New Elevator!

[Grumblenuts]

A pulley is a simple machine that uses a rope or chain wrapped around a grooved wheel to change the direction of force, making it easier to lift heavy objects.

Correct.

 

[trevorjohnson83]

In a pulley system, the different sizes of pulley wheels can be seen as analogous to gears with different diameters. By using larger or smaller pulley wheels, the speed or force applied can be modified.

 

[Grumblenuts]

Correct.

 

[trevorjohnson83]

You could use the system in the OP to put weight on piezocrystals take the weight off, and apply it again without losing the energy of the lifting the weight again. I suppose you would lose some energy depending on how heavy the gear system is but in a car you create pressure then just dump the pressure out. I bet this combined with my Hoberman ball/spring brake storage a typical car could get 200 miles to the gallon.

 

[trevorjohnson83]

the pressure in a large piston could be used to open a spring, then the pressure fills a second piston storing the energy in a smaller spring. The second spring can act on levers and gears to store its energy in the first spring. You can spread the energy out through pistons until the pressure is low in all the pistons. Can you fill in some hypothetical numbers and see how much energy could be stored this way?

Assistant
Poe


To explore your concept of energy storage using pistons and springs, let's set up a hypothetical scenario with some numbers. We’ll break down the energy storage process step by step.

Scenario Setup​

1. Piston A (Large Piston):
   • Diameter: 0.1 m (10 cm)
   • Area: AA=π(d2)2=π(0.12)2≈0.00785 m2AA=π(2d)2=π(20.1)2≈0.00785m2
2. Pressure in Piston A:
   • Pressure: 500,000 Pa (500 kPa)
   • Force: FA=P×AA=500,000×0.00785≈3925 NFA=P×AA=500,000×0.00785≈3925N
3. Stroke Length of Piston A:
   • Stroke Length: 0.2 m
   • Work Done: WA=FA×Stroke Length=3925×0.2≈785 JWA=FA×Stroke Length=3925×0.2≈785J
4. Piston B (Smaller Piston):
   • Diameter: 0.05 m (5 cm)
   • Area: AB=π(0.052)2≈0.00196 m2AB=π(20.05)2≈0.00196m2
5. Pressure Transfer to Piston B:
   • When Piston A compresses, it transfers pressure to Piston B. Assuming the pressure remains constant at 500,000 Pa:
   • Force on Piston B: FB=P×AB=500,000×0.00196≈980 NFB=P×AB=500,000×0.00196≈980N
6. Stroke Length of Piston B:
   • Stroke Length: 0.2 m
   • Work Done on Piston B: WB=FB×Stroke Length=980×0.2≈196 JWB=FB×Stroke Length=980×0.2≈196J

Spring Energy Storage​

1. Energy Stored in Springs:
   • Spring Constant (Piston A's Spring): Assume kA=1000 N/mkA=1000N/m
   • Compression Distance: Assuming full stroke of 0.2 m,
   • Energy in Spring A: EA=12kAx2=12×1000×(0.2)2≈20 JEA=21kAx2=21×1000×(0.2)2≈20J
2. Spring for Piston B:
   • Spring Constant (Piston B's Spring): Assume kB=500 N/mkB=500N/m
   • Compression Distance: 0.2 m,
   • Energy in Spring B: EB=12kBx2=12×500×(0.2)2≈10 JEB=21kBx2=21×500×(0.2)2≈10J

Total Energy Storage​

• Total Work Done: The energy transferred through the system from Piston A to Piston B would be the sum of stored energies in both springs.
• Total Energy:
  Etotal=EA+EB≈20+10=30 JEtotal=EA+EB≈20+10=30J

Energy Distribution​

• If you spread the energy across multiple pistons and springs, you could potentially increase the total stored energy, depending on the arrangement and efficiency of the system.
• The energy could be gradually released by decompressing the springs and transferring it back through the pistons or to an external load.

