Gravity Energy Storage

Calliban · 2021-11-11 14:47:17

Gravity energy storage using a raised weight to pressurise a hydraulic fluid, which drives a hydraulic motor coupled to a generator; would be a useful tool for grid frequency control. It can respond very rapidly to changes in load and provides the crucial minutes needed to start gas turbines if wind turbines drop off load rapidly, as often they do. In such cases, you don't need hours of storage. You need small amounts of energy storage that can respond rapidly at high power levels and achieving high power density, whilst the gas turbine picks up speed. And a mechanical system like this could last for many decades of intense cycling. It wouldn't wear out in the way battery systems would.

I think hydraulic gravity storage could certainly have a place in our energy mix. In the absence of spinning reserve, we need systems that can stretch time constants to allow backup powerplants time to respond. This may be just the thing.

tahanson43206 · 2021-11-11 14:58:03

For Calliban re #51

Thanks for picking up on the posting about the company that is offering gravity power storage.

If you've not already noticed, that company is close to you (from my perspective).

I'd be ** really ** interested if you were to call them to see if they'd like to engage your services as a consultant.

(th)

SpaceNut · 2021-11-11 16:19:07

Any potential mass using gravity are going to aid in the overall powering of mars but we need it not expending energy to get the mass to the height that we need to get movement from it to make power generation occur.

The potential in the first post makes use of mass that must get moved to be used to the bottom of a canyon or gully where as the moving of a mass to the top of that would mean expending energy to make use of it.

A home use of gravity from a rain storm would collect the water into the pale which under mass collected would get it to move down an inclined line. Which means the collection volume needs to be capable to hold the rain until its being needed to generate burst or peak power supplement. If enough rain was to be had then you could generate while collecting.

The generator should use permanent magnets as part of the design as that makes the design simpler as it requires no power while idle and not moving.

https://www.magnax.com/magnax-blog/axia … ize-weight

The optimized rotation is what needs to be known for the generators design to achieve the power that you desire.

2kw of generation is what most homes on earth would use for the average through out the day which means you need a storage device to not lose the effort of continuous run of the generator gives. I know that my average draw of power is 500 w when averaged per hour but you would be hard to find any appliance that can start operation with that for a limit let alone run after it did.

https://generatorist.com/power-consumpt … appliances
Appliance run wattage additional need to start
Refrigerator / Freezer 700 W 2,200 W

Calliban · 2021-11-11 17:52:40

Thanks SpaceNut, the link you provide is highly informative. I knew that inductive loads could crash the frequency of a power system. I wasn't aware of how significant the problem was. But looking down that list, some of the items create some really huge power spikes when they start. Are there not smoothing capacitors built into electric appliances to deal with issues like this? I am little puzzled by some of the items on the list. Some items that I thought have though of as being resistive loads are listed with high starting power. I wonder why that is?

I think one way to deal with this problem is to ensure that the generator is rated for maximum peak power (including surges) and then install enough momentum in the system to be able to deal with the current surges without crashing frequency. In most cases, I don't think hydraulic reservoirs would respond quickly enough. More likely, you simply need enough angular momentum in the wind turbine or generator to overcome starting torque. Hydraulics could assist through governor operated valves, but more likely we are looking at a flywheel directly coupled to the rotating system. In a wind turbine, the blades themselves have a certain amount of angular momentum. I don't know how the power electronics of a PV system could cope with surges like this. I can only assume that there are smoothing capacitors in place prior to DC to AC conversion.

For sudden large but short duration loads, that are above what a home wind turbine can generate, a diesel generator could be brought online for supplementary power. To allow sufficient time for this to happen, a combination of inherent angular momentum in a wind turbine, flywheel energy storage and hydraulic motor, could prevent frequency crash for the time it takes a DG to activate.

SpaceNut · 2021-11-11 18:11:22

These are A/C connected where if you are designing a DC units they do not have the spike or surge current requirement.
Most of the spike is due to inductive charging of parts that move such as motors and capacitors that are at rest discharged.
Stove elements need to heat up for optimal use by temperature circuits to turn them on and off as they approach operating temperature.
It is why so many thing have a standby current circuit built in such as a TV and other stuff.

So a brown out line voltage detector supplement system to keep the line from further sag...

The power supply units for back up or UPS are sort of that type of circuit but these are intended for sustained power but they could be altered to make surge power on demand as part of the appliances system.

The Musk power wall acts this way as well...

SpaceNut · 2021-11-11 21:19:30

http://ijcem.in/wp-content/uploads/2017 … EE-014.pdf

https://coldwargenerator.com/wp-content … erator.pdf

tahanson43206 · 2021-11-12 07:12:54

Thanks to Calliban and SpaceNut for adding valuable insights, links and suggestions to this topic!

For SpaceNut ... while the idea of capturing rain water to fill the energy storage system is interesting, it is NOT what this topic is about.

This topic is about storing energy you receive now and want to use later.

The idea of capturing rain water is (I think) what Calliban is talking about with hydraulic systems (such as a dam above a power station).

But whatever the mechanism for acquiring "new" energy may be, the point of the exercise is to save it for use later.

Ideally, we would like to save the energy in such a way that the value of the savings is not lost due to leakage.

The points about being able to address equipment starting loads are important to keep in mind when designing a system, whether for home use, or for industrial applications.

Both the Generac and Kohler systems I am using as a conceptual model are able to handle starting loads of a typical home. They do this by automatically monitoring the voltage being delivered, and adjusting fuel flow into the internal combustion engine (typically a 2 cylinder 4-Cycle design) as needed.

The system produced by the company whose offerings start this topic use cables to transfer kinetic energy from a falling mass to generators.

The rate of descent of the suspended mass can be adjusted rapidly to meet increased demand, although ** how ** that would be managed is a mystery to me. That would be an impressive piece of engineering, if the mechanism is a cable suspended mass.

A grandfather clock (and most pendulum clocks) use an escape mechanism to advance the progress of a weight an incremental distance.

Something similar might be employed in a much larger system, but whatever the device may be, it must be robust to withstand the shock loads to be expected.

With any luck, this topic may attract contributions that show the physical mechanisms needed to design a gravity energy storage system.

Thanks again for contributions to this topic already posted!

(th)

SpaceNut · 2021-11-12 10:01:22

The "Generac and Kohler systems" are power outage back ups, where the fuel is the storage of energy to be used later. While these fuels could be made by the gravity on earth its not likely for mars due to the lesser levels.

We have a topic that always leaks in Methane backup not nuclear vs solar that's got the concept of making fuels using it as storage when energy excess is available.

