Calculate require force for moving in circular motion

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Calculate require force for moving in circular motion

Postby newrobert on November 17th, 2014, 11:57 pm 

Hi,

There is big roller coaster loop 100% round and vehicles are moving in this circular path, I want to calculate how much force will be required to move these vehicles.

Circle path diameter: 100 meter
Time: 1 round completed in 15 minutes
1 vehicle mass: 1,000 kg
Total Vehicles: 50

All vehicles are tied with each other just like train or roller coaster cars.

What force will be required to move these vehicles at required speed i.e. 1 round completed in 15 minutes. What's equation and how to solve it.

Please help me

Robert
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Re: Calculate require force for moving in circular motion

Postby Marshall on November 18th, 2014, 12:05 pm 

Hi Robert, I'm in the dark about some details, like how much friction you are assuming, and what role gravity plays, if any.

We can start talking, if you like, by making the simplest possible assumptions----no friction, level track, no role played by gravity.

So then the circumference is 628 meters and the time to make one circuit is 15 minutes or 900 seconds.

Roughly speaking the train is going 0.7 meters per second. That seems awfully slow for a roller coaster ride.
It is less than 2 mph.

If you assume no friction then once you have pushed the train to get it started it will role around the track forever without stopping, at the same speed. I'm ignoring Corioloris force which favors counterclockwise rotation in the northern hemisphere. It is to tiny to bother with.

I think the main force in the picture is that exerted by the track INWARDS. to keep the train cars going in a circle.

Each car masses 1000 kg, and it is being accelerated INWARDS towards the center of the circle with an acceleration which is V2/R

So each car is being subjected to a force F = Ma = 1000 kg * (.7 m/s)^2/100 m
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Re: Calculate require force for moving in circular motion

Postby newrobert on November 18th, 2014, 12:15 pm 

Hi Marshall,

Thank you for replying to my post. In fact these numbers are not actual and just used to understand. I am fixing formula in excel and will enter same formula with actual numbers. But I can give you correct numbers:

Track Type: Roller coaster loop type but 100% circular

Diameter: 107 meter
Circumference: 336 meter (pie 3.14159)
1 Round completed in: 15 minutes
Time: 15 x 60 = 900 seconds
Train mass: 835,628 kg

Now I want to know how much power I need to keep this train moving @ 15 minutes per round?

Can you please guide me how to solve this? Which equation will use first. I searched and found that Power = Work / Time but how to get work?

Can you please solve this step by step. At the end I need to know how much power electric motor is required to keep this train moving at required speed i.e. 15 minutes per round.

Later I will add roughly 10% of friction etc.

Can't say anything about gravity because I am not professional. :)
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Re: Calculate require force for moving in circular motion

Postby Marshall on November 18th, 2014, 12:36 pm 

Yikes! Is this one of the commercial "for profit" college courses? The problem does not make sense to me!
Are you paying then money to be given problems like this? Excuse me for asking *embarrassed*

Maybe some body else here, Faradave or Canadyspeak or several others, can help. they often help and they're good at this kind of thing.

But I'm confused. It would seem to me that the WHOLE THING IS THE FRICTION. If you want to know the power required to keep it going, ALL that power is doing is overcoming friction.

As a rule the friction is linearly proportional to the speed. It would depend very slightly on the INWARDS force exerted by the track, but roughly speaking that inwards force does not cost you any electric power.

In any case the key to the problem would be in the coefficients of friction. And they don't seem to have given you those. I'm at sea. Completely confused by the problem.
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Re: Calculate require force for moving in circular motion

Postby newrobert on November 18th, 2014, 12:53 pm 

Hi man,

I am happy to see that there is someone here to help beginners or who need help, really thank you for this.

In fact I have one amusement park project where I required to lift heavy water at 250+ feet height, in other words simply 100+ meter. Water is required for some amusement games/rides and water will drop in ground pool again from where it requires to be lifted up again. So traditional pumps cost is very very heavy and impossible to bear as compare to water volume.

Since I don't have any physics background but during last couple of days I did some research and use to with some basic terms and I set formulas in Excel.

First I found "water wheel" of ancient times used to lift water and I planned to create big 100 meter water wheel like ancient time, see my thread here viewtopic.php?f=77&t=27996

But I can't judge how much mass of wheel structure will be i.e. rim + spokes etc so without rim + spokes mass, can't guess how much powered will be required to rotate wheel.

