Gravity Powered Generator

[DavidJames]

I want to consider an idea for a rock mine operation to generate electricity for 10 or more years. There's a constant row of trucks that dump loads off a cliff about a 100 foot drop to a lower level.

 

Would it be possible to build a chain driven tower with huge buckets on the chain drive that are spaced apart around the chain drive so that each truck dumps its huge load into a shoot that slides rock into the bucket at the top of the cliff, and about 7 full buckets would be full at all times on the desending side (empty buckets would be returning on the other side), and the chain drive would be turning a drive shaft at about a steady 18 rpm (same as a large wind turbine) and the drive shaft would be connected to a wind turbine type gearbox, which drives a large generator @ 1600 - 1800 rpm?

 

As an example, a 2 MegaWatt wind turbine "input torque says 980 - 1.250 kNm" (whatever that is), so I'm wondering how much tonnage of rock would have to be on the desending side of the chain tower at all times to match that? I'm seeking a common denominator to be able to roughly estimate, for now, how big a generator we could install to serve the mine operations and feed back electric into the grid.

 

I know a 2 Mega Watt wind turbine might require much more tonnage to turn it, but any size system would be better than wasting all this tremendous kinetic energy (thousands of tons daily).

 

I know this is tough, since this is a new concept, and it may be out of your line to roughly estimate, but if you can't help me, who in the forum might be able to help?

 

Thanks, David James

[swansont]

The maximum energy you generate will be mgh, i.e. for each kilogram and each meter it drops, you generate 9.8 Joules. If it's dropping at constant speed (e.g. because it's turning a turbine) the power will be mgv. 1 kilogram falling at 1 m/s generates 9.8 Watts, for as long as it can fall.

 

The speed of the buckets will be dictated by the mechanics of the turbine, but for simplicity let's assume 2 m/s and round up g from 9.8 to 10. The mass you want on the descending side is m = P/gv or 100,000 kg, i.e 100 metric tons. A faster drop means less mass.

 

Of course, all of this ignores the energy you expend bringing the rocks in and dispersing them at the bottom. IOW, there's a reason we do this with water and not rocks.

[DavidJames]

The maximum energy you generate will be mgh, i.e. for each kilogram and each meter it drops, you generate 9.8 Joules. If it's dropping at constant speed (e.g. because it's turning a turbine) the power will be mgv. 1 kilogram falling at 1 m/s generates 9.8 Watts, for as long as it can fall.

 

The speed of the buckets will be dictated by the mechanics of the turbine, but for simplicity let's assume 2 m/s and round up g from 9.8 to 10. The mass you want on the descending side is m = P/gv or 100,000 kg, i.e 100 metric tons. A faster drop means less mass.

 

Of course, all of this ignores the energy you expend bringing the rocks in and dispersing them at the bottom. IOW, there's a reason we do this with water and not rocks.

 

The maximum energy you generate will be mgh, i.e. for each kilogram and each meter it drops, you generate 9.8 Joules. If it's dropping at constant speed (e.g. because it's turning a turbine) the power will be mgv. 1 kilogram falling at 1 m/s generates 9.8 Watts, for as long as it can fall.

 

THE CHAIN TOWER WILL BE CONSTANTLY TURNING WITH A CONSISTENT TOTAL WEIGHT OF ALL THE LOADED BUCKETS ON THE DESCENDING SIDE WHICH WILL SLOWLY TURN A DRIVE SHAFT AT THE BOTTOM OF THE TOWER AT A STEADY 18RPM (SAME AS A WIND TURBINE OPERATION THAT GOES TO A GEARBOX).

 

WE WILL KEEP THE BUCKETS LOADED WITH THE SAME WEIGHT CONSTANTLY, WITH WHATEVER CONSTANT TOTAL WEIGHT DISTRIBUTED EVENLY IN EACH OF THE BUCKETS TO CREATE TREMENDOUS GRAVITATIONAL POWER TO DO THE REQUIRED WORK OF TURNING THE DRIVE SHAFT. ONE BUCKET WILL BE LOADED AT THE TOP OF THE TOWER (AT THE TOP OF THE CLIFF) WHILE ONE BUCKET WILL BE DUMPING ITS LOAD INTO A TRUCK AT THE BOTTOM OF THE 100 FT CLIFF AS THE BUCKET GOES UP THE ASCENDING SIDE OF THE TOWER.

 

The speed of the buckets will be dictated by the mechanics of the turbine, but for simplicity let's assume 2 m/s and round up g from 9.8 to 10. The mass you want on the descending side is m = P/gv or 100,000 kg, i.e 100 metric tons. A faster drop means less mass.

