Wheel Momentum Study
Determining wheel stability on spacers during fork truck acceleration/deceleration
In the wheel momentum diagram, the direction of forward motion is taken to the right. A wheel of radius R is supported at symmetric points A and B. The horizontal separation between A and B is D and the depth of the arc below A and B is h.
The wheel has an effective weight of W and is acted upon by forces AX,
AY, BX, and BY. For forward acceleration from rest, rotation tends to occur about point A. At the instant acceleration is about to begin, the forces at B become zero. Thus AY equals the weight W and AX both accelerates the wheel forward and produces rotation about the center of the wheel C.
For deceleration, the forces at A are zero when rotation is about to
begin, BX provides the deceleration and rotation about C, and BY
equals the weight W. Deceleration and acceleration are symmetric so this analysis for acceleration can be applied to deceleration as well.
When rotation is about to begin, the sum of the moments about C is zero. Then
AX (R-h) = AYD/2, where R-h is the moment arm for AX and D/2 is the
moment arm for AY. Since AY = W, we find that
AX = WD = minimum force necessary for a wheel to move out of wheel channel.
2(R-h)
Example 1:
Moving a full skid of 30 lb., 16-inch (17.5 inches) wheels by fork truck, we wish to determine the force necessary to rotate the bottom layer of wheels out of the wheel channels.
The weight imparted on the bottom spacer per wheel is calculated as follows:
3 load-bearing channels on the spacer for 14,15 and 16-inch wheels; 4 wheels/channel
12 wheels per layer X 3 layers = 36 wheels @ 30 lbs. each = 1,080 lbs.
Adding the weight of three spacers @ 20 lbs. each = 60 lbs + 1,080 lbs. = 1,140 lbs. total weight.
1,140 lbs./12 wheels on the bottom spacer = 95 lbs. weight per wheel on bottom spacer.
Thus, in our equation W = 95 lbs.
In the current design of the 14_16 inch polymer wheel spacer, D = 7.125 inches, h = 0.75 inches, R = 8.75 inches (17.5 inches/2), and W =
95 lbs. These values indicate that AX = 42.3 lbs. is the minimum force to produce rotation of the wheel.
For purposes of this report, we will now use a usable fork truck speed of 15 miles per hour or 22 ft./second. In order to determine acceleration (change in velocity), we will use Newton’s second law of motion
F = m x a
Where
F = force (AX)
m = mass (determined from weight W)
a = acceleration
Thus
a = AX /m
where m = weight (W) divided by the force of gravity, defined as 32
ft./second2
In our example, then
m = 95 lbs./ 32 ft./second2 = 2.97 ft. lbs./ second2
a = 42.3 lbs./ 2.97 ft. lbs./ second2 = 14.3 ft./second2
The minimum time for acceleration to 15 mph we use the following formula
V = at or t = V/a
Where t = minimum time for acceleration to 15 mph
V = velocity = 22 ft./second (15 mph)
a = acceleration
In our example, then
t = 22 ft./second divided by 14.3 ft./second2 = 1.53 seconds
Acceleration less than this value will result in wheels being displaced from the wheel holding channels of the spacer. At acceleration (or deceleration) of 1.53 seconds or above, the wheels will remain stable in the wheel holding channels of the spacer.
Example 2:
Moving a full skid of 40 lb., 21-inch (22.75 inches) wheels by fork truck, we wish to determine the force necessary to rotate the bottom layer of wheels out of the wheel channels. The wheel channels in this example are 8 inches across.
The weight imparted on the bottom spacer per wheel is calculated as follows:
2 load-bearing channels on the spacer for 17, 18, 19, 20 and 21-inch wheels;
4 wheels/channel
8 wheels per layer X 3 layers = 24 wheels @ 40 lbs. each = 960 lbs.
Adding the weight of three spacers @ 20 lbs. each = 60 lbs + 960 lbs. = 1020 lbs. total weight.
1,020 lbs./8 wheels on the bottom spacer = 127.5 lbs. weight per wheel on bottom spacer.
Thus, in our equation W = 127.5 lbs.
In one of the current designs of the 17_21-inch polymer wheel spacer, D = 8 inches, h = 0.73 inches, R = 11.375 inches (22.75 inches/2), and W
= 127.5 lbs. These values indicate that AX = 47.9 lbs. is the minimum force to produce rotation of the wheel.