So later I got idea of "rolling coaster loop". Instead of "water wheel" I planned to create big "O" of 100 meter diameter and install small cars with water buckets hanging at both sides. Even this train complete 1 round in 10-15 minutes that's enough for us. I mean buckets will be filled from ground pool and drop at 100m+ height pool by installing a small mechanisim. Water like 400k-500k gallon per round.

If train is moving on roller coaster track like clock wise then from right side empty buckets will come down and from left side water filled buckets will go up and drop water.

So this is the whole story and I need solution of this. How much power (KW) I need to rotate this train at required speed. If power will be less i.e. less than 200 KW then I can increase speed.

I dropped water wheel idea because in that huge weight of big wheel structure is also included in rotation along with water weight that's extra load on motor. But on Roller coaster loop idea there is only water moving not huge iron structure.

Can you please help me to find out this? Complete equation, which I can set in Excel and then change variables to see where there is good combination of water and power.

Thanks in advance for helping me.

Robert
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Re: Calculate require force for moving in circular motion

Postby Marshall on November 18th, 2014, 5:11 pm 

newrobert » Tue Nov 18, 2014 9:53 am wrote:...
In fact I have one amusement park project where I required to lift heavy water at 250+ feet height, in other words simply 100+ meter. Water is required for some amusement games/rides and water will drop in ground pool again from where it requires to be lifted up again. So traditional pumps cost is very very heavy and impossible to bear as compare to water volume.
...
complete 1 round in 10-15 minutes that's enough for us. I mean buckets will be filled from ground pool and drop at 100m+ height pool by installing a small mechanisim. Water like 400k-500k gallon per round.


It sounds like you want to raise water 250 feet at a rate of 400,000 gallon to 500,000 gallon per 15 minutes.

I cannot really help. Not an engineer. I can calculate the kW of power needed if the pump or wheel is perfectly efficient---you can add friction losses later.

250 feet is 76 meters
that amount of water weighs between 1.5 and 1.9 million kg, let's call it 1.5 for definiteness

Lifting one kg up by one meter takes about TEN JOULES of energy

So for a machine working with perfect efficiency you need to supply 15 Million Joules to lift that amount of water up by one meter. That is whether it is a wheel or a pump. The work is the same, if both are efficient.

But you want to lift it 76 meters, so multiply 15 by 76 and you get 1140 million joules

And you want to do that in 900 second, so that is 1140/900 million joules per second.

That is just a bit over one million joules per second or one million WATTS.

So what I get is about 1000 kW

Or if I had to be more accurate, 1140/900 = 1.27, so it would be 1270 kW

You need at least that amount of power, whether you use a water-lifting wheel or some kind of industrial pump.
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Re: Calculate require force for moving in circular motion

Postby newrobert on November 19th, 2014, 12:12 am 

Thank you for replying.

In fact I don't want this. There is difference in calculation when one object is moving in circular motion or straight up. Water pumps moves water up in straight way so the formula you mentioned will work there.

When one thing is in "Unified Circular Motion" then formula will change. This is very strange that whenever I tell that I want to make a big 'water wheel' then he straight goes to simple linear calculation and ignore if wheel is in 'Unified circular motion' or in 'Rotational Motion".

You forget water, etc. You just see that a train moving on roller coaster track has 400k or 500k weight. That's train own metal weight or passengers on it or water. We take it as total mass.

I want to know scientifically how much power I required, so if motor power will be close to traditional pumps then I will tell this management of amusement park with calculations. :)

Roller Coaster
Loop diameter: 107 meter
Radius: 53.5 meter
Circumference: 336 meter
50% circumference: 168 meter (I mean length between bottom and top)
Time: 1 round in 15 minutes
Time: 15 x 60 = 900 seconds
Mass: 863,228 kg (full train)

These are input parameters.

Now you don't even think about water, get into roller coaster and help me to find that how much power will be required from moving train down-top. I searched on Internet and found these kinds of questions from different schools, colleges, etc. So consider this as school question.

Kinetic Energy = K.E. (J) moving object from bottom to top
Mass: 849,428 kg (1 side train mass)
Velocity/Speed: .374 m/s (336/900)

K.E.: 1/2 x 849,428 x (.374x.374)
K.E: 59,249 Joule

So we need 59,249 Joule of energy to move this object, right?