 

I WANT THE SPEED OF THE DRIVE SHAFT/SPROCKET TO MATCH THE SPEED OF A WIND TURBINE (ONE TURBINE I SAW ONLY REQUIRES 18RPM ON THE DRIVE SHAFT THAT GOES INTO A GEARBOX AND THEN TO A GENERATOR). BUT, OUR PROJECT WILL ALLOW A CONSTANT SPEED WITH NO VARIATION IN RPM AS WITH A WIND TURBINE, BECAUSE THE CHAIN WILL BE CONSTANTLY TURNING THE DRIVE SHAFT/SPROCKET UNDER THE SAME WEIGHT OF THE TOTAL DESCENDING LOAD OF ALL BUCKETS COMBINED.

 

Of course, all of this ignores the energy you expend bringing the rocks in and dispersing them at the bottom. IOW, there's a reason we do this with water and not rocks.

 

ACTUALLY, THIS WILL SAVE TIME AND MONEY, SINCE THE TRUCKS NOW DUMP THEIR LOADS OVER THE CLIFF AND THEN THE ROCK HAS TO BE RELOADED AT THE BOTTOM OF THE CLIFF FOR TRANSPORT TO PROCESSING. THIS IS ALL BEING DONE PRESENTLY IN ORDER TO KEEP ALL THE TRUCKS AT THE TOP OF THE CLIFF FROM HAVING TO TAKE A VERY SLOW 10 MILE ROUND TRIP THROUGH WINDING HILLS TO GET TO THE BOTTOM LEVEL OF THE PROPERTY WHERE THE ROCK IS TRANSPORTED OUT TO PROCESSING.

 

I'M AN AMATEUR, SO YOUR CALCS ARE TOUGH FOR ME TO FIGURE OUT. CAN YOU SAY IT IT DUMMIE LANGUAGE. WE CAN BUILD THE PROJECT WITH ANY SIZE BUCKETS AND WE COULD LOAD THEM WITH ANY SIZE LOADS NECESSARY, SO YOU CAN FIGURE OUT A SCENARIO USING YOUR OWN NECESSARY CONSTANT TOTAL DESCENDING WEIGHT THAT TURNS THE SHAFT AT THE BOTTOM OF THE TOWER AT 18RPM. WE NEED TO KNOW IF IT WOULD GENERATE ENOUGH ELECTRICITY OVER A 10 OR 12 YEAR PERIOD TO MAKE IT WORTH WHILE.

 

THAT'S WHY I GAVE THE TORQUE INPUT NECESSARY TO RUN A 2 MEGA WATT TURBINE AT 18RPM. THE SPECS ON THE WIND TURBINE SAID 980 - 1,250 kNm and RATIO 40:1 - 125.1

 

tHANK YOU,

David James

[swansont]

No need to SHOUT. Use the quote function; enclose the quoted text

Hello

shows up as

Hello

 

———

 

This is actually as simple as the physics calculations get. mg is the weight of the material (m is mass, g is the acceleration due to gravity) h is the height of the cliff. v is the speed. The power output is the weight* the speed.

 

 

18 rpm can only be turned into a linear speed if the gearing and shaft size are known — you multiply the radius by the angular speed to get linear speed. But that can probably be engineered to give you a range of speeds. To get about 2 MW you need 100 metric tons (100,000 kg) in the system at all times at 2 m/s, or 10 metric tons moving at 20 m/s. The combination of the two numbers (mass in kg, speed in m/s) gives you the maximum power output (in Watts).

 

100 feet is about 30 meters, so at 2 m/s it takes 15 seconds for a bucket to descend, so you have to fill a bucket every 15 seconds. At 20 m/s, you're filling a bucket every 1.5 seconds.

[DavidJames]

No need to SHOUT. Use the quote function; enclose the quoted text [noparse][/noparse] shows up as

 

———

 

This is actually as simple as the physics calculations get. mg is the weight of the material (m is mass, g is the acceleration due to gravity) h is the height of the cliff. v is the speed. The power output is the weight* the speed.

 

MOST EXCELLENT INFORMATION HERE. MAN, IT'S GREAT TO TALK TO A GENIUS. TO THINK YOU HAVE MOST OF THIS STUFF AT THE TIP OF YOUR MEMORY BANK, TOO!

 

M/S - WHAT IS THE S IN M/S, SINCE V IS SPEED?

 

 

18 rpm can only be turned into a linear speed if the gearing and shaft size are known — you multiply the radius by the angular speed to get linear speed. But that can probably be engineered to give you a range of speeds. To get about 2 MW you need 100 metric tons (100,000 kg) in the system at all times at 2 m/s, or 10 metric tons moving at 20 m/s. The combination of the two numbers (mass in kg, speed in m/s) gives you the maximum power output (in Watts).