The fuel use rates are in the topic

tahanson43206 · 2021-11-12 10:54:57

For Calliban re #53 ...

Gravity energy storage using a raised weight to pressurise a hydraulic fluid, which drives a hydraulic motor coupled to a generator; would be a useful tool for grid frequency control.

Mixing all the topics that are active at the moment (ie, RobertDyck about a device to help Covid patients that involves oxygen) and SpaceNut's comments about ventilators, and several contributions to ** this ** topic, I have a new vision of how a gravity energy storage system might work.

Imagine ** two ** cylinders side-by-side, descending into the Earth as far as needed to meet the power requirements.

One contains the cylindrical counterweight. A cable attached to the counter weight passes over a wheel at the top of the shaft which is mounted on a shaft that connects to the motor generator.

I'm now picking up on Calliban's "hydraulic" suggestion, but noting that gases under pressure can serve similar purposes.

Also picking up on Calliban's diaphram idea from ** another ** post ... I now have a piston (from another of Calliban's posts) that is mounted in the second shaft, attached to the same cable that is holding the counterweight.

The piston shaft is filled with compressed air, sufficient to pull the counterweight up from where ever it is in the counterweight shaft.

The counterweight would be secured with chocks at the top of it's shaft, until it is needed to deliver power.

The purpose of the air filled shaft is to permit regulation of the movement of the counterweight under the influence of gravity.

To accomplish this, the air in the compressed air shaft would be released at whatever rate is needed to deliver required power.

The response time for the air release valve would be short, so the load variations that SpaceNut has described can be met nearly instantaneously.

***
In thinking about the counterweight ... there is a lot of lead coming out of service in pipelines around the US these days.

Lead would be even better as a material for the counterweight than iron, and iron would be better than concrete.

In practice, an iron frame might be loaded with lead to the required total mass for the customer requirements.

***
There might be sound produced when air is released. That might well be a problem in a residential neighborhood.

Air exhausted from the pressure chamber would absorb heat from the surroundings. ** That ** might be a problem.

(th)

SpaceNut · 2021-11-12 17:13:38

Pressure control is achieved in hydraulic systems by metering the flow of a fluid into or out of a constrained volume to create force such as a jack for an automobile.

https://www.hi-force.com/en-uk/blog-det … te-a-force

https://www.hydraulicspneumatics.com/te … c-pressure

A pump adds energy to oil in any of the three ways, as described by Bernoulli’s equation:
Hydraulics pneumatics Com Sites Hydraulics pneumatics com Files Dec mc Eq1
where P is the pressure,
ρ is the density,
v is the velocity,
g is the acceleration due to gravity, and
h is the elevation.

More equations on the page

https://www.machinerylubrication.com/Re … re-vs-flow

Of course this is moving the fluid to make pressure from the reserviour to the location to be moved.

Hydraulic fluids would have a different set of operating temperatures than that of water or co2 for mars use and that is where science comes into play for which is the better to use. The first slide in the topic used sand as no matter how cold its still going to move and work.

Sensing functions that are computer controller-ed used to produce a response for more or less power seems to be good to have.

SpaceNut · 2021-11-12 20:12:08

Thinking about the small amount of movement that gravity makes to making lots of rotations for the generator means a gear box to convert it into useful electrical energy.

Something that works in the opposite is those power wheel toy ride on vehicles. They motors rotate at high rpm but the wheel which is about a foot in diameter propels the toy at 6 miles an hour unless in 12 volt mode which makes it go fast. These are permanent magnet motors so if you rotate them you get power out of them. motor rotates at 30,000 rpm and the gear has 10 teeth to move the next gear inside the box makes it change. This would when you turn it produce AC voltage.

51D2cMJEj1L._SL1001_.jpg

The convertor section would have many of these traits. with the input from motion being applied to the wheel side (white colored area) of the box with the motor becoming a AC generator.

tahanson43206 · 2021-11-13 07:30:53

For SpaceNut re #61

Thank you for this helpful addition to the design of a (hopefully practical) gravity energy storage system for home power backup.

Calliban has provided some numbers to help to start the optimization process. I discovered last night that the curve that describes the mathematical points for height of the counter weight and depth of the drop cylinder is a parabola with a definite sweet spot.

In addition, since Caliban proposed concrete for the counterweight, using iron or lead (or a combination of lead in an iron jacket) offers an opportunity to achieve the desired power delivery goal using less height of the cylinder and depth of the drop tube.

The gearing mechanism you've shown provides a way to draw power from the counter weight as it descends in the drop tube.

In thinking about the use of compressed air in the system to regulate the descent of the counter weight in the drop tube, it occurred to me that the problem of noise that would occur if the air were released to the atmosphere can be solved by keeping the air inside the system.

I am thinking about creating a diagram, such as the ones that Void and GW Johnson have shown.

The diagram would (if achieved) show a loop of pipe with the counter weight on one side, and the compressed air on the other.

The compressed air would hold the counter weight at the top of the drop column. When power is to be recovered from the system, air would be released from the compressed air pipe into the top of the drop column. The counter weight would descend under the influence of gravity, and a thin cable attached to the top of the counter weight would turn a windlass which would drive the power take-off mechanism as described by SpaceNut.

Energy would be stored in the system by compressing air in the compressed air column.

No air would escape to the atmosphere, so the noise that will surely be produced by flow of air from the compressed column to the uncompressed column will be confined to the location of the control valve.

While the system I'm thinking about would be designed to deliver 2 Kw for 24 hours, it could be sized for shorter intervals and less power.

Edit: Eagle eyed readers will surely note the weak point in this concept ... the counter weight must maintain an air tight seal against the wall of the drop tube. But that is not possible except in theory. There would inevitably be leakage so the storage efficiency of the system would be reduced.

In addition, conversion of rotary motion of the windlass shaft to electrical power would inevitably involve losses of energy.

However, since the model for comparison is an internal combustion engine (Generac or Kohler as examples) the drop tower design might prove competitive in efficiency, since it does not depend upon thermal processes.

(th)

tahanson43206 · 2021-11-13 13:42:14

Here is a plot of trades between height of the counter weight cylinder in a 1 meter diameter shaft, and the elevation in meters.

The plot shows that even with the limitations of concrete and the 1 meter diameter, it is feasible to build a storage system able to deliver 2 Kw for 24 hours.

Thanks to Calliban for providing the starting numbers to work with.

We should be able to improve the design by making the counter weight of lead encased in iron, and increasing the diameter of the shaft slightly.