Work (J) = K.E. x Height

Total circumference is 336 meter and distance of bottom-top (1 side) is 168 meter

Work (J) = (K.E.) 59,249 x (distance) 168 meter
Work (J) = 59,249 x 168 = 9,958,211 Joule

It means 9,958,211 Joule is required to move this mass at 168 meter.

Power: Work / time

1 round is completed in 900 seconds (15 minutes) but 50% round is completed in 450 seconds (7.5 minutes) so here time will be 450 seconds?

Power: 9,958,211 / 450 seconds = 22,129 watts (22.12 KW) excluding all frictions?

Am I making mistake, a 22KW motor can move this train?

Keeping in mind roller coaster train of this mass, please correct me where I am wrong?

Robert
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Re: Calculate require force for moving in circular motion

Postby zetreque on November 19th, 2014, 2:07 am 

newrobert:

I commend the effort and thinking put into trying to make your project more efficient.

I am not as technical to help you either, but perhaps if we re-evaluated your problem it might be easier for others to understand. Making it as specific and condensed as possible will help others understand to help you. This is what I gathered from browsing the post.

Problem: Determine most efficient way to lift water to a height of 250+ feet at a rate of (GPM).

Possible Solutions:
1. Upright waterwheel
2. Slanted waterwheel.

Theory:
Motors to turn a large wheel won't cost as much to operate as traditional water pumps.

Needed to answer problem: calculation to determine energy to operate said waterwheels to compare to traditional water pumps.



I can see where this might work out. I know that traditional water pumps can be expensive depending on the water quality. Replacing impellers all the time. Have you looked into VFD (variable frequency drive) motors? (although you did say it's a constant rate so VFD's might not be that much improvement) If you also had the materials just laying around anyway, and obviously space isn't an issue that improves the economics.

My first impression however is; I can't see a giant water wheel being that much more efficient than water pumps. Perhaps over a long time, or if you had some other sources of energy that contributed to it like horses walking around in a track pulling it, and if the ride never has to be torn down and rebuilt. Such a wheel would also have a maintenance cost. I suppose that's why you are seeking exact calculations to prove it. :)

It's an interesting problem, and would be neat if you could reclaim some of that energy from the water falling back down. If it's a water slide, that might be kind of hard to do though.

Another factor in determining the amount of energy to rotate your waterwheel is going to be the amount and size of "buckets". If you wanted to use a really low power motor it's going to be under least load when it empties a bucket, and possibly struggle under the most load when it fills a bucket unless you could time them perfectly.

May I suggest some google search terms? "ferris wheel power calculation" or like "ferris wheel power consumption" and combinations. Maybe call some other parks that have ferris wheels and make an inquiry.
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Re: Calculate require force for moving in circular motion

Postby CanadysPeak on November 19th, 2014, 10:57 pm 

You should listen to Marshall. Since you have put friction (inefficiency) aside for the moment, you have a conservative problem, that is, it is path independent. Thus, it doesn't matter what you use to do the lifting at 100 % efficiency, and it doesn't matter whether it is straight or circular. You may have some trouble with those ideas, so I urge you to do some experiments. I have built several large linear and circular machines. I have seen the equations work out. I have used a circular pump utilizing water cups probably several thousand times in my life.

There is no magic in engineering. It's all careful attention to detail.
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Re: Calculate require force for moving in circular motion

Postby newrobert on November 19th, 2014, 11:11 pm 

CanadysPeak » November 19th, 2014, 10:57 pm wrote:You should listen to Marshall. Since you have put friction (inefficiency) aside for the moment, you have a conservative problem, that is, it is path independent. Thus, it doesn't matter what you use to do the lifting at 100 % efficiency, and it doesn't matter whether it is straight or circular. You may have some trouble with those ideas, so I urge you to do some experiments. I have built several large linear and circular machines. I have seen the equations work out. I have used a circular pump utilizing water cups probably several thousand times in my life.

There is no magic in engineering. It's all careful attention to detail.


Thank you for posting and I also listened to Marshall.

1. I know there will be friction and inefficiency.
2. For a moment please ignore that water is loaded on train, you just assume that there is mass on train that is moving in circular path like roller coaster. Mass is equally distributed on train. Then what power will be required to calculate power.

Are my calculations above are correct?