 

THIS CAN ALL BE WORKED OUT WITH SPECIFIC ENGINEERING. FROM YOUR CALCS, IT APPEARS WE HAVE TO WEIGH THE COST OF BUILDING THE TOWER AND PRODUCING ABOUT 10% OFTHE POWER OF A 2 MW UNIT. A 2 MW UNIT WILL GENERATE ENOUGH POWER FOR OVER 300 HOMES, SO 10% OF THAT, ROUGHLY SPEAKING OF COURSE, WOULD DO JUST FINE AND WOULD EVEN ALLOW FOR A SIZEABLE OVERAGE FEED BACK INTO THE GRID.

 

100 feet is about 30 meters, so at 2 m/s it takes 15 seconds for a bucket to descend, so you have to fill a bucket every 15 seconds. At 20 m/s, you're filling a bucket every 1.5 seconds.

 

I'M NOT SURE WHAT THAT CALC MEANS TIL I GET THE "S", BUT BELIEVE IT OR NOT, THE TRUCKS CAN UNLOAD AT A PRETTY FAST RATE, EVEN FOUR AT A TIME, SO THEIR LOADS WOULD GO INTO A HORIZONTAL "CONVEYOR BELT WELL" AND ALLOW US TO MANAGE WEIGHTS GOING INTO THE BUCKETS.

 

USING YOUR CALCS (I think), JUST ROUGHLY SPEAKING, WE'D HAVE ABOUT 2.5 MINUTES TO LOAD A BUCKET (building a unit only 10% the size of the 2MW), WHICH WOULD BE ALMOST A MINUTE MORE THAN WE'D NEED FOR A SMALLER UNIT. ANYWAY, THIS IS WHERE THE SPECULATION HAS TO BE TURNED INTO ACTUAL CALCULATION BY AN ENGINEER, IN ORDER TO ADDRESS YOUR ASSUMPTIONS.

 

NOT SURE WE'RE ALTOGETHER ON THE SAME TRACK, BUT WE'RE CLOSE. IF THE BUCKETS ARE 18 TO 20 FEET APART, AND I WANT ONLY 18RPM ON THE SHAFT, AND SAY THE SHAFT IS 2 TO 4" diameter (the shaft drive sprocket teeth being turned by the chain should be close to a non-factor with diameter, and the bigger the shaft size needs to be, we'd have to space the fewer buckets further apart on the chain drive and make them carry a heavier load), since I need at least one minute to load a bucket. ANYWAY, THE DRIVE SHAFT GOES THROUGH THE BOTTOM AND/OR TOP CHAIN SPROCKETS (which means the drive shaft is at right angle to the tower, - WE SHOULD BE ABLE TO MAKE IT HAPPEN, SINCE EVERY REVOLUTION USES A FOOT OR LESS OF THE CHAIN DRIVE, GIVING ME 18RPM between bucket dumps. BUT, THESE ARE ALL SUCH A SPECULATIVE CALCS AT THIS POINT ANYWAY, AND THERE'S LIKELY SOME CONFUSION BETWEEN WHAT EACH IS RECEIVING IN UNDERSTANDING OF THE OTHER'S STATEMENTS.

 

I NEEDED SOMEONE TO GIVE ME A DOSE OF REALITY BEFORE EVEN VENTURING OUT TO GET PRELIMINARY ASSESSMENT FROM AN ENGINEER ON SUCH A STRANGE IDEA. HEY, WHAT ARE YOU DOING FOR THE NEXT DAY OR TWO? ANYWAY, YOU'VE SHOWN ME THAT A POSSIBLE SHORTAGE OF TIME BETWEEN DUMPS MIGHT POSE A PROBLEM, SO MAYBE PART OF THE ANSWER WOULD INCLUDE USING A UNIT THAT CAN OPERATE ON THE LOWEST POSSIBLE RPM WITH A BIGGER GEAR BOX TO GIVE US THE MINIMUM 1500-1800 RPM TO RUN A GENERATOR.

 

ANYWAY, IF YOU DON'T ANSWER THIS, THANK YOU EVER SO MUCH FOR ALL THE INFO. WISH I HAD HALF YOUR EDUCATION. JUST GREAT.

 

I'LL LET YOU KNOW WHAT, IF ANYTHING, HAPPENS ONCE AN ENGINEER PERFORMS SOME SPECIFIC PRLIMINARY CALCS THIS YEAR OR NEXT?

 

THANKS AGAIN, David James

[Iggy]

A trained physicist (which I am not) should check my reasoning, but...

 

If you are operating it continuously then you should be able to divide the power you want by the energy you get from dropping a kilogram that distance and get the amount of mass in kilograms that you would need to add to the cycle every second.

 

If it is 2 megawatts and 30 meters then divide 2,000,000 watts / 294 J. (294 comes from m*g*h). So you'd need to add 6803 kilograms per second (at least -- you never have perfect efficiency) to get 2 megawatts. That would be 587,779 metric tons per day -- which is... I think... thousands of dump trucks of rock per day.