For purposes of this report, we will now use a usable fork truck speed of 15 miles per hour or 22 ft./second. In order to determine acceleration (change in velocity), we will use Newton’s second law of motion
F = m x a
Where
F = force (AX)
m = mass (determined from weight W)
a = acceleration
Thus
a = AX /m
where m = weight (W) divided by the force of gravity, defined as 32
ft./second2
In our example, then
m = 127.5 lbs./ 32 ft./second2 = 3.98 ft. lbs./ second2
a = 47.9 lbs./ 3.98 ft. lbs./ second2 = 12.03 ft./second2
To determine the minimum time for acceleration to 15 mph we use the following formula
V = at or t = V/a
Where t = minimum time for acceleration to 15 mph
V = velocity = 22 ft./second (15 mph)
a = acceleration
In our example, then
t = 22 ft./second divided by 12.03 ft./second2 = 1.83 seconds
Acceleration less than this value will result in wheels being displaced from the wheel holding channels of the spacer. At acceleration (or deceleration) of 1.83 seconds or above, the wheels will remain stable in the wheel holding channels of the spacer.
Example 3:
Moving a full skid of 40 lb., 21-inch (22.75 inches) wheels by fork truck, we wish to determine the force necessary to rotate the bottom layer of wheels out of the wheel channels. The wheel channels in this example are 9 inches across which mandates a spacer depth of 3 inches rather than 2 ½ inches.
The weight imparted on the bottom spacer per wheel is calculated as follows:
2 load-bearing channels on the spacer for 17, 18, 19, 20 and 21-inch wheels;
4 wheels/channel
8 wheels per layer X 3 layers = 24 wheels @ 40 lbs. each = 960 lbs.
Adding the weight of three spacers @ 20 lbs. each = 60 lbs + 960 lbs. = 1020 lbs. total weight.
1,020 lbs./8 wheels on the bottom spacer = 127.5 lbs. weight per wheel on bottom spacer.
Thus, in our equation W = 127.5 lbs.
In one of the current designs of the 17_21-inch polymer wheel spacer, D = 9 inches, h = 0.983 inches, R = 11.375 inches (22.75 inches/2), and W = 127.5 lbs. These values indicate that AX = 55.21 lbs. is the minimum force to produce rotation of the wheel.
For purposes of this report, we will now use a usable fork truck speed of 15 miles per hour or 22 ft./second. In order to determine acceleration (change in velocity), we will use Newton’s second law of motion
F = m x a
Where
F = force (AX)
m = mass (determined from weight W)
a = acceleration
Thus
a = AX /m
where m = weight (W) divided by the force of gravity, defined as 32
ft./second2
In our example, then
m = 127.5 lbs./ 32 ft./second2 = 3.98 ft. lbs./ second2
a = 55.21 lbs./ 3.98 ft. lbs./ second2 = 13.87 ft./second2
To determine the minimum time for acceleration to 15 mph we use the following formula
V = at or t = V/a
Where t = minimum time for acceleration to 15 mph
V = velocity = 22 ft./second (15 mph)
a = acceleration
In our example, then
t = 22 ft./second divided by 13.87 ft./second2 = 1.58 seconds
Acceleration less than this value will result in wheels being displaced from the wheel holding channels of the spacer. At acceleration (or deceleration) of 1.58 seconds or above, the wheels will remain stable in the wheel holding channels of the spacer.
Conclusion:
The largest wheels being moved are the most unstable, so in this paper we have compared movement of 16-inch wheels with that of 21-inch wheels. Because misjudgment by the fork truck operator in moving pallets of wheels can be costly in terms of wheel damage, we have elected to design our spacer for larger wheels (17,18, 19, 20 and 21) with a 9-inch wide load-bearing channel rather than an 8-inch wide load-bearing channel. This necessitated a change of spacer depth from 2 ½ inches to 3 inches for the spacer for larger wheels. The depth of the spacer for smaller wheels remains at 2 ½ inches.
The minimum time for acceleration/deceleration is now almost identical, i.e. 1.53 seconds for the spacer for smaller wheels as compared to 1.58 seconds for the spacer for larger wheels. Acceleration or deceleration less than these values will result in wheels being displaced from the wheel holding channels of the spacers during fork truck movement.
Had the spacer for larger wheels been designed with an 8-inch channel, the minimum time for acceleration/deceleration would be 1.83 seconds. In other words, the fork truck operator would have to accelerate/decelerate slower moving a pallet of 21-inch wheels than he would moving a pallet of 16-inch wheels. And we all know that in the real world, this adjustment in acceleration/deceleration would not take place.
Wm.C. Lobe, Senior Project Manager
Lobe & Associates, LLC.
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