Robert
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Re: Calculate require force for moving in circular motion

Postby Marshall on November 19th, 2014, 11:43 pm 

This part is wrong. The work done in moving something is not equal to the K.E. multiplied by the distance.
It is in this case equal to the force times the distance, the lifting force required to raise the thing, basically its weight.

==mistake==
Work (J) = K.E. x Height

Total circumference is 336 meter and distance of bottom-top (1 side) is 168 meter

Work (J) = (K.E.) 59,249 x (distance) 168 meter
Work (J) = 59,249 x 168 = 9,958,211 Joule

It means 9,958,211 Joule is required to move this mass at 168 meter.
==endquote==

also when you multiply a joules quantity by a distance in meters, the units have to be multiplied too
so the answer has to be in joule-meters (not a usual type of quantity)

basically you need a force quantity, in NEWTONS, which you multiply by distance. Learn to multiply units, learn that newton*meters = joule.
That is what a joule is, by definition in the metric system, a newtonmeter.

A kilogram weighs 10 Newtons (approx.) so to raise one kilogram up 168 meters takes this much work:

10 Newton x 168 meter = 1680 Newtonmeter = 1680 joule.
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Re: Calculate require force for moving in circular motion

Postby newrobert on November 20th, 2014, 2:24 am 

Thank you Marshall for replying to this post.

In fact if you can see my above post in which I did calculation with "Red" headings, I just made mistake in "red heading" but did correct calculation. I mean I also calculated correct Joule.

Work (J) = 59,249 x 168 = 9,958,211 Joule

Here I want to tell you some thing about myself. I am 40 years old and I read general physics 25 years before and now forget every thing. I am a student now. From last 2 weeks while working this project I searched over Net and tried to understand all this again. So I will appreciate if you can explain things in such way that leads to calculating power. Sorry if I am shown to be dumb and asking such things again and again which are basic things for professionals like you. :)

K.E (J) = 59,249 Joule
Work (J) = 59,249 x 168 = 9,958,211 Joule

Is this the next equation for my project?

1 round is completed in 15 minutes so half circle (bottom - top) will be completed in 7.5 minutes, will I take here 7.5 minutes?

Power = Work (J) / Time
Power = 9,958,211 / 450
Power: 22,129 watts

Motor inefficiency: 10% (enough or add more)
All frictions: 10% (enough or add more)
Total Power Added: 20%

Net Power: 22,129 x 4425.80 = 26,554 Watts

So roughly this train can be moved by a 26.5 KW motor at required speed? Or I miss some step?

Please help.


Robert
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Re: Calculate require force for moving in circular motion

Postby CanadysPeak on November 20th, 2014, 7:28 am 

newrobert » Wed Nov 19, 2014 11:11 pm wrote:
CanadysPeak » November 19th, 2014, 10:57 pm wrote:You should listen to Marshall. Since you have put friction (inefficiency) aside for the moment, you have a conservative problem, that is, it is path independent. Thus, it doesn't matter what you use to do the lifting at 100 % efficiency, and it doesn't matter whether it is straight or circular. You may have some trouble with those ideas, so I urge you to do some experiments. I have built several large linear and circular machines. I have seen the equations work out. I have used a circular pump utilizing water cups probably several thousand times in my life.

There is no magic in engineering. It's all careful attention to detail.


Thank you for posting and I also listened to Marshall.

1. I know there will be friction and inefficiency.
2. For a moment please ignore that water is loaded on train, you just assume that there is mass on train that is moving in circular path like roller coaster. Mass is equally distributed on train. Then what power will be required to calculate power.

Are my calculations above are correct?

Robert


For a circular motion, the mass is not the same. You have the wheel PLUS water going up and the wheel only coming down. The power will be calculated based on the rate at which you raise the water. At 100 % efficiency, the actual mechanism is insignificant. The amount of power is the same in all instances.
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Re: Calculate require force for moving in circular motion

Postby newrobert on November 22nd, 2014, 2:09 pm 

I understand that difference is that filled buckets are going up and empty buckets are coming down. So if roller coaster is moving clockwise then it means from right side empty buckets are coming down and from left side filled buckets are going up.

Left Side = Kinetic Energy (filled buckets moving up, around 300 ton weight)
Right Side = Potential Energy (empty buckets coming down, might be 10 ton weight of empty buckets only)

So I think a solution of this so how to balance this.

