Calculating Torque and Motor Power to rotate a Ferris wheel

[shaks]

Hi Guys,

I am pleased to join this forum to learn something from experienced guys and thanks in advance for helping me to understand physics.

I am trying to design a Ferris wheel with maximum SPEED and maximum PASSENGER capacity. I want to know formula to calculate MOTOR POWER IN KW to rotate that Ferris wheel. Here are general specs of the wheel.

Diameter: 106 Meter
Radius: 53

Wheel total weight: 1000 Ton
Exclude Wheel standing structure: 30%
Rotating Wheel weight: 700 ton (Everything including spindle, hub, bearing, spoke cables, rim, cabin, riders weight, etc)
Minutes Per Round: 10 (1 round completed in 10 minutes)
Round Per Minute: 0.10
Wheel Speed: .55 meter per second

For this calculation if only rim, spoke cables, cabin/rider weight is required then you can make rough estimate what might be the weight of spindle, hub, bearing, etc. which are creating force. You can make an assumption and then calculate on that assumption. I will fix formula in Excel as per given here and then adjust weight.

I want to know
1. What is total torque generated by wheel (Newton Meters / Joules)
2. What power motor in HP or KW I needed to rotate this wheel

Once I get formula then I can set that formula in Excel and try to change speed to find good results.

Please help me.

Thanks

Shaks

[studiot]

Hopefully this is not a real wheel for real passengers, but only an exercise?

 

In which case why is it not in homework help?

[shaks]

Hopefully this is not a real wheel for real passengers, but only an exercise?

 

In which case why is it not in homework help?

 

Sorry man, please move there.

[studiot]

First thing is to get your units straight.

 

You have quoted distances in metric but weight in imperial.

 

Do you know the difference between weight and mass, it is important for rotating bodies.

 

So do you want to work in metric or imperial?

 

Second have you made any effort yourself to solve this question, if so please show where you are stuck.

[shaks]

Thank you for replying and helping me here.

 

I am not professional scientist and just trying to learn what is need to calculate in relation to ferris wheel.

 

Total Mass: 600,000 kg (rotating weight i.e. rim, spoke cables, cabin weight, riders weight only)
Spoke cable mass: 180,000 kg
Rim/rider/cabin Mass: 420,000 kg
Radius: 53 meter
Speed: 0.55 m/s

Interia-Rim(I) = 1,179,780,000 kg/m2 (420,000*53*53) (rim weight * radius square)
Interia-Spoke(I) = 168,540,000 kg/m2 (1/3*(180,000*53*53)) (1/3 * (spoke cables weight * radius square))
Total Interia of wheel (I) = 1,348,320,000 kg/m2 (1,179,780,000 + 168,540,000) (Interia Rim + Interia Spoke)

Angular Speed (w) = 0.01044 rad/s (.554/53) (speed / radius)

Mass of wheel (m) = 600,000 kg (420,000 + 180,000) (rim weight + spoke cables weight)

Angular Momentum (L) = 14,082,304 kg m2/s (1,348,320,000 * 0.01044) (Wheel Interia * Angular Speed)

KE-Joules (Rotational) = 73,540 NM (1/2 * (1,348,320,000 * (0.01044*0.01044) )) (1/2 * (Wheel Interia * Angular Speed Square))
KE-Joules (Usual) = 91,925 NM (1/2 * (600,000 * .554 * .554)) (1/2 * (Wheel total mass * wheel speed square)
KE-Total (Rot+Usual) = 165,465 NM (73,540 + 91,925) (Kinetic energy rotational + kinetic energy usual)

Force = 16,873 kg meter (165,465 / 9.80665) (total kinetic energy / gravity)
Newton = 3,122 N (165,465 / 53) (Total kinetic energy / radius)

@studiot, please guide me where I am wrong.

Shaks

Hi,

 

I tried to make calculations about ferris wheel on basis of one example of bicycle. Lecture 22, "Rotational Kinetic Energy" starting from page 2. Please see here http://rockpile.phys.virginia.edu/arc00/arch22.pdf

 

I am sure, I am not calculating correct, please guide me where I am wrong.

 

Thank you

 

Shaks

[Enthalpy]

Looks good to me up to the rotational energy.

 

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The "usual" kinetic energy is just an alternative way to compute the same rotational energy. Your estimate is less accurate because it would put all the spoke mass at the rim, hence attribute it too much speed. Sicne the spokes are rather light, you get approximately the same result, which is a way to double-check it.

 

The kinetic energy is properly computed in your line inertia-spoke. The factor 1/3 results from the mean value of the squared speed: it's an integral of the squared speed, taken from zero radius hence zero speed up to the maximum radius.

 

Both approaches, by the angular speed and the linear speed, should have given the same result (and they do approximately here). There is no reason to add both results.

 

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No torque nor force can be deduced can be deduced from the kinetic energy the way you did, which apparently comes from falling objects.

 

One need for torque is to accelerate the wheel up to 0.55m/s or 0.1turn/hour. You have to decide how much time you accept for the acceleration. The way passengers embark may decide that. The time to achieve the angular speed (rad/s) gives an angular acceleration (rad/s²) which, multiplied by the inertia (kg*m²), tells a torque (N*m). The torque multiplied by the angular speed (rad/s) is a mechanical power (W) obtained after some losses.

 

Just like the kinetic energy can be computed, approximately or exactly, through the angula speed (rad/s) or the linear speed (m/s), one could also compute the force (N) needed to accelerate (m/s²) the wheel (kg) to a certain speed (for instance 0.55m/s) within a desired time (s). This force multiplied by the speed is a mechanical power (W).

 

Both methods (torque times angular speed, versus force times linear speed) should give the same power, because the radius multiplies the force to define the torque, but divides the linear speed to define the angular speed.

 

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The other need for torque is because the wheel isn't balanced. You have to estimate by how much it's unbalanced through its construction and assembling (not obvious), and more easily, by how much the passengers unbalance the wheel. This depends on how the passengers board and unboard; if this happens to be the main contribution, and if this contribution is considered big, you might put constraints on the wheel operation. I'd try to avoid such constraints.

[studiot]

Enthalpy has offered some pretty good comments, to which I would add that

 

 

Force = 16,873 kg meter (165,465 / 9.80665) (total kinetic energy / gravity)

 

Force is measured in Newtons, not kg-metres.

 

Are you thinking of rim or centre drive as it will affect many aspects of the design.

 

In particular Enthalpy has noted the concentration of the mass along the rim, which is fine for rim drive, but the structural requirments for centre drive will require a more massive framework for the 'spokes'.

 

Also the method of drive will be different for centre and rim drive.