If there are errors in this projection of the post by Calliban, I won't be surprised and will gladly correct them.

The earlier post envisioned a block of concrete 12 meters on a side dropped for 3 meters.

I took the volume of the cube, changed the shape to a cylinder, and allowed the drop distance to increase as shown in the plot.

Edit: It appears the sweet spot is in the vicinity of a height of 46 meters and drop distance of 166 meters.

The total shaft depth is 46 + 166 or 210 meters in this case. A refinement of the calculation would include total shaft depth.

(th)

tahanson43206 · 2021-11-15 07:05:10

This post is assigned to show the specifications for an energy storage system using concrete for the counterweight.

The numbers start with the work of Calliban, earlier in the topic.

The requirement for output is 2 Kw for 24 hours

The following was developed offline and is posted here on 2021/11/18:

The objective for all four of the planned posts in this series is to provide 2 kilowatts of power without interruption for 24 hours.

The power level of 2 Kw is an average. Actual loads will vary above and below the mean, so the system must be able to handle fluctuations of load.

Post 1 # is based upon work of NewMars member Calliban.

It assumes a concrete counterweight.

Reference: http://newmars.com/forums/viewtopic.php … 52#p187152

Original: http://newmars.com/forums/viewtopic.php … 50#p187150

Calliban:
2kW for 24hours is 48kWh = 172.8MJ. 1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. Concrete bound granite rock has a density about 3000kg/m3. So 1m3 raised by 3m = 90KJ. The total volume of rock needed would be just shy of 2000m3, or a cube 12.4m aside. A lot larger than an average house and not very easy to achieve.

Following the lead of Calliban:

https://www.rapidtables.com/calc/electr … lator.html

2 kilowatts for 24 hours is 48 kilowatt hours

https://www.rapidtables.com/convert/ene … Joule.html

48 kilowatt hours is 172.8 Mj

Note: 1 kilowatt hour is 3.6 Mj

The density of concrete varies. Per Google:

Begin Quotation:
Density (Unit weight) of Concrete - Bulk Density - Civil Engineering
civiltoday.com › Materials › Concrete
The density of normal concrete is 2400 kg/m3 (150 pcf or lb/ft3) and the density of lightweight concrete is 1750 kg/m3(110 pcf or lb/ft3) · Typical density of ...
End Quotation.

The density of granite varies.

Per Google:

Begin Quotation:
The average density of granite is 2.75 g·cm−3 with a range of 1.74 g·cm−3 to 2.80 g·cm−3. The word granite comes from the Latin granum, a grain, in reference to the coarse-grained structure of such a crystalline rock.
Granite
www.cs.mcgill.ca › ~rwest › wikispeedia › wpcd › Granite
End Quotation.

The density of iron is given as:

Begin Quotation from Google:
Density of Iron in 285 units and reference information - aqua-calc
www.aqua-calc.com › Substances
Iron weighs 7 873 kg/m³ (491.49533 lb/ft³) · Iron weighs 7.873 gram per cubic centimeter or 7 873 kilogram per cubic meter, i.e. density of iron is equal to 7 ...
End Quotation from Google.

Note that cast iron is reported to have a lower density:

Begin Quotation from Google:
Cast Iron
7300

Density of Metals - The Engineering Mindset
theengineeringmindset.com › density-of-metals
End Quotation.

Begin Quotation from Google:
Density of Metals
Material
Density (ρ) kg/m3
Copper
8,944
Gold
19,320
Iron
7,860
Lead
11,343
Density of Metals - The Engineering Mindset
theengineeringmindset.com › density-of-metals
About Featured Snippets
End Quotation.

Repeating Calliban re Concrete cube:
Calliban:
2kW for 24hours is 48kWh = 172.8MJ. 1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. Concrete bound granite rock has a density about 3000kg/m3. So 1m3 raised by 3m = 90KJ. The total volume of rock needed would be just shy of 2000m3, or a cube 12.4m aside. A lot larger than an average house and not very easy to achieve.

Volume computed for 172.8/.09 is 1914 cubic meters.

Cube root of 1914 is 12.4 and change

https://www.calculatorsoup.com/calculat … linder.php

Given radius of .5 and height of 144, volume is 113 (units) cubed

Given radius of .5 and height of 100, volume is 78 and change

Distance to drop is between 52 (146 height) and 166 (46 height): 76.38~

After consultation with www.WolfRam.com, we have assurance that the equation for the current problem is X * Y = Constant.

The constant in this case is 7638

The solution is: Square Root of 7638 * two, or 174.8 meters

The height of a shaft able to deliver 2 Kw for 24 hours, using a counterweight made of concrete/granite in the form of a cylinder 1 meter in diameter, is 175 meters. The height of the cylinder is ½ of 175 meters, and the drop distance is ½ of 175 meters.

(th)

tahanson43206 · 2021-11-15 07:05:29

This post is assigned to show the specifications for an energy storage system using iron for the counterweight.

The numbers start with the work of Calliban, earlier in the topic.

The requirement for output is 2 Kw for 24 hours

The following was prepared offline:

The objective for all four of the planned posts in this series is to provide 2 kilowatts of power without interruption for 24 hours.

The power level of 2 Kw is an average. Actual loads will vary above and below the mean, so the system must be able to handle fluctuations of load.

Post 2 of 4 is based upon work of NewMars member Calliban.

It assumes an iron counterweight.

Reference: http://newmars.com/forums/viewtopic.php … 52#p187152

Original: http://newmars.com/forums/viewtopic.php … 50#p187150

Calliban:
2kW for 24hours is 48kWh = 172.8MJ. 1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. Concrete bound granite rock has a density about 3000kg/m3. So 1m3 raised by 3m = 90KJ. The total volume of rock needed would be just shy of 2000m3, or a cube 12.4m aside. A lot larger than an average house and not very easy to achieve.

Following the lead of Calliban:

https://www.rapidtables.com/calc/electr … lator.html

2 kilowatts for 24 hours is 48 kilowatt hours

https://www.rapidtables.com/convert/ene … Joule.html

48 kilowatt hours is 172.8 Mj

Note: 1 kilowatt hour is 3.6 Mj

The density of iron is given as:

Begin Quotation from Google:
Density of Iron in 285 units and reference information - aqua-calc
www.aqua-calc.com › Substances
Iron weighs 7 873 kg/m³ (491.49533 lb/ft³) · Iron weighs 7.873 gram per cubic centimeter or 7 873 kilogram per cubic meter, i.e. density of iron is equal to 7 ...
End Quotation from Google.