What if we can manage to put 100-200 ton weight on top-right side of roller coaster (from where empty buckets are coming down). For example one 20 ton concrete block is fixed in a steel cage (hanging with strong steel wires,just not let him fall) and there are heavy tyres (like plane) those can bear load of concrete and then these tyres are transferring potential energy (by of 20 ton concrete block) to roller coaster chain/belt by tyre friction? Just like many aumusement park rides are tyre driven (frictional drive).

So at left side there is weight of water and at top-right side heavy concrete blocks are trying to balance weight.

Will this balance something?

Robert
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Re: Calculate require force for moving in circular motion

Postby CanadysPeak on November 22nd, 2014, 3:12 pm 

newrobert » Sat Nov 22, 2014 2:09 pm wrote:I understand that difference is that filled buckets are going up and empty buckets are coming down. So if roller coaster is moving clockwise then it means from right side empty buckets are coming down and from left side filled buckets are going up.

Left Side = Kinetic Energy (filled buckets moving up, around 300 ton weight)
Right Side = Potential Energy (empty buckets coming down, might be 10 ton weight of empty buckets only)

So I think a solution of this so how to balance this.

What if we can manage to put 100-200 ton weight on top-right side of roller coaster (from where empty buckets are coming down). For example one 20 ton concrete block is fixed in a steel cage (hanging with strong steel wires,just not let him fall) and there are heavy tyres (like plane) those can bear load of concrete and then these tyres are transferring potential energy (by of 20 ton concrete block) to roller coaster chain/belt by tyre friction? Just like many aumusement park rides are tyre driven (frictional drive).

So at left side there is weight of water and at top-right side heavy concrete blocks are trying to balance weight.

Will this balance something?

Robert


Not unless you can talk somebody else in carrying those 200 ton weights up to the top for you. In order to balance, you would have to let them descend to the ground. Then you gotta get them back up.

The circular shape is a constant source for inventions claiming free work, but it cannot work. If you analyze this carefully, using vector calculus, you will see that the amount of net work you have to put in equals the amount of potential energy the water gains in going up, no matter what shape path.

Now, it turns out the water wheel design is pretty efficient, as pumps go. That's why it's been around as long as it has. It is still used throughout much of the world even today. It is fairly efficient, simple, economical, and doesn't require a developed infrastructure.

But, you can't get free work from it.
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Re: Calculate require force for moving in circular motion

Postby Marshall on November 22nd, 2014, 3:14 pm 

CanadysPeak » Thu Nov 20, 2014 4:28 am wrote:For a circular motion, the mass is not the same. You have the wheel PLUS water going up and the wheel only coming down. The power will be calculated based on the rate at which you raise the water. At 100 % efficiency, the actual mechanism is insignificant. The amount of power is the same in all instances.


Robert did you know that for many years people tried to devise "perpetual motion machines" to provide FREE POWER? This possibility is denied by the law of conservation of energy

Stripped to bare essentials your idea is that it would take less kW to raise a given amount of water on a CURVED path than to raise it on a STRAIGHT path.

focus on the water and ignore the machinery. You have the water going up along a semicircular arc. if this took less kW than pumping it straight then we could get FREE POWER to meet all our energy needs.

A piston pump can be run BACKWARDS to extract work from water coming DOWN a pipe, operating just in the reverse of the pump using work to push the the piston and force the water UP the pipe. Ignoring friction losses and reversing the cycle you get exactly as much power, letting the water come down, and drive the piston, as it would take to push the piston and force the water up.

Suppose your idea worked and it takes 2 kW to force water (at some rate) straight up to a reservoir, and only 1 kW to force it up along a curved semicircle path.

Then we could USE ONE KW to get the water up to the reservoir on a curve, and run the other pump back and generate TWO KW by letting the water come down straight and running the pump cycle in reverse.

So the power company could SELL the extra KW and generate power perpetually without using any fuel.

This would be defrauding Nature because the energy would be coming out of nowhere, free. Nothing would be used up.

But this doesn't happen. No matter what machinery you use it takes the same energy to raise some water on a curved path as it does on a straight path.

(ignoring mechanical inefficiency, the extra friction of using a longer pipe, and similar details)
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Re: Calculate require force for moving in circular motion

Postby newrobert on November 23rd, 2014, 4:50 am 

Dear Marshall / CanadysPeak,

Thank you for replying to this post for my better understanding.

My calculation in this thread
In this thread as I calculated and you people did not find any mistake on these calculations i.e. this method of calculation is correct.