 

Half the power would require half the material, twice would require twice, etc.

 

Of course, your hole would fill up right quick.

Edited September 18, 2011 by Iggy

[swansont]

A trained physicist (which I am not) should check my reasoning, but...

 

If you are operating it continuously then you should be able to divide the power you want by the energy you get from dropping a kilogram that distance and get the amount of mass in kilograms that you would need to add to the cycle every second.

 

If it is 2 megawatts and 30 meters then divide 2,000,000 watts / 294 J. (294 comes from m*g*h). So you'd need to add 6803 kilograms per second (at least -- you never have perfect efficiency) to get 2 megawatts. That would be 587,779 metric tons per day -- which is... I think... thousands of dump trucks of rock per day.

 

Half the power would require half the material, twice would require twice, etc.

 

Of course, your hole would fill up right quick.

 

That's jibes with my calculation. I had rounded g to 10, but 10,000 kg in 1.5 seconds is 6666 kg/sec. If a standard dump truck can carry ~25 metric tons, that's ~270 trucks per second. 130,000 in an eight-hour day. Even for an off-road hauler, which has a little over 10x the carrying capacity, you're talking >20 loads per second.

 

Edit: See correction below.

[insane_alien]

That's jibes with my calculation. I had rounded g to 10, but 10,000 kg in 1.5 seconds is 6666 kg/sec. If a standard dump truck can carry ~25 metric tons, that's ~270 trucks per second. 130,000 in an eight-hour day. Even for an off-road hauler, which has a little over 10x the carrying capacity, you're talking >20 loads per second.

 

eh?

 

for 6666kg/sec its only 1 25 ton truck every 3.75 seconds still over 100000 trucks an eight hour day.

 

your shaft is going to get filled pretty damn quick.

 

lets assume it's 5mx5mx30m thats 750m^3 of rock. but lets go ahead and assume best case scenario, a single solid block of lead thats fits perfectly. thats 8505 metric tonnes.

 

dropping this in would yield 2.502*10^9 J (using 9.81 as g) and at a consumption of 2MW you get a grand total of 1251seconds of power.

 

so your shaft gets filled (under best case scenario) in under 21 minutes.

 

so you'll need just over 69 similar shafts a day to maintain power output.

 

and this isn't accounting for inefficiencies or the fact that as the shaft fills, the energy per kilogram will drop off proportionally.

[swansont]

eh?

 

for 6666kg/sec its only 1 25 ton truck every 3.75 seconds still over 100000 trucks an eight hour day.

 

your shaft is going to get filled pretty damn quick.

 

lets assume it's 5mx5mx30m thats 750m^3 of rock. but lets go ahead and assume best case scenario, a single solid block of lead thats fits perfectly. thats 8505 metric tonnes.

 

dropping this in would yield 2.502*10^9 J (using 9.81 as g) and at a consumption of 2MW you get a grand total of 1251seconds of power.

 

so your shaft gets filled (under best case scenario) in under 21 minutes.

 

so you'll need just over 69 similar shafts a day to maintain power output.

 

and this isn't accounting for inefficiencies or the fact that as the shaft fills, the energy per kilogram will drop off proportionally.

 

Gah, I dropped the "k" and had kg where it should have been tons. All my numbers are off by a factor of 1000. ~.27 trucks per second, 130 per shift.

 

I think the rock gets hauled away at the bottom, so filling is not an issue. But that's 130 more truck loads.

[insane_alien]

I think the rock gets hauled away at the bottom, so filling is not an issue. But that's 130 more truck loads.

 

one would still assume a finite space inside. maybe 50 times the capacity of the shaft which would give maybe 17 hours of power.

 

which of course begs the question, why not just take the fuel from the trucks and burn it in a gas turbine generator?

[DavidJames]

A trained physicist (which I am not) should check my reasoning, but...

 

If you are operating it continuously then you should be able to divide the power you want by the energy you get from dropping a kilogram that distance and get the amount of mass in kilograms that you would need to add to the cycle every second.

 

WHEN YOU SAY "DROP A KILO THAT DISTANCE", I'M NOT TALKING FREE FALL HERE. i THINK YOU MIGHT HAVE MISSED THE CHAIN DRIVEN TOWER PART OF THE EQUATION, WHERE BUCKETS OF ROCK ATTACHED TO THE REVOLVING TOWER CHAIN ARE CONSTANTLY DESCENDING SLOWLY BECAUSE THE BUCKETS ARE BEING FILLED CONSTANTLY AT THE TOP OF THE CLIFF. AT THE BOTTOM OF THE CLIFF AND TOWER A CHAIN SPROCKET ATTACHED TO A HORIZONTAL DRIVE SHAFT GOES INTO A GEARBOX WHICH GOES TO A GENERATOR.

