Force required to pull a cruise liner or cargo ship

[KuriousKid]

TL;DR Summary

Force required to pull a cargo ship using a rope, connected to Kite.

I was wondering how much force is required to pull the Cargo ship using a Kite as shown in below image and specifications given in table? This picture is taken from a seawing aka airseas website. I do not know the water / air / skin resistance values or coefficients for the cargo.

All I know is F = ma

Tonnage	235,579 GT [Gross Ton]
Length	400 m (1,312 ft 4 in)
Beam	61.5 m (202 ft)
Draught	17 m (55 ft 9 in)

 

[Baluncore]

The vessel will be turned by the kite, unless you attach the kite to the centre-of-drag of the vessel.

The force required will depend on the speed required through the water, and on the wind direction and strength.

 

[anorlunda]

The force required will depend on the speed required through the water, and on the wind direction and strength.

Plus the current direction and strength.

@KuriousKid ,
I believe that this idea has been explored before, but not as the sole means of propulsion, but rather as a way to use less fuel. In other words, both kite and engine at the same time.

But the kite arrangement must compete with conventional masts and sails. Remember that if the kite falls into the water, the ship will run it over. Conventional masts and sails don't have that specific problem.

 

[KuriousKid]

@Baluncore @anorlunda Thank you for your insights.
I Understand the issue of just using Kite. But I'm trying to understand (only) how we can calculate the force required for complete Cargo to be pulled only by Kite and no Engine force (ignoring the issues comes with it).

For the sake of calculation we can consider normal speed of Cargo around 25 knots and maybe we can ignore air resistance too as it will be almost around 800 times less than water resistance, considering the density difference [but again it depends on how tall is ship and so on].

 

[Baluncore]

For the sake of calculation we can consider normal speed of Cargo around 25 knots ...

What is the normal operating power of the engines in a similar ship, at that speed?

 

[jbriggs444]

@Baluncore @anorlunda Thank you for your insights.
I Understand the issue of just using Kite. But I'm trying to understand (only) how we can calculate the force required for complete Cargo to be pulled only by Kite and no Engine force (ignoring the issues comes with it).

For the sake of calculation we can consider normal speed of Cargo around 25 knots and maybe we can ignore air resistance too as it will be almost around 800 times less than water resistance, considering the density difference [but again it depends on how tall is ship and so on].

To be clear, we have a tethered air foil? With a really good lift to drag ratio so that even with 25 knots of relative headwind, we can still eke out positive thrust from, let us say, 15 knots of absolute crosswind?

Which means that not only must the sail have a good lift to drag ratio, the keel must also have a good lift to drag ratio.

I am not saying that it is impossible. Sailboats manage the trick with conventional sails. But the pictures on the web site depict a ship being pulled by the kite. That mode will not work unless you have a tailwind that is faster than the ship speed.

The ship class we seem to be talking about is the Emma Maersk. 108,920 horsepower (about 81 megawatts).

At 25 knots (about 13 meters per second), that translates to a force of $\frac{81}{13}$ or about 6.2 megaNewtons. A metric ton is about 10,000 Newtons. So that comes to about 620 metric tons. The claim I see is that this tethered airfoil generates about 100 tons of "traction" by flying in a back and forth pattern to sweep out an area greater than its nominal size.

This idealized performance would result in a fuel savings of about 16 percent -- as long as one has a sufficiently speedy tail wind. The web site seems to be short on detailed specifications. [Better than 16 percent if we figure in the propellor. The propellor will not be 100% efficient. The tether will be]

Note that $F=ma$ deals with acceleration. We are talking about constant speed cruising here. It is all about thrust and drag. Mainly hull form drag, I expect, but I am not a naval architect.

Edit: One more thing I gleaned from the web site...

The wing zips around at "over 100 km/h". Which would be about 50 knots. If we assume a lift to drag ratio of about 10 to 1, this means that the wing needs at least five knots of relative wind to function at peak efficiency. So we need 30 knots of tailwind to get peak performance on our 25 knot container ship.

On the Beaufort scale, that is:

7​

28-33

Near Gale

Sea heaps up, waves 13-19 ft, white foam streaks off breakers

Whole trees moving, resistance felt walking against wind

 

[KuriousKid]

@jbriggs444
I believe if you go higher than 300 m, you get considerable higher wind speeds than what is it at sea level. So we cannot relate sea level wind speed with speed at 300m above sea level. But you are right at Cruise speed... so we may get 5-6 m wind speed difference. So it might be impossible to get speed of 25 knots by only wind speed. So that way our calculation would be for around 5-6 m/s cruise speed.