Note that cast iron is reported to have a lower density:

Begin Quotation from Google:
Cast Iron
7300

Density of Metals - The Engineering Mindset
theengineeringmindset.com › density-of-metals
End Quotation.

Begin Quotation from Google:
Density of Metals
Material
Density (ρ) kg/m3
Copper
8,944
Gold
19,320
Iron
7,860
Lead
11,343
Density of Metals - The Engineering Mindset
theengineeringmindset.com › density-of-metals
About Featured Snippets
End Quotation.

Repeating Calliban re Concrete cube:
Calliban:
2kW for 24hours is 48kWh = 172.8MJ. 1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. Concrete bound granite rock has a density about 3000kg/m3. So 1m3 raised by 3m = 90KJ. The total volume of rock needed would be just shy of 2000m3, or a cube 12.4m aside. A lot larger than an average house and not very easy to achieve.
End Quotation.

To adjust for density of iron: Using above: Iron 7860 kg/m3

Energy required is 172.8 MJ. (1.728e+10)
1 kg raised 1 meter is 10 Joules (1.0e+1)
1.728e+8 / 1.0e+1 >> 1.728e+7
172.8 MJ / 10 Joules per kilogram gives 1.728e+7 kg lifted 1 meter
1.728+7 / 7.86e+3 is >>
Iron is given as having a density of 7860 kg / Cubic Meter
2.198473282e+3 is given as the quotient of 1.728e+7/7.86e+3
2198.473283715 is given as the quotient of 1.728e+7/7.86e+3

This number (2198) may be for a lift of 1 meter
Given Calliban's example, dividing by 3 gives: 732.8 (733 cubic meters)

1914/733 >> 2.61 (and change)

The ratio of iron to concrete density is: 2.62 (approximately)

https://calculator.name/cube-root/733

7860/3000 >> 2.62 so the ratios match.

For comparison, Calliban's computation gave this:
Volume computed for 172.8/.09 is 1914 cubic meters.

Cube root of 1914 is 12.4 and change

Cube root of 733 is 9.01 and change

A cube of iron would be 9 meters on a side, compared to 12+ for concrete.

Question: how did I get the constant of 7638? A:height * 3 meters
7638 / 3 is 2546 meters (height of cylinder of concrete with radius .5)

https://www.calculatorsoup.com/calculat … linder.php

Given radius of .5 and height of 934, volume is 733.56 (units) cubed

Using the same method as previously, height * drop of 3 >> 934 * 3

The constant is then: 2802

The square root of 2802 is 53 * 2 >> 106

Distance to drop iron is 53 meters in a shaft of 106 meters

After consultation with www.WolfRam.com, we have assurance that the equation for the current problem is X * Y = Constant.

The constant in this case is 2802

The solution is: Square Root of 2802 * two, or 106 meters

That number compares to 175 meters for concrete.

The height of a shaft able to deliver 2 Kw for 24 hours, using a counterweight made of iron in the form of a cylinder 1 meter in diameter, is 106 meters. The height of the cylinder is ½ of 106 meters, and the drop distance is ½ of 106 meters.

(th)

tahanson43206 · 2021-11-15 07:05:49

This post is assigned to show the specifications for an energy storage system using lead for the counterweight.

The numbers start with the work of Calliban, earlier in the topic.

The requirement for output is 2 Kw for 24 hours

The following was prepared offline:

The objective for all four of the planned posts in this series is to provide 2 kilowatts of power without interruption for 24 hours.

The power level of 2 Kw is an average. Actual loads will vary above and below the mean, so the system must be able to handle fluctuations of load.

Post 3 of 4 is based upon work of NewMars member Calliban.

It assumes counterweight made of lead.

Reference: http://newmars.com/forums/viewtopic.php … 52#p187152

Original: http://newmars.com/forums/viewtopic.php … 50#p187150

Calliban:
2kW for 24hours is 48kWh = 172.8MJ. 1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. Concrete bound granite rock has a density about 3000kg/m3. So 1m3 raised by 3m = 90KJ. The total volume of rock needed would be just shy of 2000m3, or a cube 12.4m aside. A lot larger than an average house and not very easy to achieve.

Following the lead of Calliban:

https://www.rapidtables.com/calc/electr … lator.html

2 kilowatts for 24 hours is 48 kilowatt hours

https://www.rapidtables.com/convert/ene … Joule.html

48 kilowatt hours is 172.8 Mj

Note: 1 kilowatt hour is 3.6 Mj

The density of lead is given as: 11,343 kg/m3

Begin Quotation from Google:
Density of Metals
Material
Density (ρ) kg/m3
Copper
8,944
Gold
19,320
Iron
7,860
Lead
11,343
Density of Metals - The Engineering Mindset
theengineeringmindset.com › density-of-metals
About Featured Snippets
End Quotation.

Repeating Calliban re Concrete cube:
Calliban:
2kW for 24hours is 48kWh = 172.8MJ. 1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. Concrete bound granite rock has a density about 3000kg/m3. So 1m3 raised by 3m = 90KJ. The total volume of rock needed would be just shy of 2000m3, or a cube 12.4m aside. A lot larger than an average house and not very easy to achieve.
End Quotation.

To adjust for density of lead: Using above: Lead 11,343 kg/m3

Energy required is 172.8 MJ. (1.728e+10)
1 kg raised 1 meter is 10 Joules (1.0e+1)
1.728e+8 / 1.0e+1 >> 1.728e+7
172.8 MJ / 10 Joules per kilogram gives 1.728e+7 kg lifted 1 meter
1.728+7 / 1.1343e+4 is >>
Lead is given as having a density of 11,343 kg / Cubic Meter
1.5234e+4 is given as the quotient of 1.728e+7/1.12343e+4
1523.4 is the same value in traditional notation.

This number (1523.4) may be for a lift of 3 meter
Given Calliban's example, dividing by 3 gives: 507.8 (508 cubic meters)

https://calculator.name/cube-root/508

For comparison, Calliban's computation gave this:
Volume computed for 172.8/.09 is 1914 cubic meters.

Cube root of 1914 is 12.4 and change

Cube root of 733 is 9.01 and change

Cube root of 512 is 8. Cube root of 508 is 7.9+

A cube of lead would be 8 meters on a side, compared to 12+ for concrete
A cube of iron would be 9 meters on a side, compared to 12+ for concrete.