Mass moved up in 1 round: 863,228 kg
Mass moved up of one side train: 849,428 kg (water weight + buckets weight)
Mass equal to gallon: 441,499 gallon
50% of total water: 220,750 gallon (we calculated one side here @ 50% time i.e. 7.5 minutes)

Water lifted up: 220,750 gallon (equals to 849,428 kg mass)
Water lifting Time: 7.5 minutes (50% of 1 round time, from bottom-top)
Water lifting height: 107 meter (net height of water lifted)
Per Minute Water Lifted: 220,750 gallon / 7.5 minutes = 29,433 GPM
Motor Power: 26.50 KW (excluding frictions) or we can consider this 50 KW

So as per calculations here 26.50 KW motor is rotating this roller coaster and in 7.5 minutes 220,750 gallon water will be lifted at 107 meters in 7.5 minutes @ 29,433 GPM.

Simple Calculations
If same mass is lifted at same height then here are simple calculations.

Mass: 849,428 kg
Height: 107 meters
Time: 450 seconds (7.5 minutes x 60)
Speed: 107/450 = .24 m/s

Force: Mass x Gravity = 849,428 x 9.81 = 8,332,889 N
Work: Force x Height = 8,332,889 x 107 = 891,619,089 Joule
Power: Work / Time = 891,619,089 / 450 = 1,981,376 Watts (1,981 KW)

Water Pumping Calculations
Here are water pumping formula that I found on Internet.

Water Flow: 29,433 GPM
Height: 107 meter
Height: 351 Feet
Power Required: (Feet x gpm)/3960 = (351 x 29,433)/3960 = 2,609 HP = (1,945 KW)

1. What's wrong in our calculation?
2. I did not know anything about "perpetual motion machines" but I will now I will know more about these.
3. You understand correctly that I think that "due to circular motion less energy can be consumed to lift same amount of water as compare to traditional water". There is no "centripetal force" involved in simple water pump and this force can help in saving some energy.
4. I want to understand this scientifically. To make a water wheel or roller coaster how much energy will be required.

Marshall / CanadysPeak, can you please give your comments on Point # 1 + 3 + 4.

I am making a diagram of what's in my mind regarding recent idea of adding some potential energy at right side of roller coaster from where empty water buckets are coming down.

Robert
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Re: Calculate require force for moving in circular motion

Postby CanadysPeak on November 23rd, 2014, 8:05 am 

newrobert » Sun Nov 23, 2014 4:50 am wrote:Dear Marshall / CanadysPeak,

Thank you for replying to this post for my better understanding.

My calculation in this thread
In this thread as I calculated and you people did not find any mistake on these calculations i.e. this method of calculation is correct.

Mass moved up in 1 round: 863,228 kg
Mass moved up of one side train: 849,428 kg (water weight + buckets weight)
Mass equal to gallon: 441,499 gallon
50% of total water: 220,750 gallon (we calculated one side here @ 50% time i.e. 7.5 minutes)

Water lifted up: 220,750 gallon (equals to 849,428 kg mass)
Water lifting Time: 7.5 minutes (50% of 1 round time, from bottom-top)
Water lifting height: 107 meter (net height of water lifted)
Per Minute Water Lifted: 220,750 gallon / 7.5 minutes = 29,433 GPM
Motor Power: 26.50 KW (excluding frictions) or we can consider this 50 KW

So as per calculations here 26.50 KW motor is rotating this roller coaster and in 7.5 minutes 220,750 gallon water will be lifted at 107 meters in 7.5 minutes @ 29,433 GPM.

Simple Calculations
If same mass is lifted at same height then here are simple calculations.

Mass: 849,428 kg
Height: 107 meters
Time: 450 seconds (7.5 minutes x 60)
Speed: 107/450 = .24 m/s

Force: Mass x Gravity = 849,428 x 9.81 = 8,332,889 N
Work: Force x Height = 8,332,889 x 107 = 891,619,089 Joule
Power: Work / Time = 891,619,089 / 450 = 1,981,376 Watts (1,981 KW)

Water Pumping Calculations
Here are water pumping formula that I found on Internet.

