Hey all,
I’ve been working on an idea that combines a scissor mechanism (XX shape) and a pantograph to achieve vertical lift from lateral motion without relying on traditional falling weights. Here’s a breakdown of the concept:
The Scissor Mechanism: The mechanism starts at the bottom of a V-shaped guide, with small wheels on the ends of the X legs that climb the angled walls. The weight, which normally would pull downward, is instead guided laterally.
The Pantograph Effect: A pantograph is added to control the movement of the weight. By keeping the weight lateral rather than vertical, we harness the leverage created by the expanding scissor arms.
Mechanical Leverage: As the scissor expands laterally due to the pantograph-guided weight, it drives upward motion. The pantograph helps prevent the weight from dropping significantly, allowing the lateral motion to provide the necessary force for the vertical lift.
No Violation of Physics: This mechanism does not break any laws of physics, as the lateral movement still requires energy and the system follows the principles of leverage and force multiplication.
Current Status: I've run some calculations, and while the forces involved don’t break any rules, this design seems like it might be able to achieve a vertical lift higher than the starting position! The combination of geometry and mechanics is key, and optimizing the design could further increase its efficiency.
Weightlifting
Moderator: scott
- killemaces
- Enthusiast
- Posts: 167
- Joined: Wed Jan 28, 2009 9:54 am
- Location: Norway
Weightlifting
Questions are my greatest tool, i am only the mechanic
Rune 2009
Rune 2009
Re: Weightlifting
Looking forward to more updates!
Re: Weightlifting
Bonjour Tue-mache,
Un croquis serait le bien venue et plus explicite!
Hello Tue-mache,
A sketch would be the welcome and more explicit!
Un croquis serait le bien venue et plus explicite!
Hello Tue-mache,
A sketch would be the welcome and more explicit!
Last edited by SHADOW on Sun Sep 29, 2024 5:15 am, edited 1 time in total.
La propriété, c'est le vol!
P.J. PROUDHON
P.J. PROUDHON
- killemaces
- Enthusiast
- Posts: 167
- Joined: Wed Jan 28, 2009 9:54 am
- Location: Norway
Re: Weightlifting
The math thanks to Ai:
To show the math behind this theoretical mechanism, let’s calculate how the system works based on leverage, geometry, and mechanical advantage.
Assumptions:
Input weight:
100
kg
100kg (about
980
N
980N of force due to gravity).
Scissor mechanism (X-shape) starts at the bottom.
Pantograph mechanism moves laterally and is connected to the scissor.
We assume the mechanical advantage (MA) of the scissor mechanism is 3, which is typical in lifting setups that use such a principle. This is derived from how leverage in a scissor lift increases lift but requires larger lateral forces.
Step 1: Mechanical Advantage (MA)
The mechanical advantage of a scissor lift is based on the relationship between the distance the weight moves vertically and how far the scissor arms move laterally. If the scissor arms have an MA of 3, it means for every 1 unit of force applied laterally, the vertical lift is 3 times that.
MA
=
D
vertical
D
horizontal
=
3
MA=
D
horizontal
D
vertical
=3
Step 2: Force Calculations
With
100
kg
100kg, the force due to gravity is:
F
gravity
=
m
⋅
g
=
100
kg
⋅
9.8
m/s
2
=
980
N
F
gravity
=m⋅g=100kg⋅9.8m/s
2
=980N
The total force needed to lift this weight is
980
N
980N.
Given the mechanical advantage (MA) of 3, the scissor mechanism will multiply the input force, meaning the input lateral force needed to lift the weight will be:
F
input
=
F
gravity
MA
=
980
N
3
=
326.67
N
F
input
=
MA
F
gravity
=
3
980N
=326.67N
So,
326.67
N
326.67N is the lateral force needed to lift the
100
kg
100kg weight through the scissor mechanism.
Step 3: Vertical Lift Calculation
Now, we can calculate how much vertical lift we get given the scissor mechanism and lateral movement. Assuming the pantograph moves laterally (without vertical movement), we focus on how far the scissor mechanism can expand vertically.