Question: how did I get the constant of 7638? A:height * 3 meters
7638 / 3 is 2546 meters (height of cylinder of concrete with radius .5)

https://www.calculatorsoup.com/calculat … linder.php

Given radius of .5 and height of 647, volume is 508.15 (units) cubed
Given radius of .5 and height of 934, volume is 733.56 (units) cubed

Using the same method as previously, height * drop of 3 >> 647 * 3
Using the same method as previously, height * drop of 3 >> 934 * 3

The constant is then 1941. The square root of 1941 is 44 and change
The height of the shaft is 44*2 or 88 meters.

Distance to drop lead is 44 meters in a shaft of 88 meters

After consultation with www.WolfRam.com, we have assurance that the equation for the current problem is X * Y = Constant.

The constant in this case is 1941

The solution is: Square Root of 1941 * two, or 88 meters

That number compares to 175 meters for concrete.

The height of a shaft able to deliver 2 Kw for 24 hours, using a counterweight made of lead in the form of a cylinder 1 meter in diameter, is 88 meters. The height of the cylinder is ½ of 88 meters, and the drop distance is ½ of 88 meters.

(th)

tahanson43206 · 2021-11-15 07:06:22

This post is assigned to show the specifications for an energy storage system using lead encased in iron for the counterweight.

The numbers start with the work of Calliban, earlier in the topic.

The requirement for output is 2 Kw for 24 hours. Proportion of lead to iron: 90% lead in 10% iron case. Radius is increased from .5 Meter to .6 Meter.

The following was prepared offline. The goal is to find a solution with shaft depth of 80 meters or less.

This document is 20211121 Large 18 x 24 Using Calliban Example for all four variations.

The objective for all four of the planned posts in this series is to provide 2 kilowatts of power without interruption for 24 hours. The materials considered are: concrete/granite, iron, lead and an alloy of lead and iron.

The power level of 2 Kw is an average. Actual loads will vary above and below the mean, so the system must be able to handle fluctuations of load.

Post 1 # is based upon work of NewMars member Calliban.

It assumes a concrete/granite counterweight.

Reference: http://newmars.com/forums/viewtopic.php … 52#p187152

Original: http://newmars.com/forums/viewtopic.php … 50#p187150

Begin quotation of Calliban:
2kW for 24hours is 48kWh = 172.8MJ. 1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. Concrete bound granite rock has a density about 3000kg/m3. So 1m3 raised by 3m = 90KJ. The total volume of rock needed would be just shy of 2000m3, or a cube 12.4m aside. A lot larger than an average house and not very easy to achieve.
End Calliban quotation.

2021/11/20: Analysis of example from Calliban:

2kW for 24 hours is 48kWh = 172.8MJ << This is the energy to be stored
1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. << Physics

Concrete density (about) 3000kg/M3 << The amount is a bit high but convenient
So 1 cubic meter of concrete raised by 3 meters is 3000 kg * 30J/kg gives 90,000 Joules or 90 KJ

To perform the same calculation for iron or lead or a combination, multiply the mass per M3 by 30
For iron, density is given as 7860 kg/m3. 7860 * 30 >> 235,800 Joules or 235.8 KJ
For lead, density is given as 11,343 kg/m3. 11,343 * 30 >> 340,290 Joules or 340.3 KJ
For an alloy of 10% iron and 90% lead, density is given as 10,862 so 10,862 * 30 >> 325,860 Joules 325.86 KJ

Following the lead of Calliban:

https://www.rapidtables.com/calc/electr … lator.html

2 kilowatts for 24 hours is 48 kilowatt hours

https://www.rapidtables.com/convert/ene … Joule.html

48 kilowatt hours is 172.8 Mj

Note: 1 kilowatt hour is 3.6 Mj

The density of concrete varies. Per Google:

Begin Quotation:
Density (Unit weight) of Concrete - Bulk Density - Civil Engineering
civiltoday.com › Materials › Concrete
The density of normal concrete is 2400 kg/m3 (150 pcf or lb/ft3) and the density of lightweight concrete is 1750 kg/m3(110 pcf or lb/ft3) · Typical density of ...
End Quotation.

The density of granite varies.

Per Google:

Begin Quotation:
The average density of granite is 2.75 g·cm−3 with a range of 1.74 g·cm−3 to 2.80 g·cm−3. The word granite comes from the Latin granum, a grain, in reference to the coarse-grained structure of such a crystalline rock.
Granite
www.cs.mcgill.ca › ~rwest › wikispeedia › wpcd › Granite
End Quotation.

The density of iron is given as:

Begin Quotation from Google:
Density of Iron in 285 units and reference information - aqua-calc
www.aqua-calc.com › Substances
Iron weighs 7 873 kg/m³ (491.49533 lb/ft³) · Iron weighs 7.873 gram per cubic centimeter or 7 873 kilogram per cubic meter, i.e. density of iron is equal to 7 ...
End Quotation from Google.

Note that cast iron is reported to have a lower density:

Begin Quotation from Google:
Cast Iron
7300

Density of Metals - The Engineering Mindset
theengineeringmindset.com › density-of-metals
End Quotation.

Begin Quotation from Google:
Density of Metals
Material
Density (ρ) kg/m3
Copper
8,944
Gold
19,320
Iron
7,860
Lead
11,343
Density of Metals - The Engineering Mindset
theengineeringmindset.com › density-of-metals
About Featured Snippets
End Quotation.

Repeating Calliban re Concrete cube:
Quotation from Calliban:
2kW for 24hours is 48kWh = 172.8MJ. 1kg raised 1m = 10J. Let's us assume a 3m raise. That's 30J/kg. Concrete bound granite rock has a density about 3000kg/m3. So 1m3 raised by 3m = 90KJ. The total volume of rock needed would be just shy of 2000m3, or a cube 12.4m aside. A lot larger than an average house and not very easy to achieve.
End Quotation.

Begin Concrete using estimated density of 3000kg/m3:
Concrete density (about) 3000kg/M3 << The amount is a bit high but convenient
So 1 cubic meter of concrete raised by 3 meters is 3000 kg * 30J/kg gives 90,000 Joules or 90 KJ
The energy of a cubic meter of concrete raised 3 meters is 90 KJ which is .09 MJ
Volume computed for 172.8 MJ /.09 MJ is 1914 cubic meters.
End Concrete

Cube root of 1914 is 12.4 and change [12.419697 per calculatorsoup.com]
Height of a cylinder of radius .5 and volume 1914 is: 2437 per calculatorsoup.com
The product of 2437 and 3 is 7311 (This is the corrected constant for concrete)
The square root of 7311 is 85.5 and change
The total shaft height for a concrete counter weight is 85.5 * 2 or 171 meters.