 

If it is 2 megawatts and 30 meters then divide 2,000,000 watts / 294 J. (294 comes from m*g*h). So you'd need to add 6803 kilograms per second (at least -- you never have perfect efficiency) to get 2 megawatts. That would be 587,779 metric tons per day -- which is... I think... thousands of dump trucks of rock per day.

 

I'M NOW THINKING THAT ANY EQUATION THAT CONSIDERS "DISTANCE" IS NOT VALID HERE, SINCE THE DISTANCE IS NOT A FACTOR TO MAKING THE MACHINE TURN THE DRIVE SHAFT. IT'S A MACHINE THAT IS CONSTANTLY TURNING A LOW RPM DRIVE SHAFT THAT GOES INTO A GEARBOX, AND ALL THE WORK IS DONE BY CONSTANT GRAVITATIONAL FORCE ON TURNING THE SPROCKET/SHAFT AS THE CHAIN REVOLVES UP AND DOWN THE TOWER IN ONE DIRECTION.

 

Half the power would require half the material, twice would require twice, etc.

 

NOT GETTING THIS ONE.

 

Of course, your hole would fill up right quick.

 

THE HOLE WOULD NOT FILL UP BECAUSE THERE IS A LINE OF RECEIVING TRUCKS AT THE BOTTOM AND EACH TRUCK RECEIVES ONE BUCKET LOAD THAT IS DUMPED.

[Fuzzwood]

Please go study some elementary physics before you start shouting again. Every time when there is a calculation given you are like: I DON'T UNDERSTAND WHAT YOU ARE SAYING AND HERE IS MY PSEUDO PERPETUAL MOBILE AGAIN.

[csmyth3025]

I think I understand your type of operation and your motivation. Right now you're dumping a lot of rock down a cliff face from your mining source so that trucks at the base of the cliff can transport the material to your processing facility. The cliff is, essentially, a shortcut to avoid hauling the rock an additional 10 miles overland to reach the same ultimate destination.

 

Your intention is to recycle some of the kinetic energy of the falling rock in order to generate electricity in place of (or to supplement) wind-generated power that might be proposed for your facility as a cost savings and/or source of revenue.

 

I get the feeling that you would like some small assurance that if you propose this idea to your boss or to the company engineer, you won't be laughed out of the office or (at best) told "it won't work".

 

Heavy duty bucket elevators are widely available through industrial machinery suppliers. Your idea is to use them to drive a generator as the rock goes down instead of using an electric motor to raise the rock up to a higher elevation.

 

Gearing the elevator drive sprockets to match a suitable generator is a simple mechanical problem. Again, an industrial machinery supplier would probably have an off-the-shelf gear drive that would serve your purpose.

 

With the proper engineering I believe this can be done and significant usable energy can be generated. The question is (first) whether your employer is willing to consider the idea , and (second) whether he's willing to spend the money to have an engineer look at the proposal. Only an engineer working on your specific installation can provide you with an estimate of the capital costs involved and the possible cost savings (electrical generation capacity) that can be expected.

 

If your idea is shot down don't be discouraged. In many companies the folks in the front office are predisposed to dismiss suggestions from anyone who spends his workday in mud-smeared coveralls.

 

Personally, I think your idea shows original thinking. I like it. I think it deserves a proper engineering analysis.

 

Chris

 

PS - Generally, you don't need to use your cap-lock or bold letters. In forums, typing in capital and bold letters is interpreted as yelling out your reply (the people you're replying to think you're yelling at them). Also, if you plan to use the forum very much it would be a good idea to figure out how to use the "quote" button at the top of your post box to separate what someone else has said from what your reply is.

 

Edited to improve clarity

Edited September 18, 2011 by csmyth3025

[DavidJames]

one would still assume a finite space inside. maybe 50 times the capacity of the shaft which would give maybe 17 hours of power.

 

iT'S NOT "INSIDE" A SHAFT, BUT RATHER THE LOADS ARE PRESENTLY DUMPED OFF A 100 TO 150 FOOT MOUNTAIN CLIFF, AND THE LOADS TUMBLE DOWN TO A LOWER LEVEL INTO AN ANGULAR CHUTE THAT IS MAN CONTROLLED TO ALLOW A LINE OF TRUCKS TO EACH RECEIVE A LOAD FROM THE CHUTE AND TAKE OFF WITH THEIR LOAD FOR VARIOUS TYPES PROCESSING.

 

IN OTHER WORDS, TRUCKS AT THE TOP OF THE CLIFF DRIVE TO THE CLIFF'S EDGE FROM OTHER AREAS AT THE TOP OF THE MOUNTAIN MINE AREA.