If the lateral distance the scissor mechanism moves is
D
horizontal
D
horizontal
, the corresponding vertical lift would be:
D
vertical
=
3
⋅
D
horizontal
D
vertical
=3⋅D
horizontal
Let’s assume a
D
horizontal
=
10
cm
D
horizontal
=10cm movement of the pantograph. Then:
D
vertical
=
3
⋅
0.1
m
=
0.3
m
=
30
cm
D
vertical
=3⋅0.1m=0.3m=30cm
Thus, the weight could theoretically be lifted 30 cm from its starting point.
Step 4: Friction Considerations
For this to work, friction between the scissor mechanism, the bearings, and the sliding surfaces must be low enough to avoid cancelling out the mechanical advantage. If the frictional forces exceed the available input force, the system will fail to lift the weight.
Let’s assume a friction coefficient
μ
μ of 0.1 (which is low, but possible with good bearings). The frictional force for each joint or surface in contact would be:
F
friction
=
μ
⋅
F
normal
F
friction
=μ⋅F
normal
If the normal force acting on the scissor mechanism is roughly equal to the gravitational force:
F
friction
=
0.1
⋅
980
N
=
98
N
F
friction
=0.1⋅980N=98N
This frictional force would need to be subtracted from the input force, meaning the effective force for lifting becomes:
F
effective
=
326.67
N
−
98
N
=
228.67
N
F
effective
=326.67N−98N=228.67N
With this force, the system could still work, but the lift would be slightly reduced due to frictional losses. However, the system would still generate lift, albeit less efficiently.
Conclusion:
In this scenario, with a mechanical advantage of 3, a lateral input force of
326.67
N
326.67N (from a
100
kg
100kg weight moving horizontally) can theoretically lift the scissor mechanism by around 30 cm.
Friction reduces this force somewhat, but with low friction (0.1), the mechanism could still lift, just slightly less than 30 cm.
Therefore, under these conditions and calculations, the system can provide a lift, but it will be limited by friction and efficiency.
To show the math behind this theoretical mechanism, let’s calculate how the system works based on leverage, geometry, and mechanical advantage.
Assumptions:
Input weight:
100
kg
100kg (about
980
N
980N of force due to gravity).
Scissor mechanism (X-shape) starts at the bottom.
Pantograph mechanism moves laterally and is connected to the scissor.
We assume the mechanical advantage (MA) of the scissor mechanism is 3, which is typical in lifting setups that use such a principle. This is derived from how leverage in a scissor lift increases lift but requires larger lateral forces.
Step 1: Mechanical Advantage (MA)
The mechanical advantage of a scissor lift is based on the relationship between the distance the weight moves vertically and how far the scissor arms move laterally. If the scissor arms have an MA of 3, it means for every 1 unit of force applied laterally, the vertical lift is 3 times that.
MA
=
D
vertical
D
horizontal
=
3
MA=
D
horizontal
D
vertical
=3
Step 2: Force Calculations
With
100
kg
100kg, the force due to gravity is:
F
gravity
=
m
⋅
g
=
100
kg
⋅
9.8
m/s
2
=
980
N
F
gravity
=m⋅g=100kg⋅9.8m/s
2
=980N
The total force needed to lift this weight is
980
N
980N.
Given the mechanical advantage (MA) of 3, the scissor mechanism will multiply the input force, meaning the input lateral force needed to lift the weight will be:
F
input
=
F
gravity
MA
=
980
N
3
=
326.67
N
F
input
=
MA
F
gravity
=
3
980N
=326.67N
So,
326.67
N
326.67N is the lateral force needed to lift the
100
kg
100kg weight through the scissor mechanism.
Step 3: Vertical Lift Calculation
Now, we can calculate how much vertical lift we get given the scissor mechanism and lateral movement. Assuming the pantograph moves laterally (without vertical movement), we focus on how far the scissor mechanism can expand vertically.