Water Flow: 29,433 GPM
Height: 107 meter
Height: 351 Feet
Power Required: (Feet x gpm)/3960 = (351 x 29,433)/3960 = 2,609 HP = (1,945 KW)

1. What's wrong in our calculation?
2. I did not know anything about "perpetual motion machines" but I will now I will know more about these.
3. You understand correctly that I think that "due to circular motion less energy can be consumed to lift same amount of water as compare to traditional water". There is no "centripetal force" involved in simple water pump and this force can help in saving some energy.
4. I want to understand this scientifically. To make a water wheel or roller coaster how much energy will be required.

Marshall / CanadysPeak, can you please give your comments on Point # 1 + 3 + 4.

I am making a diagram of what's in my mind regarding recent idea of adding some potential energy at right side of roller coaster from where empty water buckets are coming down.

Robert


OK, this time I will state my points more forcefully, not to criticize you, but because you are not listening:
1. Your calculations are unclear. They cannot be checked.
2. There is a standard manner, accepted worldwide, of doing engineering calculations. After 50+ years, I am unwilling to do those calculations otherwise.
3. You appear to be using simple arithmetic to do vector calculus. That is simply wrong, unless you know which assumptions and shortcuts you can take.
4. I will tell you how an engineer would look at this problem.
Assume that friction is negligible.
Assume you have a water wheel of diameter d.
Assume that the water wheel has n buckets, each carrying a weight of water w.

In one rotation, the wheel lifts nw of water to a height of d. Since work is the dot product of force and displacement, the work here is W = nwd.

Power is the rate of doing work, or W/t. If the time of one rotation is T, the power required here is
P = nwd/T.

Note that the weight of the wheel does not enter into this calculation because we assumed friction to be negligible.

Note that I used one revolution as my calculation basis. This is standard practice; you should also do this. If you do not understand why the use of one revolution is standard, that's OK. Do it anyway.

Note that I solved the problem without using any numeric values, only literals. This is also a standard way of doing engineering calculations. Someone else can quickly glance at my formulas and see if I have made a mistake. Once the problem is solved thusly, plugging in the actual numbers is trivial.
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Re: Calculate require force for moving in circular motion

Postby Marshall on November 23rd, 2014, 12:32 pm 

Robert, I pointed out your mistake back on Wednesday. 19 November. I quoted you and said "mistake"
Here I show this, and highlight the wrong formula where you say
"work = K.E. x distance".
As far as I know you are the only person in the history of mankind that has ever used that formula. It is totally wrong. and your number you got by it (9,958,211 joules) is totally wrong.
That means that the kW number you got from it, namely 26.50 kW is totally wrong.

It is off by a factor of more than 70. Your number is roughly 70 times too small because you used a no-good formula.

I am disappointed in you that you did not drop this back on 19 November when I took the trouble to point out your mistake.


Marshall » Wed Nov 19, 2014 8:43 pm wrote:This part is wrong. The work done in moving something is not equal to the K.E. multiplied by the distance.
...
==mistake==
Work (J) = K.E. x Height

Total circumference is 336 meter and distance of bottom-top (1 side) is 168 meter

Work (J) = (K.E.) 59,249 x (distance) 168 meter
Work (J) = 59,249 x 168 = 9,958,211 Joule

It means 9,958,211 Joule is required to move this mass at 168 meter.
==endquote==
...
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Re: Calculate require force for moving in circular motion

Postby Marshall on November 23rd, 2014, 12:56 pm 

Robert, after I pointed out your mistake on Wednesday you did not listen but you instead immediately REPEATED the mistake in the next post:

newrobert » Wed Nov 19, 2014 11:24 pm wrote:...I mean I also calculated correct Joule.

Work (J) = 59,249 x 168 = 9,958,211 Joule

...:)

K.E (J) = 59,249 Joule
Work (J) = 59,249 x 168 = 9,958,211 Joule


This number is over 70 times too small and the formula "work = K.E. x distance" is worthless.
You then continue and you get the absurd answer for the power of some 22 kW, which you adjust to 26.55 kW

Power = Work (J) / Time
Power = 9,958,211 / 450
Power: 22,129 watts

Motor inefficiency: 10% (enough or add more)
All frictions: 10% (enough or add more)
Total Power Added: 20%

Net Power: 22,129 x 4425.80 = 26,554 Watts


My point is that you were not RESPONSIVE. People took the trouble to point out mistake and you did not listen. You just went on declaring the same false stuff over and over.

Our normal policy is to BAN folks who totally do not listen and are UNRESPONSIVE. That is, who act DEAF.

