Optimizing Angle for Material Transfer System Efficiency

  • Thread starter Thread starter physicsodyssey
  • Start date Start date
  • Tags Tags
    Angle
AI Thread Summary
The discussion focuses on optimizing the angle for a material transfer system, specifically questioning the suggested 5-degree angle for effective operation. This angle is believed to facilitate a hauling effect that allows the tray to revolve back, overcoming friction in the pivot bearing. Participants emphasize the need for mathematical approaches to determine the optimal angle, considering parameters like mass, arm lengths, and counterweights. The equation presented aims to calculate the minimum load required to unbalance the arm and enable movement along the conveyor arc. However, the conversation highlights a lack of specific data necessary for calculating speed and shock absorption requirements, indicating that further details are needed for precise optimization.
physicsodyssey
Messages
11
Reaction score
6
I am trying to design a material transfer system ( the system at duration 0:57 to 1:34 ).
The general layout is http://i.imgur.com/ajQ05qy.png
It was suggested that angle be 5 degree. The reason is for hauling effect so the tray can revolve back. can you explain this? why 5 degrees? supporting references will be of great help.
Thank You.
 

Attachments

  • fzg.PNG
    fzg.PNG
    6.7 KB · Views: 450
Engineering news on Phys.org
physicsodyssey said:
why 5 degrees?
Whatever inclination of the plane of the arm and counterweight is adequate to overcome the friction of the pivot bearing and yield a cycle time between loading and unloading the tray that is suitable for the process.
 
There can be a range for it but how to approach it mathematically ?
Could you please elaborate it in detail? How would you approach the problem ? Do you need any data? m1=3kg & m2=10kg
I want to clear this concept and your help is highly appreciated.
 
m 1 = Mass of empty pan in kg
Pw = Weight of parts in N
m 2 = Mass of counter weight needed to be attached in kg
θ = Angle of inclination of the conveyor track
x = Length of tray arm from axis of rotation in mm
y = Length of counter weight arm from axis of rotation in mm
a = Diameter of bearing housing in mm
Height of counterweight from datum = 900 mm
Height of pan from datum = 913 mm
x = Length of tray arm from axis of rotation in mm =1000mm
y = Length of counter weight arm from axis of rotation in mm = 300mm
m1=3kg(assumption)
m1(h + a sin θ) g x = m2 y h g
m2 = 3 x 900 x 913 /300 x 900 =9.13kg
How to proceed for assumption of θ?
Can you explain me the following equation?
Pwg ≥( m1 * x^2 + m 2 * y^2 ) sin2 θ / [ x(h-xsin2 θ)]
 
physicsodyssey said:
Pwg ≥( m1 * x^2 + m 2 * y^2 ) sin2 θ / [ x(h-xsin2 θ)]
This is supposed to be giving you the minimum load (part weight) to unbalance the arm so it will swing and carry the part around the conveyor arc from station 1 to station 2 and hold it there. Unless the tray/pan and counterweight arms aren't 180 degrees opposed, there's no reason for that load to depend on the magnitude of theta other than that it be off vertical, so the form of the equation is somewhat arcane.

There isn't enough information to to calculate the speed of the movement, or the shock absorbing requirements at the ends of the travel range. Speed increases as theta increases, but without knowing particulars of bushings/bearings on the pivot, viscosities of lubricants, moments of inertia of the loaded and unloaded assembly there's no way of coming up with numbers.
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...
Back
Top