 

THESE TRUCKS WOULD HAVE TO DRIVE 10 OR MORE MILES AROUND SLOW WINDING ROADS TO GET THEIR LOADS OFF THE MOUNTAIN AND TO THE BOTTOM OF THE MOUNTAIN, SO THEY USE THIS DOUBLE UNLOAD AND RELOAD SYSTEM INSTEAD.

 

which of course begs the question, why not just take the fuel from the trucks and burn it in a gas turbine generator?

 

I GUESS EVERYBODY IS CONFUSED. I'M TRYING TO CAPTURE THOUSANDS OF TONS OF POTENTIAL GRAVITATIONAL POWER/FORCE, WHICH IS NOW BEING DUMPED OFF THE SIDE OF A MOUNTAIN AND THEREBY BEING WASTED.

 

I WANT THE TOWER TO BE OPERATING ON THE GRAVITATIONAL FORCE OF TONS OF ROCK THAT SLOWLY, SLOWLY DESCENDS IN BUCKETS ATTACHED TO A PERPENDICULAR CHAIN TOWER THAT HAS A CHAIN SPROCKET AT THE TOP OF THE TOWER AND A CHAIN SPROCKET AT THE BOTTOM. THE SPROCKET AT THE BOTTOM WOULD BE ATTACHED TO A DRIVE SHAFT GOING THROUGH IT AND INTO A GEARBOX THAT MULTIPLIES THE SPEED FAST ENOUGH TO GENERATE ELECTRICITY, ON THE SAME PRINCIPLE AS THE WIND TURBINE (12-18 RPM CONVERTED INTO 1600 - 1800 RPM).

 

WE WANT TO SEE THE REVOLVING CHAIN AND BUCKETS PURPOSELY REVOLVE AS SLOW AS POSSIBLE, AND THE SLOWNESS MUST BE THE DIRECT RESULT OF THE RESISTANCE FROM DRIVING THE GEARBOX AND GENERATOR.

 

AN ENGINEER ONCE TOLD ME THAT EVEN IF YOU HAVE A DRIVE SHAFT THAT ONLY TURNS 1 RPM, BUT IS AN ALMOST UNSTOPPABLE POWER, THAT POWER CAN BE HARNESSED BY MULTIPLYING IT THROUGH A SYSTEM OF GEARS. AGAIN, IT'S ON THE SAME PRINCIPLE AS A MODERN WIND TURBINE THAT CAN ONLY TURN THE MAIN DRIVE SHAFT AT 12 TO 18 RPM, AND THIS LOW, LOW RPM IS MULTIPLIED A HUNDRED TIMES BY THE SHAFT GOING THROUGH A MASSIVE GEARBOX AT THE TOP OF THE WIND TURBINE.

 

MY PROBLEM IS BASICALLY THIS: WE HAVE NEARLY UNLIMITED STORED GRAVITATIONAL ENERGY AT THE TOP OF THE MOUNTAIN, BUT WE MUST INVENT A PRACTICAL METHOD OF CONVERTING IT INTO ELECTRICITY. BILLIONS OF TONS OF KINETIC ENERGY IS BEING WASTED BY THROWING IT OFF A CLIFF. I WANT TO CAPTURE THIS ENERGY ON THE WAY DOWN, AND THE BEST WAY I SEE TO DO IT IS ON A REVOLVING CHAIN TOWER WITH BUCKETS ATTACHED TO THE CHAIN AT ABOUT 20 FEET APART.

 

WE MUST SLOW THE CHAIN TOWER DOWN AS SLOW AS POSSIBLE, BY USING THE RESISTANCE WE'D RECEIVE FROM A MASSIVE GEARBOX AND GENERATOR, SIMILAR TO THE SYSTEMS USED ON WIND TURBINES.

 

I GUESS MY ANSWER MIGHT BE FOUND BY DISCOVERING THE AMOUNT OF WIND FORCE NECESSARY TO TURN THREE 150 FOOT BLADES ON A 2 OR 3 MW WIND TURBINE.

 

THEN I NEED TO FIGURE OUT A SYSTEM THAT CAN GENERATE ELECTRICITY WITH THE LOWEST POSSIBLE RPM TO GIVE US ADEQUATE LOADING TIME AT THE TOP OF THE MOUNTAIN.

 

I KNOW IT CAN BE DONE, BUT MOST EVERYBODY IS COMING UP WITH THESE CALCULATIONS THAT RELY ON A VERY FAST REVOLVING TOWER, WHICH IS IMPOSSIBLE TO EMPLOY HERE.

 

THINK "GEARBOX WITH LOW, LOW RPM INTAKE"

[insane_alien]

WE WANT TO SEE THE REVOLVING CHAIN AND BUCKETS PURPOSELY REVOLVE AS SLOW AS POSSIBLE, AND THE SLOWNESS MUST BE THE DIRECT RESULT OF THE RESISTANCE FROM DRIVING THE GEARBOX AND GENERATOR.