Begin iron using reported density of 7860kg/m3:
Compute mass of a cubic meter of iron:
Iron density 7860kg/M3
So 1 cubic meter of iron raised by 3 meters is 7860 kg * 30J/kg gives 235,800 Joules or 235.8 KJ
The energy of a cubic meter of iron raised 3 meters is 235.8 KJ which is .2358 MJ
Volume computed for 172.8 MJ /.2358 MJ is 732.8 cubic meters.
End iron

Cube root of 732.8 is 9 and change [Per calculatorsoup.com, 9.01561077]
Height of a cylinder of radius .5 and volume 732.8 per calculatorsoup.com is: 933 meters
Per calculatorsoup.com, a cylinder of radius .5 and height 933 has a volume of: 732.776486
The product of 933 and 3 is 2799. (this is the corrected constant for iron)
The square root of 2799 is 52.9 and change. The total shaft height for an iron counterweight is: 106 meters

Begin lead using estimated density of 11,343kg/m3:
Compute mass of a cubic meter of lead:
Iron density (about) 11,343kg/M3
So 1 cubic meter of lead raised by 3 meters is 11343 kg * 30J/kg gives 340,290 Joules or 340.29 KJ
Volume computed for 172.8 MJ /.34029 MJ is 508 cubic meters. (rounding 507.8)
End Lead

Cube root of 507.8 is 7.976 and change [ 7.9764935 per calculatorsoup.com]
Height of a cylinder of radius .5 and volume 508 per calculatorsoup.com is: 647 meters
calculatorsoup.com gives a volume of 508.152612 for a cylinder of radius .5 and height of 647
The product of 508 and 3 is 1524. This is the corrected constant for lead.
The square root of 1524 is 39 and change, so a shaft height would be 39*2 or 78 meters
This is below the target height of 80 meters, per SpaceNut report of a water well drilled on local property.

10% iron at 7860 kg/m3 >> 786 kg
90% lead at 11,343 kg/m3 >> 10754.1
Sum of 786 and 10754.1 >> 11540.1
But the density of a mixture should be less than lead by itself
How to compute the density of a mixture of two metals?

Per indium.com:
Begin Quotation:
The density of the alloy is its mass (100g) divided by its volume (11.76 cc) or 100/11.76 g/cc = 8.50 g/cc. In general then, the equation to calculate density is 1/Dalloy = Mass Fraction Metal 1/Dmetal 1 + Mass Fraction Metal 2/Dmetal 2.
End Quotation.

786/7860 + 10754.1/11343 >> 1 (ie, .1 + .9 is 1)

https://www.handymath.com/cgi-bin/densi … bmit=Entry

Density of alloy is computed as 10861.7

Begin Mixture of alloy (lead in iron casing) using estimated density of 10861.7kg/m3:
Compute mass of a cubic meter of alloy: 10861.7kg
So 1 cubic meter of alloy raised by 3 meters is 10862 kg * 30J/kg gives 325,860 Joules or 325.86 KJ
The energy of a cubic meter of alloy raised 3 meters is 325.86 KJ which is .326 MJ
Volume computed for 172.8 MJ /.326 MJ is 530 cubic meters.
End Mixture

Cube root of 530 is 8.1 and change [8.09267234 per calculatorsoup.com]
8 cubed is 512. 8.1 cubed is 531.441 (per Calculator)
Height of a cylinder of radius .6 and volume 531 per calculatorsoup.com is: 470 meters
Per calculatorsoup.com, the volume of a cylinder of .6 and height of 470 is 531.557477
The product of 470 and 3 is 1410. The square root of 1410 is 37.549966711
Twice the square root of 1410 is 75 meters and change.

*** The next section converts a cube to a cylinder

https://www.calculatorsoup.com/calculat … linder.php

Given radius of .5 and height of 144, volume is 113 (units) cubed

Given radius of .5 and height of 100, volume is 78 and change

Distance to drop is between 52 (146 height) and 166 (46 height): 76.38~

After consultation with www.WolfRam.com, we have assurance that the equation for the current problem is X * Y = Constant.

Computation of Shaft Height for Concrete:

The constant in the case of concrete is 7638 (2546 * 3)
The constant is the product of the height of a cylinder of a chosen material and 3 meters drop.

The solution is: Square Root of 7638 * two, or 174.8 meters

The height of a shaft able to deliver 2 Kw for 24 hours, using a counterweight made of concrete/granite in the form of a cylinder 1 meter in diameter, is 175 meters. The height of the cylinder is ½ of 175 meters, and the drop distance is ½ of 175 meters.

Computation of Shaft Height for Iron:

The constant in the case of concrete is 7638 (2546 * 3)
The constant is the product of the height of a cylinder of a chosen material and 3 meters drop.
The constant in the case of iron is:

The solution is: Square Root of 7638 * two, or 174.8 meters

The height of a shaft able to deliver 2 Kw for 24 hours, using a counterweight made of concrete/granite in the form of a cylinder 1 meter in diameter, is 175 meters. The height of the cylinder is ½ of 175 meters, and the drop distance is ½ of 175 meters.

Computation of Shaft Height for Lead:

The constant in the case of concrete is 7638 (2546 * 3)
The constant is the product of the height of a cylinder of a chosen material and 3 meters drop.
The constant in the case of lead is:

The solution is: Square Root of 7638 * two, or 174.8 meters

The height of a shaft able to deliver 2 Kw for 24 hours, using a counterweight made of concrete/granite in the form of a cylinder 1 meter in diameter, is 175 meters. The height of the cylinder is ½ of 175 meters, and the drop distance is ½ of 175 meters.

Computation of Shaft Height for Alloy at radius .5 Meter:

The constant in the case of concrete is 7638 (2546 * 3)
The constant in the case of Alloy is:
The constant is the product of the height of a cylinder of a chosen material and 3 meters drop.

The solution is: Square Root of 7638 * two, or 174.8 meters

The height of a shaft able to deliver 2 Kw for 24 hours, using a counterweight made of concrete/granite in the form of a cylinder 1 meter in diameter, is 175 meters. The height of the cylinder is ½ of 175 meters, and the drop distance is ½ of 175 meters.
Computation of Shaft Height for Alloy at radius .6 Meter:

The constant in the case of concrete is 7638

The solution is: Square Root of 7638 * two, or 174.8 meters

The height of a shaft able to deliver 2 Kw for 24 hours, using a counterweight made of concrete/granite in the form of a cylinder 1 meter in diameter, is 175 meters. The height of the cylinder is ½ of 175 meters, and the drop distance is ½ of 175 meters.

