Anti-Capsulation Calculation with Lifting AGV/AMR

The lifting AGV is smaller than the upper pallet and forms an inverted triangle structure. Therefore, during emergency braking, due to the effect of inertial force, the pallet + material has the risk of overturning. In addition, the lifting plate will carry a pallet for rotational movement. If the centrifugal force is too large, there will be a risk of overturning.

Capscaling Calculations During Start-Up and Braking

L: Load length; h: Load height

L: The distance from the center of gravity to the edge of the lifting plate; h: The height from the height of the lifting plate;

R: The radius of the center of gravity from the geometric centerline.

When starting, the condition that there is no overturning occurs is: Mg×l>ma×h, then a＜gl/h. As long as you know the acceleration, you can judge whether there is a overturning problem. In addition, this formula can also be used to calculate the minimum size value of the lifting plate.

During emergency braking, the condition for not capsizing is:

Mg×l>ma×h, then a＜gl/h,

 as long as you know the deceleration, you can judge whether there is a problem of capsizing.

Braking time: Braking time: T=10v/3gu;

Brake acceleration: A1=v/t.

Overturning When the Lifting Plate Rotates

When rotating, the condition that no rotation centrifugation occurs is

mw×wr＜mgu, i.e. r＜gu/w²   (gravity acceleration × friction coefficient/(rotation acceleration × rotation acceleration))

Case 1: Calculate Whether There Is a Risk of Overturning

1. When the overturning torque calculation is started, the conditions for not overturning are:

a＜gl/h=10×515/400=12.875m/s²;

Check the AGV product manual, the acceleration a=0.6m/s², so it will not overturn when starting.

When braking,

Braking time: t=10v/3gu=10×1.8/(3×10×0.5)=1.2s;

Brake acceleration: a1=v/t=1.8/1.2=1.5m/s＜12.875m/s².

Therefore, it will not overturn when braked.

2. Rotation offset calculation

During rotation,

r＜gu/w² =10×0.5/(1.22×1.22) =4m,

The 2.3m pallet must satisfy the center of gravity position within 4m of the geometric center. (The rotation speed of 70°/s is 1.22rad/s), so there will be no rotational overturning.

In summary, in case 1, there will be no risk of overturning

Calculate the AGV’s Own Parameters

1. Calculation of lifting plate size is known: the size of the pallet and material is 1100×1100×1500mm; the center of gravity of the material is calculated according to its geometric center, and the distance between the center of gravity of the material and the AGV lifting plate is 750mm. Assuming that the distance between the center of gravity of the material and the edge of the AGV lifting plate is L; (1500mm is the height)

The condition that there will be no overturning is

F1×L>F2×750mm, then mg L>ma×750

Then L>750×a/g

In the rising formula, there is only the positional parameter acceleration a. As long as the acceleration a is calculated, the minimum value of L can be obtained.

Theoretical calculation Ft=Mv=(3/10)Mgut  (the proportion of the total load on the driver);

M: AGV+total load weight;

u: The friction coefficient between the driving wheel and the ground is 0.5;

v: The speed of AGV under load (full load speed 0.5m/s);

t: Braking time.

Then Mv= Mgut×3/10,

t=10v/3gu=10÷0.5/(3×9.8×0.5)=0.34

Then a=(v-v0)/t=(0.5-0)/0.34=1.47

So in theory:

The distance from the center of gravity of the material to the AGV hoisting edge L>750×1.47/9.8=112.5mm

The dimensions of the AGV lifting disc in the length direction should be greater than 225mm (2×L=225mm)

The above calculation is based on the position where the center of gravity of the material is located in the middle of the AGV lifting plate. Considering the bias, the size of the AGV lifting plate in the length direction should be appropriately increased.

Considering the stability of operation and the existing structure, the center of gravity of the material should be within the range of 200mm of the AGV lifting plate, so

AGV lifting plate should be greater than

2×(200+112.5)=625mm

In the width direction, the AGV top plate should be larger than

2×200=400mm.

2. How much external force can cause the tray to be offset

With the above formula, we can know that the conditions for the tray to be offset are:

a＞gu;

From the previous calculation results, we can see that a=1.47, g=9.8N/kg

After actual measurement, it can be seen that the friction coefficient between the pallet and the rubber plate is 0.6, then F3＜F4

The tray will not be offset on the AGV. If the offset occurs, the required external force

 F4=mgu=500×9.8×0.6=2940N.