If the lateral distance the scissor mechanism moves is
D
horizontal
D
horizontal
, the corresponding vertical lift would be:
D
vertical
=
3
⋅
D
horizontal
D
vertical
=3⋅D
horizontal
Let’s assume a
D
horizontal
=
10
cm
D
horizontal
=10cm movement of the pantograph. Then:
D
vertical
=
3
⋅
0.1
m
=
0.3
m
=
30
cm
D
vertical
=3⋅0.1m=0.3m=30cm
Thus, the weight could theoretically be lifted 30 cm from its starting point.
Step 4: Friction Considerations
For this to work, friction between the scissor mechanism, the bearings, and the sliding surfaces must be low enough to avoid cancelling out the mechanical advantage. If the frictional forces exceed the available input force, the system will fail to lift the weight.
Let’s assume a friction coefficient
μ
μ of 0.1 (which is low, but possible with good bearings). The frictional force for each joint or surface in contact would be:
F
friction
=
μ
⋅
F
normal
F
friction
=μ⋅F
normal
If the normal force acting on the scissor mechanism is roughly equal to the gravitational force:
F
friction
=
0.1
⋅
980
N
=
98
N
F
friction
=0.1⋅980N=98N
This frictional force would need to be subtracted from the input force, meaning the effective force for lifting becomes:
F
effective
=
326.67
N
−
98
N
=
228.67
N
F
effective
=326.67N−98N=228.67N
With this force, the system could still work, but the lift would be slightly reduced due to frictional losses. However, the system would still generate lift, albeit less efficiently.
Conclusion:
In this scenario, with a mechanical advantage of 3, a lateral input force of
326.67
N
326.67N (from a
100
kg
100kg weight moving horizontally) can theoretically lift the scissor mechanism by around 30 cm.
Friction reduces this force somewhat, but with low friction (0.1), the mechanism could still lift, just slightly less than 30 cm.
Therefore, under these conditions and calculations, the system can provide a lift, but it will be limited by friction and efficiency.
Questions are my greatest tool, i am only the mechanic
Rune 2009
Rune 2009
Re: Weightlifting
I think you have made a little blunder, or I am missing something...
My belief is that a pantograph has a non-linear mechanical advantage, but if that is our limitation, there are ways around it....
With a mechanical advantage of 3:1, you are lifting 100kg, with the force normally required to lift 33kg. This means you apply 33kg worth of force for 3x the distance the 100kg moves (++)...
So if your horizontal distance is 10cm, you will only lift the 100kg 3.3cm, discounting friction etc.
My belief is that a pantograph has a non-linear mechanical advantage, but if that is our limitation, there are ways around it....
With a mechanical advantage of 3:1, you are lifting 100kg, with the force normally required to lift 33kg. This means you apply 33kg worth of force for 3x the distance the 100kg moves (++)...
So if your horizontal distance is 10cm, you will only lift the 100kg 3.3cm, discounting friction etc.
- killemaces
- Enthusiast
- Posts: 167
- Joined: Wed Jan 28, 2009 9:54 am
- Location: Norway
Re: Weightlifting
You're right about the pantograph having a non-linear mechanical advantage, where force is traded for distance. In the 3:1 example, you'd need to move the mechanism 3 times the horizontal distance to lift the weight by a third of that. So, if the horizontal movement is 10 cm, the vertical lift would be around 3.3 cm, assuming minimal friction. While the lift height is limited by this trade-off, the key point is that it will still lift, and adjustments to the design could improve efficiency.
Questions are my greatest tool, i am only the mechanic
Rune 2009
Rune 2009
Re: Weightlifting
.killemaces:
You're right about the pantograph having a non-linear mechanical advantage, where force is traded for distance
Copilot:
Incorporating segments at the golden ratio (Phi, approximately 1.618) into a pantograph can indeed affect its mechanical advantage. The golden ratio is known for its unique properties and can introduce non-linearities into the system. This means that the output force or motion may not change proportionally with the input force or motion, potentially enhancing the precision and adaptability of the pantograph
It’s odd how he has added Phi into the Storkbill on the toypage. It is added not in the vertical section every section but every second section.
What goes around, comes around.