The computations for concrete, iron, lead and an alloy of lead and iron are updated earlier in this post.

(th)

Calliban · 2021-11-15 07:51:34

Lead is a relatively expensive material. For such a low energy density storage medium you want to use a cheap form of mass - like rock or dirt. Because the energy stored per unit mass is low, the embodied energy per unit mass should also be low. Probably the cheapest thing of all is water. And water has the advantage of being a fluid. Which takes us back to pumped storage. You could build a miniature pumped storage plant. But 48kWh of storage isn't going to be easy. It means a big reservoir that would cost as much to build as the house it serves and would consume a lot of land. I would suggest a diesel generator or some other heat engine for that task. Your raised weight hydraulic storage can give you the time needed to start the generator. Or alternatively, it could meet short duration high power loads.

One area where gravity storage could come into its own is in powering a mass driver on Lunar or Mars. To launch 1kg of material from lunar surface to escape velocity with a 70% efficient mass driver, requires about 2.5kWh of energy. But this modest amount of energy must be delivered at very high power levels. If a mass driver is 500m long, then it will need to receive 2.5kWh in 0.4 seconds. That is a power of 22.5MW. A hydraulic accumulator, maybe gravity based, would be a good way of storing and releasing modest amounts of energy at very high power. The accumulator would be charged by centrifugal pumps that receive steady power from a nuclear reactor or solar array and would discharge in rapid pulses. Capacitors could be used to further compress the pulse. A 1MWe nuclear reactor feeding the accumulator, would allow one launch roughly every 10 seconds. For an application like this, we would couple the hydraulic accumulator to a piston that drives a linear generator. To generate the pulse, we would use a solenoid to open a rotating valve. Pressure would rapidly build up behind the piston, which would then be mechanically unlocked, forcing a powerful electromagnetic through the generator coil. This would charge capacitors, compressing the power spike into a rectangular profile pulse.

Other applications where hydraulic energy storage would be useful are anywhere where pulsed power is needed. Inertial confinement fusion, for example. Direct energy weapons. Mechanical drilling through rock.

Last edited by Calliban (2021-11-15 08:31:52)

tahanson43206 · 2021-11-15 09:24:45

For Calliban re #68

Thank you for another substantial (and inspiring) contribution to the topic!

One detail caught my eye .... your assertion that lead is an expensive material ...

The status of lead in the United States is not all that good right now, due to its proven ability to destroy the lives of children who are permeated with lead due to the use of lead pipes in water lines. In the next few years, all those lead pipes (in the United States and perhaps other nations) will be removed, and all that lead will need a new (tightly restricted) home.

The materials of which the cylindrical shaft is constructed is open for review as well... I've been thinking of iron as the most suitable material, but large sewer pipes are made locally of carbon based materials, and that might be an excellent sink for carbon extracted from the atmosphere over coming decades.

I have published figures showing that a storage system can be constructed to deliver the required sustained backup power using concrete. I am expecting to find that the size of the facility can be reduced by upgrading the counterweight to either iron or lead, or a combination of the two.

The overall cost of the facility will be one of the objectives I hope to add to the working documents created today.

***
You and I have discussed launching from the Moon in the past .... there are numerous posts in the archive that discuss the topic in some depth.

I believe that there is NO (ZERO/NADA) demonstration of linear motor acceleration over a distance greater than the length of an aircraft carrier deck.

While a gravity based energy storage system sounds eminently practical for the Lunar application, I am deeply skeptical that there is anyone living on Earth today who has more than a vague idea of how to implement an actual Lunar mass driver.

The demonstration of an actual (original design) mass driver at Princeton decades ago was for a distance of a few meters.

That design used capacitors to hold the charge that fed the magnets.

In thinking about your (to me interesting) variation on the idea, I am wondering if your gravity electron pump might prove less expensive to construct than the capacitor based design. The capacitor based design that I saw was using state-of-the-art industrial capacitors (ca 1985) and those are NOT likely to be manufactured in the quantity needed on the Moon. In contrast, my (limited to be sure) understanding of your idea seems (to me until corrected) feasible to construct of existing Lunar materials on the Moon.

If you are so inspired, please feel free to expand your presentation skills to include the use of images.

Void has shown the way, as has GW Johnson.

I would like to see diagrams of the mass driver you have in mind, including detail of the storage subsystem, and the power delivery stations.

(th)

Calliban · 2021-11-15 10:49:53

I will look into this. I had downloaded an Imgur ap for this purpose.

Using a linear generator to produce pulsed power, requires that we accelerate a superconducting magnet using expanding gas or liquid and then rapidly convert its kinetic energy into electric power by passing it through a conducting coil, probably constructed from aluminium. The result is a pulse of power as the magnet decelerates and its field lines cut through the coils, inducing current. Alternatively, we could do away with the need for a moving solenoid and use the falling mass to force a liquid metal, like sodium, through an MHD generator. That is probably the simplest design overall.

For the mass driver: At the speeds we are talking about (up to 2.4km/s, 5400mph, Mach 7.5), I would propose using a sliding contact on the mass driver bucket being accelerated, to energise the coils ahead of the bucket. That avoids the need for rapid switching. These speeds have already been achieved by rocket sleds, so we know that vehicles mounted on sliding rails will stand up to the frictional forces. The shots must be timed to allow the rails and bucket sliding contacts sufficient time to cool by radiation.

In terms of steering the payloads after launch: O'Neil raised the idea of using magnetic fields to fine tune the trajectory of the projectiles. As we are launching from the same point on the lunar surface, the local gravity map will not change between shots. The only variables are the gravitation influence of other planets, which will exert tiny but significant forces on the payload. However, O'Neil envisaged that it should be possible to deliver packages on lunar dirt to a bag at L5 with an accuracy of a few tens of metres, moving at 60m/s on impact.

Last edited by Calliban (2021-11-15 11:01:20)

tahanson43206 · 2021-11-15 11:42:36

For Calliban re #79

Thanks for taking a look at the imgur ap! I recognize I was (and am) imposing upon one of our most productive members, but I thought/hoped the risk would be worth taking, if we can (somehow) get a better sense of how some of your ideas fit together.

Regarding the sliding switch idea... The benefit that occurs to me immediately is that the design would compensate for variations in the mass of individual buckets. In (what I understand to be) the "traditional" design approach, fixed distances are set between stations, and a frequency of alternating current is supplied to achieve maximum pull/push effect when a bucket passes by a station. Maximum effectiveness for this technique would depend upon consistent mass of the buckets (or so I'm imagining).