The ship engine capacity of 81 Mega Watt might not be used in total just for propulsion. Maybe part of it is used for other purposes. Not sure. But this pulling force will give us right answer to the question. How much engine power is used to push the cargo.

I also believe the pulling is much more efficient than pushing by propeller from under the water, because LOT of power is lost in friction and not all power is translated into moving the Cargo ship. Whereas in pulling condition, all power is used in moving the Cargo ship minus drag forces.

 

[Filip Larsen]

I believe if you go higher than 300 m, you get considerable higher wind speeds than what is it at sea level.

Even at moderate heights an increase can be expected [1]. Since the kite uses lift to extract power from the wind I guess it is fair to consider the available off-shore wind power density as a function of height and wind speed at a reference height [2]. According to [3] it seems one can expect around 10 times more wind power at 100m compared to 10m.

[1] https://en.wikipedia.org/wiki/Log_wind_profile
[2] https://en.wikipedia.org/wiki/Wind_profile_power_law#Wind_power_density
[3] http://www.inference.org.uk/withouthotair/cB/page_266.shtml

 

[Baluncore]

Once the vessel has accelerated, later that day, all the available main engine power may not be needed while cruising. Let's assume 50%, so 40 MW goes to keep it moving. Assume the propeller is 50% efficient, that gives an estimate of vessel drag, from water and air, of 20 MW.

Work done is force * distance; so we can estimate the total drag forces from the 40 MW power and the speed of 25 knots.
First, convert to SI units. Velocity, 25 knots = 12.86 m/s.
Then compute the force, 40 MW / 12.86 m/s = 3.11 MN.

The force required to pull a cruise liner or cargo ship will be the same as the force needed to push it. My initial estimate is 3 million Newtons.

Big diesel propulsion engines are directly coupled to the propeller shaft.
Electrical power usually comes from two or more separate diesel generator sets.

 

[KuriousKid]

@Baluncore
I think we are going in reverse direction (doing reverse engineering) in calculating how much force is required based on our knowledge/assumptions of losses and total power available.

My intention was to calculate the minimum power required to pull the Cargo ship using the tethered rope at X m/s. I'm not even interested to know how much thrust or pull force that Kite can exert on ship [They say it can save 20-30% fuel, that's based on small vessel size, which doesn't apply to this big sized vessel we are talking].

What I'm trying it to understand how to calculate the pulling force required to move Ship from Point A to Point B with speed of 10 m/s or so. We have Mass, We have Speed. But I'm not sure about the drag force from the water / skin effect and how we can take that into account to calculate the pulling force.

Drag force = 1/2 m V^2 Area Cd = 0.5 * 235579000 * 10 * 10 * 61 * 17 * 0.04 = 488590846000 N [I hope I'm wrong somewhere]

F = m a = 235579000 * ? [We want to pull it with 10 m/s speed constantly], so not sure how to solve this.

F + Drag Force = Total force required.

 

[Baluncore]

I was wondering how much force is required to pull the Cargo ship using a Kite as shown in below image and specifications given in table?

I have not reverse engineered a kite powered ship being "pulled". I have estimated the total of all drag forces from the power it normally takes to drive a real ship, to keep the ship moving, once it is moving.

What I'm trying it to understand how to calculate the pulling force required to move Ship from Point A to Point B with speed of 10 m/s or so. We have Mass, We have Speed.

Point A to point B is irrelevant to the force, you must apply force for the whole voyage. You only need to study the force and the energy needed to keep the ship going for one second. Power in watts is the rate of flow of energy in joules per second.
As far as acceleration of the ship's mass is concerned, that is irrelevant. But the mass of the ship will displace the same mass of water, a volume of water. The dimensions and shape of that volume of hull will determine the wetted area and surface drag. The length of the hull will decide maximum hull speed due to wave generation, but like a real vessel, you will have to reduce that speed to get improved fuel efficiency.

If you think you can work out the optimum hull dimensions and calculate the drag from your hydrodynamic model, then go ahead. You could then compare your result with my quicker estimate.

 

[tech99]

The size of kite seems impracticably large. For instance, the wind jammer Herzogin Cecilie of 3242 tons required 3530 m2 of sail area. Scale that up to a modern cargo ship and the requirements for the kite are formidable.

 

[KuriousKid]

I was able to find Drag Coefficient for Cargo ship is about 0.004 and the formula is wrong above as we don't have to include the mass of ship for calculation of drag force. So it is like this

Drag Force = 0.5 * Cd * A * V^2 * Water density = 0.5 * 0.004 * 61 * 17 10 * 10 * 1.2 = 248.88 N

Am I still missing something?