Conclusion​

In this hypothetical scenario, a system using pistons and springs could store approximately 30 J of energy. By optimizing the design, increasing the number of pistons, or using materials with higher spring constants, you could enhance the energy storage capabilities significantly.



can you do the math for smaller and smaller pistons? lets say seven pistons.

Assistant
Poe


Let's explore a system with seven pistons, each smaller than the previous one, and calculate the energy storage potential at each stage. We will use a consistent approach, gradually reducing the diameter of each subsequent piston.

Setup​

1. Piston Sizes:
   • Piston 1 (Large): Diameter = 0.1 m (10 cm)
   • Piston 2: Diameter = 0.08 m (8 cm)
   • Piston 3: Diameter = 0.06 m (6 cm)
   • Piston 4: Diameter = 0.04 m (4 cm)
   • Piston 5: Diameter = 0.03 m (3 cm)
   • Piston 6: Diameter = 0.02 m (2 cm)
   • Piston 7: Diameter = 0.01 m (1 cm)

Calculations​

We will assume:

• Pressure: 500,000 Pa (500 kPa)
• Stroke Length: 0.2 m for all pistons
• Spring Constants: Each spring constant will be assigned values that increase as piston size decreases (e.g., k=1000 N/mk=1000N/m for Piston 1, k=2000 N/mk=2000N/m for Piston 2, etc.).

1. Piston Areas and Forces​

Using the formula A=π(d2)2A=π(2d)2:

• Piston 1:
  • Area A1=0.00785 m2A1=0.00785m2
  • Force F1=500,000×A1≈3925 NF1=500,000×A1≈3925N
• Piston 2:
  • Area A2=π(0.082)2≈0.00402 m2A2=π(20.08)2≈0.00402m2
  • Force F2≈2008 NF2≈2008N
• Piston 3:
  • Area A3=π(0.062)2≈0.00283 m2A3=π(20.06)2≈0.00283m2
  • Force F3≈1415 NF3≈1415N
• Piston 4:
  • Area A4=π(0.042)2≈0.00126 m2A4=π(20.04)2≈0.00126m2
  • Force F4≈630 NF4≈630N
• Piston 5:
  • Area A5=π(0.032)2≈0.00071 m2A5=π(20.03)2≈0.00071m2
  • Force F5≈355 NF5≈355N
• Piston 6:
  • Area A6=π(0.022)2≈0.00031 m2A6=π(20.02)2≈0.00031m2
  • Force F6≈155 NF6≈155N
• Piston 7:
  • Area A7=π(0.012)2≈0.00008 m2A7=π(20.01)2≈0.00008m2
  • Force F7≈40 NF7≈40N

2. Work Done by Each Piston​

Using the formula W=F×Stroke LengthW=F×Stroke Length:

• Piston 1:
  • W1=3925×0.2≈785 JW1=3925×0.2≈785J
• Piston 2:
  • W2=2008×0.2≈402 JW2=2008×0.2≈402J
• Piston 3:
  • W3=1415×0.2≈283 JW3=1415×0.2≈283J
• Piston 4:
  • W4=630×0.2≈126 JW4=630×0.2≈126J
• Piston 5:
  • W5=355×0.2≈71 JW5=355×0.2≈71J
• Piston 6:
  • W6=155×0.2≈31 JW6=155×0.2≈31J
• Piston 7:
  • W7=40×0.2≈8 JW7=40×0.2≈8J

3. Total Work Done​

Calculating the total work done across all seven pistons:
Wtotal=W1+W2+W3+W4+W5+W6+W7Wtotal=W1+W2+W3+W4+W5+W6+W7
Wtotal≈785+402+283+126+71+31+8≈1706 JWtotal≈785+402+283+126+71+31+8≈1706J

4. Energy Stored in Springs​

Assuming spring constants increase with decreasing piston size:

• Piston 1: k1=1000 N/mk1=1000N/m
  • E1=12k1(0.2)2≈20 JE1=21k1(0.2)2≈20J
• Piston 2: k2=1500 N/mk2=1500N/m
  • E2=12k2(0.2)2≈30 JE2=21k2(0.2)2≈30J
• Piston 3: k3=2000 N/mk3=2000N/m
  • E3=12k3(0.2)2≈40 JE3=21k3(0.2)2≈40J
• Piston 4: k4=2500 N/mk4=2500N/m
  • E4=12k4(0.2)2≈50 JE4=21k4(0.2)2≈50J
• Piston 5: k5=3000 N/mk5=3000N/m
  • E5=12k5(0.2)2≈60 JE5=21k5(0.2)2≈60J
• Piston 6: k6=3500 N/mk6=3500N/m
  • E6=12k6(0.2)2≈70 JE6=21k6(0.2)2≈70J
• Piston 7: k7=4000 N/mk7=4000N/m
  • E7=12k7(0.2)2≈80 JE7=21k7(0.2)2≈80J

Total Energy Stored in Springs​

Etotal springs=E1+E2+E3+E4+E5+E6+E7Etotal springs=E1+E2+E3+E4+E5+E6+E7
Etotal springs≈20+30+40+50+60+70+80≈350 JEtotal springs≈20+30+40+50+60+70+80≈350J

Conclusion​

In this hypothetical scenario with seven pistons, the total work done is approximately 1706 J, and the total energy stored in the springs is approximately 350 J. By optimizing the design further, you could increase efficiency and energy storage capacity even more.

 

[trevorjohnson83]

Here are estimates for how the 1706 J of energy could be stored in various systems, considering their typical efficiencies and energy storage capacities:

1. Flywheel Energy Storage​

• Efficiency: Approximately 85-90%
• Energy Stored: Assuming 90% efficiency:
  Energy stored=1706×0.90≈1535 JEnergy stored=1706×0.90≈1535J

2. Hydraulic Energy Storage​

• Efficiency: Typically around 70-80%
• Energy Stored: Assuming 75% efficiency:
  Energy stored=1706×0.75≈1279.5 JEnergy stored=1706×0.75≈1279.5J

3. Compressed Air Energy Storage (CAES)​

• Efficiency: Generally about 60-70%
• Energy Stored: Assuming 65% efficiency:
  Energy stored=1706×0.65≈1109 JEnergy stored=1706×0.65≈1109J

4. Battery Storage Systems​

• Efficiency: Around 80-95% (depends on the type of battery)
• Energy Stored: Assuming 90% efficiency:
  Energy stored=1706×0.90≈1535 JEnergy stored=1706×0.90≈1535J

5. Supercapacitors​

• Efficiency: Approximately 85-95%
• Energy Stored: Assuming 90% efficiency:
  Energy stored=1706×0.90≈1535 JEnergy stored=1706×0.90≈1535J

6. Thermal Energy Storage​

• Efficiency: Generally around 75-90%
• Energy Stored: Assuming 80% efficiency:
  Energy stored=1706×0.80≈1365 JEnergy stored=1706×0.80≈1365J

Summary of Energy Storage Estimates​

Storage System	Estimated Energy Stored (J)
Flywheel	1535
Hydraulic	1279.5
Compressed Air	1109
Battery	1535
Supercapacitor	1535
Thermal Energy	1365

These estimates illustrate the potential energy storage capabilities of each system, factoring in their efficiencies. The actual values may vary based on specific designs and operational conditions.

 

[trevorjohnson83]

what are the disadvantages of storing car brake energy> why isn't it widely applied? how much improvement do cars with brake storage have in mpg over the same car without the brake storage?