A variation on the sliding switch idea might be crossing an optical beam? There are probably trades in favor of the sliding contact method.

In either case, the avalanche of stored energy would (presumably) need to be poised ahead of a launch, so a computer check of every station would seem (to me at least) advisable. A signal when the bucket passes each station could be fed back to the launch computer for feedback on how well the launch went.

***
Capture of buckets at the destination has received a lot of thought. Some of the earlier ideas (I'm vaguely remembering) seemed a bit on the fanciful side, but that work goes back all the way to the 1970's.

(th)

tahanson43206 · 2021-11-16 13:01:31

for SpaceNut ... I just reread this topic from the top ...

You, Calliban, KnightdePaix and perhaps other members contributed.

At one point, you were looking seriously at creating a small water power storage system on your property, since you have a naturally flowing spring of undrinkable water available. I'm wondering if you ever went any further with that idea. The power that might be secured from a 50 foot drop with just a few gallons per minute might not be much, but it would be steady, and it could be accumulated.

(th)

SpaceNut · 2021-11-16 17:27:39

The alternator from my 2001 escape is out of the vehicle so I have a 110 amp 12v to design around since it was not the same mounting as my 2005 escape that just failed see kbd512 topic for the run down.

I have also been looking at the solar concentrated light as another thermal generation for heat, hot water and such as well. While being out in the yard working to fix my alternator problem. I have been noting where the sun hits and the time when its above the horizon. A right angle to the horizon means a vertical wall receivers for that energy that tilts as the sun rises and sets toward the west but I get light for only 90' to 120' of the day from start to where it falls in the trees by 3:30 pm. It may be light earlier than 8 am but its not above the tree line of the east.

tahanson43206 · 2021-11-18 08:33:11

For SpaceNut re #73

Best wishes for success in developing a home energy supply using available resources. Those appear to include a constant supply of fresh (undrinkable) water, sunlight and occasional breezes.

As a reminder, if you dig down far enough (ie, about 1000 meters ~ ) you will be able to draw heat from the Earth's core.

Edit: Per a post on Quora, a "best guess" for boiling water is about 4 kilometers down. If you can afford to drill that far down, then you would have an inexhaustible source of hot water (steam) to power a turbine generator. You'll need to deliver cold water (or air) into the cold side of the system, and pipe the steam from the geothermal source into the hot side. Once you've made the investment, it should last as long as you do, and the resource would increase the value of the property due to the freedom from dependence upon the local utility. Another benefit is that the steam should condense to potable water, so you would have achieved two of the main objectives for supporting a family ... a source of reliable power that would not be taxed (beyond the initial investment) and a source of potable water independent of the local supply.

It should most definitely be possible for an entrepreneur to develop this line of product/service almost anywhere on the (non-mountainous) surface of the Earth.

***

I'm posting today the results of a visit to www.WolframAlpha.com ...;

The problem shaped by the work of Calliban, to evaluate options for a vertical drop gravity storage system, turns out to be suitable for expressions as:

X * Y = Constant

In the case of the opening presentation by Calliban, earlier in this topic, the constant is 7638.

The sweet spot I'd been expecting turns out to be obvious to someone whose brain is properly tuned, which mine was obviously not!

The sweet spot occurs at precisely double the square root of the constant.

Therefore, in the case of the example provided by Calliban, the constant is 7638, the Square root is 87.3957 (etc) and the sweet spot is 174.791 (etc)

Accordingly, for a gravity storage system using concrete as the counterweight (note that concrete varies in density) the shaft depth is 175 meters.

If the diameter of the shaft is 1 meter, then the sweet spot occurs when the height of the counterweight is precisely equal to the drop distance.

The sum of the height of the counterweight and the drop distance is the total height of the shaft.

175 meters is feasible for implementation on Earth. The payoff for the investor is a system able to deliver 2 Kw for 24 hours.

However, 175 meters is the OUTER limit of what is possible.

The density of concrete (or granite as given in the example of Calliban) is less than the density of iron and less again than that of lead.

The tradeoffs at work include:

Cost of material for counterweight
Cost of pipe for shaft
Cost of wire for power take off
Costs of other hardware required

There is (no doubt) a sweet spot for cost of this facility.

The sweet spot will vary depending upon market conditions at the time of construction.

It appears that 175 meters is the outer limit of what is required.

it should be possible to reduce that distance by half or more, through a combination of:

1) Greater density of counterweight
2) Greater diameter of counterweight

Edit: the post at this address has been updated to show computations for the concrete counterweight:
http://newmars.com/forums/viewtopic.php … 44#p187344

Existing power back up companies (such as Generac and Kohler) ** should ** be interested in this option.

Much of their existing electronics can be adapted to this application without change. The generator should be usable as is, given suitable gearing as suggested by SpaceNut recently in this topic. The existing two cylinder motorcycle engine could be retained for backup past 24 hours.

Update at 14:14 local time ... Post #2 of 4 above has been updated to show calculations for iron as the counterweight material.

The sweet spot I found is 106 meters for the shaft using a cylinder of 1 Meter diameter.

The work is preliminary and I would appreciate someone checking for errors. I ** think ** the numbers are close, but am only about 80% confident.

Tomorrow I'll work up the post (#3 of 4) for lead. The shaft length should drop below 80 meters but we'll see.

(th)

tahanson43206 · 2021-11-18 13:46:00

Looking ahead to work scheduled for tomorrow, I asked Google what the market price is for lead...

what is market price of a ton of lead
All
ImagesShoppingNewsVideosMore
Tools
About 258,000,000 results (0.64 seconds)
Unit conversion for Lead Price Today
Conversion Lead Price Price
1 Ton = 1,000 Kilograms Lead Price Per 1 Kilogram 2.28 USD
Live Price of Lead per Ounce | Markets Insiderhttps://markets.businessinsider.com › commodities › lead-...

$2.28 per kilogram. or (presumably) $2,280 per ton

For comparison, iron (cast iron) is much more modest ...

About 9,600,000 results (0.61 seconds)
Updated 11/18/2021
Metal Average Price Date Updated
Brass Shells $1.63/lb Updated 11/18/2021
Light Iron $168.00/ton Updated 11/18/2021
#1 Steel $237.00/ton Updated 11/18/2021
Cast Iron $248.00/ton Updated 11/18/2021
156 more rows
Current Scrap Metal Prices - iScrap App
https://iscrapapp.com › prices

The tradeoff is between the price of the material; and the depth of the shaft.

No doubt there is a sweet spot that depends upon market conditions in the vicinity of the job site.

(th)

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