 

AN ENGINEER ONCE TOLD ME THAT EVEN IF YOU HAVE A DRIVE SHAFT THAT ONLY TURNS 1 RPM, BUT IS AN ALMOST UNSTOPPABLE POWER, THAT POWER CAN BE HARNESSED BY MULTIPLYING IT THROUGH A SYSTEM OF GEARS.

 

There is probably some confusion here, engines can produce more than their native torque by gearing down to a slower angular velocity, going in reverse means you'd need a huge amount of torque to turn your generator. this would add quite a bit of friction losses.

 

Why not have it going a bit faster so that it turns more freely and there is less mass on the system so that it does not need to be as sturdy.

 

If you can tell us how much rock is being moved down then we can tell you the theoretical maximum power output you can get (you can't set both power and mass flowrate to arbitrary values as they will be related variables.

 

Please stop using ALLCAPS it is difficult to read.

[DavidJames]

There is probably some confusion here, engines can produce more than their native torque by gearing down to a slower angular velocity, going in reverse means you'd need a huge amount of torque to turn your generator. this would add quite a bit of friction losses.

 

 

I agree with the friction problem. However, friction might be the price we have to pay to make use of all the wasted tons of potential gravitational energy now being thrown over the side of the cliff. Producing a slow, constant and nearly uniform gravitational force on the chain is of utmost importance as I see it, since speedy revolution of the tower chain would directly relate to time issues at the top loading level.

 

 

I'm looking at a fractional replacement of gravitational force. For example's sake, say each bucket had a capacity of 10 tons of rock and say we have 6 buckets totaling 60 tons of gravitational force on the descending side of the tower, with ascending empty buckets on the other side, (all buckets whether loaded or empty are evenly spaced on the chain. Say the cliff goes 140 feet from top to bottom.

 

At the bottom of the cliff, as a bucket begins turning upside down on the bottom of the chain tower where the chain loops around the sprocket to the ascending side, thereby dumping its load, while another load at the top level is being placed in the empty bucket that just came up the ascending side and over the top loop and sprocket of the chain tower. This way we'd have a truck load per bucket, with 4 loaded buckets always in between the top bucket and the bottom bucket. Any question on that or is it still hard to determine what the thing looks like? As I said before, I'm an amateur.

 

 

Why not have it going a bit faster so that it turns more freely and there is less mass on the system so that it does not need to be as sturdy.

 

 

I think higher speed of the revolving tower chain is too tough to accomplish, since it leaves little to no room for error or time delays at the top loading level. How does the 100x multiplier gearbox on the wind turbine do the job, and now it's quite a common operation? (By the way, I wonder how many tons of torque or turning force those turbine blades develop on the wind turbine drive shaft. I've looked but can't find the answer so far.)

 

 

If you can tell us how much rock is being moved down then we can tell you the theoretical maximum power output you can get (you can't set both power and mass flowrate to arbitrary values as they will be related variables.

 

 

Since I'm not privy to exact info at this point, and they don't even know about what I'm doing, why not use the 60 ton example above, with fractional replacement of lost energy (dumped bucket) being done at the rate of 10 ton every 2 minutes, with that steady 60 tons of force always descending and it runs ONLY at the speed allowed by the resistance factor from the gearbox and generator.

 

After friction considerations and other considerations, I sure hope you'll come out with more energy production than what is produced from a bicycle headlight generator (ha).

 

Also, what about a flywheel instead of a gearbox? In other words, if say a 4" drive shaft that goes through the chain tower loop and continued through the chain sprocket a short distance out to and through the center of a flywheel that directly turns a generator at a 100x rate?

 

 

Please stop using ALLCAPS it is difficult to read.

 

Thanks, I used them in an attempt to separate the two sides of conversation. Now that you say that, I know I'm not the only one who hates it when people write like that.

 

I think I understand your type of operation and your motivation. Right now you're dumping a lot of rock down a cliff face from your mining source so that trucks at the base of the cliff can transport the material to your processing facility. The cliff is, essentially, a shortcut to avoid hauling the rock an additional 10 miles overland to reach the same ultimate destination.

 

Your intention is to recycle some of the kinetic energy of the falling rock in order to generate electricity in place of (or to supplement) wind-generated power that might be proposed for your facility as a cost savings and/or source of revenue.

 

I get the feeling that you would like some small assurance that if you propose this idea to your boss or to the company engineer, you won't be laughed out of the office or (at best) told "it won't work".

 

Heavy duty bucket elevators are widely available through industrial machinery suppliers. Your idea is to use them to drive a generator as the rock goes down instead of using an electric motor to raise the rock up to a higher elevation.