Assistant
Poe


Storing car brake energy, particularly through systems like regenerative braking or mechanical storage, has several disadvantages and challenges that limit widespread application. Here’s a detailed look:

Disadvantages of Storing Brake Energy​

1. Complexity and Cost:
   • System Complexity: Integrating energy storage systems (like flywheels or batteries) with the vehicle’s braking system adds complexity to the design, requiring advanced control systems and components.
   • Cost: The initial investment for these systems can be high, which may not be justified in all vehicle models, especially lower-cost ones.
2. Weight:
   • Added Weight: Energy storage systems, particularly batteries or flywheels, add weight to the vehicle, which can negatively affect overall efficiency and performance.
3. Space Constraints:
   • Installation Space: Finding suitable space to install energy storage systems in vehicles can be challenging, especially in compact or budget models.
4. Efficiency Losses:
   • Energy Conversion Losses: Converting kinetic energy from braking into stored energy and then back into usable energy involves losses. While regenerative braking improves efficiency, it’s not 100% effective.
5. Limited Energy Recovery:
   • Energy Recovery Limitations: Not all braking energy can be captured, especially in situations where braking is sudden or not very forceful. The effectiveness of energy recovery depends on driving patterns.
6. Consumer Awareness and Acceptance:
   • Market Readiness: Many consumers are not familiar with or do not prioritize energy recovery technologies, which can impact manufacturers' decisions to implement them.

Improvement in Fuel Efficiency (MPG)​

The improvement in miles per gallon (MPG) for vehicles with brake energy storage systems varies depending on several factors, including the vehicle type, driving conditions, and the specific technology used. However, here are some general insights:

• Typical Improvements: Vehicles equipped with regenerative braking systems, like hybrid or electric vehicles, can see improvements ranging from 10% to 30% in fuel efficiency compared to conventional vehicles.
• Specific Studies: Some studies have shown that hybrid vehicles can achieve fuel efficiencies of 50% or more due to regenerative braking and other energy-saving technologies.

Conclusion​

While brake energy storage systems have the potential to improve fuel efficiency and reduce emissions, their complexity, cost, and practical limitations have hindered widespread adoption. As technology advances and consumer awareness grows, we may see more vehicles equipped with these systems in the future.



using a Hoberman ball as a gear that opens and closes, changing the torque of the gear on a fixed pipe down the center of the Hoberman ball could act as a brake gear that could store energy suddenly or in a slower manner. would this reduce the complexity of the energy storage system? and its weight?

Assistant
Poe


Using a Hoberman ball as a gear mechanism for energy storage is an interesting concept. Here’s an analysis of its potential benefits and drawbacks regarding complexity and weight:

Potential Benefits​

1. Reduced Complexity:
   • Simplified Mechanism: The Hoberman ball's ability to expand and contract could simplify the design of a braking system. Instead of multiple gears and moving parts, a single expanding mechanism could control energy storage and release.
   • Integrated Functionality: It could potentially serve dual purposes—acting as both a brake and an energy storage device—reducing the need for multiple systems.
2. Compact Design:
   • Space Efficiency: The design of a Hoberman ball allows for a compact form factor that can expand and contract, potentially making it easier to fit into limited spaces within a vehicle.
3. Dynamic Torque Adjustment:
   • Variable Torque: The ability to change the torque applied through the mechanism could allow for adaptable braking, potentially improving the efficiency of energy recovery during braking events.

Potential Challenges​

1. Weight Considerations:
   • Material Choice: Depending on the materials used to construct the Hoberman ball, it could be lightweight or heavy. While the design itself may be compact, the material could negate weight savings.
   • Additional Components: The mechanism to control the expansion and contraction may add weight, depending on how it is implemented.
2. Energy Transfer Efficiency:
   • Mechanical Losses: While the concept may reduce complexity, the efficiency of energy transfer through the Hoberman ball mechanism needs to be carefully evaluated. Any mechanical losses in the system could affect overall performance.
3. Durability and Reliability:
   • Wear and Tear: The moving parts in a Hoberman ball might experience wear over time, which could affect reliability compared to more traditional systems.
4. Design Complexity:
   • Engineering Challenges: While the concept may seem straightforward, engineering a reliable and effective energy storage system using a Hoberman ball would still require careful design and testing.

Conclusion​

Using a Hoberman ball as a brake gear mechanism could potentially reduce complexity and offer a compact solution for energy storage. However, careful consideration of weight, material choice, and mechanical efficiency is essential. Further research and development would be needed to assess its practicality and effectiveness compared to existing energy storage technologies.