Also this thread is most likely destined for trash because it repeatedly involves the mistaken idea that work=KE x distance
and it does not involve LEARNING. The main purpose of this kind of discussion is that you would learn something from it and not keep repeating the same mistaken answer.

Now it is Sunday, four days later. I must ask you please to stop.

Don't bother to apologize. Just stop posting on this thread.
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Re: Calculate require force for moving in circular motion

Postby newrobert on November 24th, 2014, 3:14 am 

This is impossible that I don't reply to your recent comments and in fact I don't need to be apologetic.

And this is my last comment in this FORUM because I assume that this forum is for people having high engineering and physics background.

1. First of all I appreciate that you and CanadaysPeak spend time and tried to help me.
2. But I don't like the word use "UNRESPONSIVE" and "DEAF" that's totally irrelevant.
3. In fact I did not get much results by spending time on this thread although I tired by best to spend time here.

REASON:
The reason I can understand why I could not get something from this thread in spite of spending much time is "Since you are very experienced person in engineering" and you talked too much about theory.

I mention in one of my early post (18 November) that I am "not professional" and I just need some calculations.

On November 18, I posted this again.

"Since I don't have any physics background but during last couple of days I did some research and use to with some basic terms and I set formulas in Excel."

I mentioned twice that I am just looking for some equations and calculations to find power and I don't have physics background. You tried to help me but I don't know how you calculated.

The way I calculated is easy to know how calculations are being done but your figures I did not understand how these are calculated.

I think, we have saved much time and this thread has been idled a long time, if you have given your input in equations like I did that's easy to understand.

You pointed out mistake, that's fine but that would have been great if you mention correct figure in same way, then I would not have repeated it again and again.

I think if you would have been teaching in class room you would have been the only professor in the history of mankind that teaches every thing in narration not in equations which are also necessary for learning.

You mention this "You appear to be using simple arithmetic to do vector calculus. That is simply wrong, unless you know which assumptions and shortcuts you can take."

Now first I have to search to learn what's "vector calculus" and then understand what you are saying.

Its was very easy to understand if you have solved this in equations so I can modify wheel speed, buckets weight to see what is acceptable.

I have 20+ years experience in IT and I am not professional of physics and engineering so definitely if some professional from Physics or Engineering did some thing (according to you) like me to learn some thing of IT field then he will be DEAF for that IT professional.

There is a certain way of teaching something to beginners and your method will be working for people who already have good engineering and physics background.

We must should respect of professionals like you and I am thankful to all people for spending time on this thread but there is much difference in your teaching and my learning method. Your mostly talk in narrations and I want equations and calculations.

Again I did not like and I am very disappointed for the word "UNRESPONSIVE" and "DEAF" used in this thread.

Good Bye
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Re: Calculate require force for moving in circular motion

Postby zetreque on November 24th, 2014, 3:56 am 

I thought it was kinda lame that everyone just glossed over my post, and you didn't even acknowledging it when I was trying to understand your problem, and I did notice that you seemed to suddenly be talking about perpetual motion which was a bit odd.

I mean how can you be talking about a weight magically appearing at the top of your wheel to send down and balance out the water coming up?

I guess if that doesn't makes sense to you why you can't continuously send down a magical weight appearing, then there is no way you could comprehend my suggestions.


So I think a solution of this so how to balance this.

What if we can manage to put 100-200 ton weight on top-right side of roller coaster (from where empty buckets are coming down). For example one 20 ton concrete block is fixed in a steel cage (hanging with strong steel wires,just not let him fall) and there are heavy tyres (like plane) those can bear load of concrete and then these tyres are transferring potential energy (by of 20 ton concrete block) to roller coaster chain/belt by tyre friction? Just like many aumusement park rides are tyre driven (frictional drive).


If you aren't' going to acknowledge my comments, why should I try to make sense of that strange idea?

Too bad this thread is headed to the trash. I thought it had potential (bad pun).
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Re: Calculate require force for moving in circular motion

Postby CanadysPeak on November 24th, 2014, 7:25 am 

I provided the equation for power as requested.

I introduced the idea of perpetual motion, since this is a classic "over unity" machine.

I did all this in normal sized, black font.

Good day, and good riddance.
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Re: Calculate require force for moving in circular motion

Postby TheVat on November 24th, 2014, 12:53 pm 

"...had potential" <g>

I was going to remark how "cranky" this thread was, and how the chat kept going in circles.
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