 

Gearing the elevator drive sprockets to match a suitable generator is a simple mechanical problem. Again, an industrial machinery supplier would probably have an off-the-shelf gear drive that would serve your purpose.

 

With the proper engineering I believe this can be done and significant usable energy can be generated. The question is (first) whether your employer is willing to consider the idea , and (second) whether he's willing to spend the money to have an engineer look at the proposal. Only an engineer working on your specific installation can provide you with an estimate of the capital costs involved and the possible cost savings (electrical generation capacity) that can be expected.

 

If your idea is shot down don't be discouraged. In many companies the folks in the front office are predisposed to dismiss suggestions from anyone who spends his workday in mud-smeared coveralls.

 

Personally, I think your idea shows original thinking. I like it. I think it deserves a proper engineering analysis.

 

Chris

 

 

Wow, what a concise overview of the issues! Everything is correct above, but I'm the one who would want to pay an engineer to do a preliminary assessment of the potential operation. Once I get a preliminary assessment I can approach them with it, since only rough cost numbers would be essential for them right now, and an engineer could do a rough workup, where the project would not likely have cost overruns by a million dollars. But yes, indeed, you hit z nail on z head.

 

Thank you for your overview, and for showing me that the reverse system of conveyor lift buckets is already in place. I do think, however, that maybe I was thinking the steadily descending massive weight would produce more energy than what the other gentlemen are figuring on, and I'm not sure more than one has actually comprehended the same issues as you have concisely explained above.

 

But what the hay, that's why I came here first to have y'al kick it around, and see if I'm altogether wet on my thoughts.

 

Cheez - thanks again Chris.

 

PS - Generally, you don't need to use your cap-lock or bold letters. In forums, typing in capital and bold letters is interpreted as yelling out your reply (the people you're replying to think you're yelling at them). Also, if you plan to use the forum very much it would be a good idea to figure out how to use the "quote" button at the top of your post box to separate what someone else has said from what your reply is.

 

Edited to improve clarity

I've already corrected that. As I said in the last letter, I thought it was a way to separate sides of the conversations. And, I hate to read it in block myself, so thanks for the advice.

 

Please go study some elementary physics before you start shouting again. Every time when there is a calculation given you are like: I DON'T UNDERSTAND WHAT YOU ARE SAYING AND HERE IS MY PSEUDO PERPETUAL MOBILE AGAIN.

 

Sorry that it sounds like I'm shouting. I am not shouting. I was using caps in place of the (quote), and I didn't pick up on this until just a while ago. No offense meant. Further, this is not a perpetual mobile, since it is being constantly fed by a finite energy source and only for a period of 10 to 12 years. I had not been responding in sequence to others' responses either, since there were so many responses and I did not expect it.

 

Are you still angry and mad and upset and ready to kick me into elementary school? I probably couldn't blame you. My sister says she'll never again write an email to anyone, because of the lack of properly intended voice inflection that does not come out in writing, and she got someone so mad at her when no such issue should have come to fruition, had voice inflection tempered her statements.

 

Thanks.

[Iggy]

A trained physicist (which I am not) should check my reasoning, but...

 

If you are operating it continuously then you should be able to divide the power you want by the energy you get from dropping a kilogram that distance and get the amount of mass in kilograms that you would need to add to the cycle every second.

 

WHEN YOU SAY "DROP A KILO THAT DISTANCE", I'M NOT TALKING FREE FALL HERE. i THINK YOU MIGHT HAVE MISSED THE CHAIN DRIVEN TOWER PART OF THE EQUATION, WHERE BUCKETS OF ROCK ATTACHED TO THE REVOLVING TOWER CHAIN ARE CONSTANTLY DESCENDING SLOWLY...

No. When I say "the energy you get from dropping a kilogram that distance" I mean the difference in gravitational potential energy between the top and bottom of the cliff. You can not get more than 294 joules from moving a kilo that distance downward no matter how things are geared or what mechanical system you are using to convert the energy. That's what mgh calculates.

 

From there it is a simple calculation to find the absolute least amount of rock you would need under the best circumstances (the best gearing -- the least friction -- the best mechanical setup) to get the power you want. Divide the power you want in watts (2 000 000 watts) by the product of gravitational acceleration and height (9.8 * 30). That will give you the mass per second (kg/s) that you need.

 

You get 6,802 kg/s. Or roughly 587,000 metric tons per day. That is the best case scenario. Twenty three thousand dump truck loads of material per day for 2 megawatts.

 

Half the power would require half the material, twice would require twice, etc.

 

NOT GETTING THIS ONE.

For 2 megawatts you need to unload at least 6,802 kilograms per second. To double that power (4 MW) you would need twice the material. To only get half the power (1 MW) you need half the material (3,401 kg/s). The power and the mass needed to get it are proportional.

 

Does this all make sense?