 

[Baluncore]

Am I still missing something?

Yes, three orders of magnitude.
A density of 1.2 = 1200 kg/m3

 

[PeroK]

I was able to find Drag Coefficient for Cargo ship is about 0.004 and the formula is wrong above as we don't have to include the mass of ship for calculation of drag force. So it is like this

Drag Force = 0.5 * Cd * A * V^2 * Water density = 0.5 * 0.004 * 61 * 17 10 * 10 * 1.2 = 248.88 N

Am I still missing something?

250 Newtons to pull a cargo ship! Really?

 

[Baluncore]

A drag coefficient of 0.004 applies to the wetted area of a ship's hull, not to the cross-section.
Length 400 m, Beam 61.5 m, Draught 17 m.
Wetted area = 400 * ( 17 + 61.5 + 17 ) = 38200 m2

Drag = 0.5 * 0.004 * 38200 * 100 * 1200 = 9.168 MN.

 

[KuriousKid]

@Baluncore You nailed it. Thank you. ​

Wetted Surface Area Formula​

The following formula is used to calculate the Wetted Surface Area.

WSA = WLL * (B+D)

Variables:

• WSA is the Wetted Surface Area (ft^2)
• WLL is the water line length (ft)
• B is the beam (ft)
• D is the draft (ft)

To calculate the wetted surface area, multiply the water line length by the sum of the beam and draft.This makes Wetted area = 400 * ( 17 + 61.5 ) = 31400 m2

So

Drag = 0.5 * 0.004 * 31400 * 100 * 1200 = 7536000 N = 7.5 MN

This means I can use about 750 N force to move Cargo liner at 0.01 m/s speed considering there is no other drag force acting on ship.

Now I have another question in this case, in case above calculations are correct.

Let's say our ship is cruising at speed of 10 m/s in sea. To maintain this speed, do we have to continuously provide same force or only part of it?

e.g. If at 10 m/s ship's engine is turned off, that means it will keep going far before stopping completely. May be 1000 meter. Which means Every 100 meter it looses 1 m/s speed. If this is the case, then can we supply only 75360 N force [this force can pull ship with 1 m/s speed], to keep it running at 10 m/s speed?

 

[Baluncore]

Let's say our ship is cruising at speed of 10 m/s in sea. To maintain this speed, do we have to continuously provide same force or only part of it?

The drag force is operating while ever the ship is moving. You must maintain the force to maintain the velocity. If you turn off the energy supply to the propeller, the ship will begin to decelerate. Remember that the drag is proportional to the square of the velocity, so in theory, the ship will never quite stop.

 

[KuriousKid]

I didn't understand this sentence "Remember that the drag is proportional to the square of the velocity, so in theory, the ship will never quite stop." [Never stop?]

 

[jbriggs444]

I didn't understand this sentence "Remember that the drag is proportional to the square of the velocity, so in theory, the ship will never quite stop." [Never stop?]

The mathematically correct description of this is with a differential equation. Let us use something simpler. An approximation which will lead to the same conclusion. We will get there...

Start with the fact that drag is proportional to the square of velocity. So $F=kv^2$ for some contstant k.

The ships mass is fixed. Call it m. If drag is the only force operating then the ships acceleration $a=\frac{kv^2}{m}$.

Suppose that the ship is going at $v=1$ meters per second. How long that it will take to slow down to $v=0.1$ meters per second?

During the time interval from when $v = v_0=1$ to when $v = v_1=0.1$, the deceleration force will be at most $F=k{v_0}^2$.

So the deceleration will be at most $\frac{F}{m} = \frac{k}{m}{v_0}^2$.

This means that the elapsed time between $v=v_0$ and $v=v_1$ must be at least $\frac{v_0-v_1}{k{v_0}^2/m}$. Since $v_1 = v_0/10$, this simplifies to $\frac{0.9m}{kv_0}$.

Now repeat the analysis for the interval between $v=v_1=0.1$ and $v=v_2 = 0.01$. If you follow through the algebraic steps, you should end up with $\frac{0.9m}{kv_1}$.

Each of these guaranteed minimum time intervals is getting ten times larger than the one before. There are infinitely many of them. The series sum does not converge. The conclusion is that there is no time at which the ship stops moving.

Even if drag were linear, the ship would still never stop moving. In that case, each element in the series sum would be constant and the series sum would diverge.

