when the angle is small at the start of lifting.The threshold power spec is 850 Watts.Mechanical advantage with unequal scissor’s lengths is only 15-35%.
9 Car Jack Potential Design Remedies Longer scissors’ lengthIncrease retract height and scissor start angleLonger pitch lead of screw and nut lengthMust also minimize friction and consider those effectsClevis mounted angle optimized drive pivot mount?
11 Claims of the Belt Design LERC S.A.BP – SAINT AMAND LES EAUX CEDEX – FranceTel: – Fax: –Internet:Page 2/4Main Advantages• Resistance to bullet impacts: a bullet impact on a pneumatic mastmanufactured from light alloy or pultruted composite will make a hole thatwill result in an air leak and in the mast collapse. It will also make a slightcrack in the matrix likely to break the tube. In a belt drive telescopic mast,a bullet impact (see picture) will make a hole without affecting the mastheight. Moreover, the woven and crossed structure (Filament Winding -FW) of the composite material prevents any crack in the tube.• Height maintained at constant level when the mast is in erection foran extended time : a pneumatic mast tends to go flat and therefore toretract, which can result in a cutting off of the radio contact.• Outstanding resistance to corrosion, chemical attacks and ageing ;• Undeformability: the tube sections show no permanent deformation even after extensive use (strengthmaintained, no ovalizing);• Best ratio between deployed and retracted heights• Lightweight and outstanding mechanical resistance• No maintenance other than wiping or brushing to clean and for the telescopic masts, replacement of thebelt without dismounting tubes (can be performed on the field).• Excellent resistance to environmental conditions (use of Epoxy resin and FW process): Sand, dirt,dust, snow, ice will not cause degradation of mast performance. On mast with the new belt system, the beltis fully inside the mast, protected against outside environment.• No air tightness to ensure : no adjustments to make height maintained at constant level ;• Manipulation with naked hands, even under cold or hot temperature ;• Adaptability to the customer’s needs : the mast structure is computer designed (SAMCEF method)• LERC proven experience: close to 60 years in the field of composite materials, 30 years in themanufacturing of tactical masts and antennas.
12 Design Concerns with Strap Concept Resistance of bending around small pinsPossibility of slipping at speed desired if not enough tensionContact area on small pinsPower required increases with tension and the number of pulley contacts for simultaneous lifting.
13 Question for you?How do you and the customers rank or weight the various features?:Low power requiredSmall area, lengthLow weightHigh reliability and strengthLow sway and deflection, high stabilityLow start heightHigh extended heightLow costHigh payload capabilityHigh speed capabilityHigh center clearance area and proximity for the cabling and basket below
14 Consider a Simple Straight Pulley Lift Work output = mgh = 45 pound payload x 20 feet lifting height = 900 lb - ft If the total system mechanical efficiency were 64% (although not realistic , for a system lifting from underneath) Work input needed = 900 / 0.64 = 1406 lb – ft = F x d = System pull force x System pull distance This must be done within 15 seconds, Power = Work done per unit time = 1406 / 15 = ft – lbf / second 550 ft – lbf / sec = 1 Hp, so for a 93% efficiency motor / (550 *0.93) = Hp = 137 Watts