One might ask if there is a goal line beyond which the ship will never progress. In the case of linear drag, the answer is yes, there is a goal line that the ship will never coast across. It would keep approaching the goal, perhaps getting asymptotically close, but will never make it there, not even when given infinite time. [If you work through it, the series sum for distance traversed is geometric and can have a finite sum despite having infinitely many non-zero terms]

In the case of quadratic drag, I believe that the answer is no, there is no goal line that the ship will never coast across, though it may take a very long time. I think that the series for this one is the harmonic series. That one goes infinite. But the sum is logarithmic in the number of terms and the time elapsed for each term increases exponentially, so it takes a long time to get to a big total. [Try throwing a ping pong ball a light year]

 

[KuriousKid]

I'm trying to compare this with coasting with Car or Bicycle on road now.
When I ride my bicycle, I apply 40 lb equivalent force to begin with. Once I get to a speed of 20 miles / hr and if the road is plane, to maintain the 20 miles/hr speed, I only have to apply a force of 2 lb now. Although the peddling RPM is higher at this time.

Here the energy supplied is almost same as I supplied it at the beginning but it is lot more easier to supply 2 lb force than 40 lb force.

So isn't it valid for Cargo ship too? When it is travelling at higher speed, to maintain that speed, we may have to supply 10 times less thrust force than to get it started. Energy supplied is same but thrust force is much lower than at start period. Am I correct in understanding or analyzing this now?

 

[PeroK]

I'm trying to compare this with coasting with Car or Bicycle on road now.
When I ride my bicycle, I apply 40 lb equivalent force to begin with. Once I get to a speed of 20 miles / hr and if the road is plane, to maintain the 20 miles/hr speed, I only have to apply a force of 2 lb now. Although the peddling RPM is higher at this time.

Here the energy supplied is almost same as I supplied it at the beginning but it is lot more easier to supply 2 lb force than 40 lb force.

So isn't it valid for Cargo ship too? When it is travelling at higher speed, to maintain that speed, we may have to supply 10 times less thrust force than to get it started. Energy supplied is same but thrust force is much lower than at start period. Am I correct in understanding or analyzing this now?

A cargo ship doesn't run on wheels. And water gives a lot more resistance to motion than air.

 

[Baluncore]

This includes ships and bicycles.
https://en.wikipedia.org/wiki/Energy_efficiency_in_transport

 

[snorkack]

I'm trying to compare this with coasting with Car or Bicycle on road now.
When I ride my bicycle, I apply 40 lb equivalent force to begin with. Once I get to a speed of 20 miles / hr and if the road is plane, to maintain the 20 miles/hr speed, I only have to apply a force of 2 lb now. Although the peddling RPM is higher at this time.

Here the energy supplied is almost same as I supplied it at the beginning but it is lot more easier to supply 2 lb force than 40 lb force.

So isn't it valid for Cargo ship too? When it is travelling at higher speed, to maintain that speed, we may have to supply 10 times less thrust force than to get it started. Energy supplied is same but thrust force is much lower than at start period. Am I correct in understanding or analyzing this now?

That´s the difference between solid friction and fluid friction.
Solids keep their shape and require a nonzero force to alter it and start moving. Fluids will move under arbitrarily small force.
The cargo ship will need a starting force when and if it has a solid attachment to overcome. Like a mudbank in contact with bottom, solid ice sheet on waterline, rope/s attached to bollards or anchor...

When none of these apply, the ship can undergo a slow drift under an arbitrarily small force, or inertia. When you leave your bicycle in a parking lot, it will have little trouble staying put where you left it - it will not slowly wander to the opposite edge of the parking lot, or drop off that edge. When you leave your boat in a port basin without solid attachments like the ones I listed (beached, frozen in, moored, anchored), you may well find it drifted away to the middle of port basin, or opposite edge, or off the port...

 

[PeroK]

When none of these apply, the ship can undergo a slow drift under an arbitrarily small force, or inertia.

It's probably not a good business model for a shipping company to put its vessels into the water and let them drift with the tides.

 

[sophiecentaur]

It's probably not a good business model for a shipping company to put its vessels into the water and let them drift with the tides.

Sounds to me just like a leisure sailing cruiser. Not only that but coastal barges (commercial and pretty big by those standards) did exactly that. 120 tons with 400m² of sail, up and down the Thames. Very reliable motive power! But that wasn't out in the open ocean.

I have to say, the kite idea would have to prove itself very superior to conventional sails. The possible lack of control would really worry me. It's only loopy young men who actually use kites at sea and they'd have a real problem convincing shareholders that it's a good idea (with or without pages of figures and calculations.)

 

[snorkack]

When you are exploring a sailing ship, it may be useful to keep track of the forces and momenta involved.
Some sails are fixed to a rigid beam among two edges (yard and mast). Some only along one edge (yard) and the opposite corners held by lines. Some sails are not attached at any edge, only to lines at corners - e. g. spinnakers - and therefore have degrees of freedom to swing around the mast. In all these cases, it is important to keep track of the stability of the sail, and consider how you avoid the sail flapping itself into unfavourable position, or recover from such position. As well as the rigid beams undertaking unwanted motions, like booms jibing.
Considering the large force desired for sail propulsion, lifting the sail atop a tall mast means giving the force a large moment, which must be resisted by the heeling stability of the underwater hull.
So instead the idea of a kite sail, whose drag force is exerted low on the ship, to keep the heeling moment low, and the sail is kept high by the lift on the sail. But how do you make sure to recover from maneuvers where the kite sail falls into sea?

 

[Arjan82]

So isn't it valid for Cargo ship too? When it is travelling at higher speed, to maintain that speed, we may have to supply 10 times less thrust force than to get it started. Energy supplied is same but thrust force is much lower than at start period. Am I correct in understanding or analyzing this now?

No, that's not how it works. Just for reference, this is the measured (measured on model scale and extrapolated to full scale) resistance for 'a' ship which is a little bit smaller than the ship that is referred to in the first post (Length 390m, beam 60m, draught 16m, displacement 245Mt) which I happen to have laying around:

This ship needs about 35MW at 20 knots in ideal conditions (no wind, no waves, clean hull). This number also depends on the propeller efficiency of course.

So the resistance increases continually with speed. Note that:

• The total resistance is higher than frictional resistance because of wave resistance. The frictional resistance here also includes a form factor resistance due to the ship hull not being a flat plate.
• The difference between frictional resistance and total resistance also increases since wave resistance increases with speed.
• This is the minimal force that you need to apply to keep the ship at that velocity. The actual force may be larger due to interaction effects (the thrust of a propeller needs to be larger due to something called 'thrust deduction', which is the part of the thrust that is lost because the resistance of the ship is increased due to the propeller doing its work and generating a low pressure area on the aft part of the ship) or due to weather effects (wind, waves) etc.

 

[Baluncore]

A kite has lift and drag when flown from a vessel.
The vessel is driven horizontally by the drag.
The vessel displacement is reduced by the lift.
Should the kite be designed for high lift, or for high drag?

 

[snorkack]

A kite has lift and drag when flown from a vessel.
The vessel is driven horizontally by the drag.
The vessel displacement is reduced by the lift.
Should the kite be designed for high lift, or for high drag?

Since the aerodynamic forces on the ship should be a small fraction of its displacement, the lift should mostly be used to keep the kite up.
Also, note that the reason a sailing ship can sail close-hauled is that it has both an efficient airfoil AND an efficient waterfoil. The underwater part of the ship must be efficient in producing lift!

 

[Arjan82]

The vessel displacement is reduced by the lift.

The ship in my post has a displacement of 245,000,000 kg. I don't think the lift of a kite is going to have any significant influence on that... Furthermore, it is not at all said that lifting the bow will reduce the drag since wave resistance can become worse and frictional resistance is hardly affected.

 

[Arjan82]

The vessel is driven horizontally by the drag.

If you use the drag of the kite for propulsion you're doing it wrong ;). The lift is much higher than the drag and the lift vector of the kite should have a significant component in the horizontal direction (and of course as much as possible in the forward direction). So you should optimize the lift over drag ratio and get a high as possible/practical amount of lift.

 

[Arjan82]

So something like this:

 

[jbriggs444]

If you use the drag of the kite for propulsion you're doing it wrong ;). The lift is much higher than the drag and the lift vector of the kite should have a significant component in the horizontal direction (and of course as much as possible in the forward direction). So you should optimize the lift over drag ratio and get a high as possible/practical amount of lift.

The advertising information for the kite in question indicates that they do just this. The kite is flown back and forth in a figure eight pattern at high speed (100 kph) so that it is primarily lift that tensions the cable rather than drag.

The claim is that this increases the "effective area" of the kite by a significant factor. Much like three bladed wind turbines effectively harvest wind energy from a region larger than just the area of the blades.

 

[Baluncore]

The underwater part of the ship must be efficient in producing lift!

If you generate lift from the hull, you generate bigger waves.
We are discussing heavily loaded displacement hulls here, not surface skimming hydroplanes or hydrofoils.