Can Leverages and Propellers Work Together to Create Efficient Movement?

[Sergio Silva]

TL;DR Summary

Why don't I consider a propeller like a type of leverage?

Hello,

My first innocent thread.

I want to be clarified about a subject that it's hanging around me. Maybe is ridiculous, but I want to know if the force that I applied in center of the propeller (I think it calls "boss") is equal to force applied in edge of the blades to get it move around.

This is a silly thing, but mechanics always use a crankshaft to power propellers, could we use chains to make centrifugal movement, displacing the engine from the center with power force increased and aerodynamic advantages?

Maybe propellers must be more resistant on the center because they haven't any support without being the one made by the circle of chains. I thought in leverages because the blades could be considered various leverages cutting the air and resistance that is somewhat a weight.

Criticise me without petty.

Thanks for the patience.

 

[hmmm27]

Criticise me without petty.

"Petty" ? or "pity". Both work.

A propeller is a screw, not a lever. Anyways, there's no reason you couldn't run it from the edge, instead of the hub. Bear in mind then you have a chain/track going over 500mph.

But, the very middle bit does almost nothing as far as thrust is concerned anyways, so why not stick an engine behind it ?

 

[DaveC426913]

There are some helicopter designs that are driven by jet emissions at the rotor tips.

They are called, astonishingly, tip jets.

https://en.m.wikipedia.org/wiki/Tip_jet

 

[DaveE]

You can have the propeller and engine offset, some do that. Although they use belts or gears, not chains. Look up "propeller reduction drive". Like this: https://www.wikiwand.com/en/Reduction_drive

 

[kuruman]

There are some helicopter designs that are driven by jet emissions at the rotor tips.

They are called, astonishingly, tip jets.

https://en.m.wikipedia.org/wiki/Tip_jet

I followed the link and the beginning of the third paragraph caught my eye:

"If the helicopter's engine fails, the tip jets on the rotor increase the moment of inertia, hence permitting it to store energy, which makes performing a successful autorotation landing somewhat easier."

The link to "moment of inertia" gives a definition in terms of the desired angular acceleration which is contrary to what I have learned and taught. To make things worse, they compare the moment of inertia to inertial mass in the linear case. The unsuspecting reader is left with the impression that a torque (or force) is needed for an object to have moment of inertia (or mass).

 

[anorlunda]

Summary: Why don't I consider a propeller like a type of leverage?

I want to be clarified about a subject that it's hanging around me. Maybe is ridiculous, but I want to know if the force that I applied in center of the propeller (I think it calls "boss") is equal to force applied in edge of the blades to get it move around.

That sounds like a question about an ordinary lever, not about airplanes or helicopters.

No, force is not the same along a lever. The short end of the lever has the most force but the longest movement. The long end has the least force and the most movement. Call force F and movement D, then for any two points 1 and 2 along the lever. F₁D₁=F₂D₂

Think of the "boss" of a propeller as the short end of a lever.

 

[Sergio Silva]

About using of a chain, remember that motorbikes use chains to transmit power to rear wheel, but yes could be malfunctions.

On applying the force in the edge of the blades, the early planes started with human help, who pull the propeller by the edge of a blade. I think this proceeding is similar of removing a flat tyre with screw shaft, but this was only because of the weight of the propeller. I think air resistance has importance here, I think it will be easy to transmit force in the edge of the blades in opposition of traditional method. But is only a guess. Could physics give me answer?

 

[Lnewqban]

About using of a chain, remember that motorbikes use chains to transmit power to rear wheel, but yes could be malfunctions.

On applying the force in the edge of the blades, the early planes started with human help, who pull the propeller by the edge of a blade. I think this proceeding is similar of removing a flat tyre with screw shaft, but this was only because of the weight of the propeller. I think air resistance has importance here, I think it will be easy to transmit force in the edge of the blades in opposition of traditional method. But is only a guess. Could physics give me answer?

Welcome, Sergio!
Very interesting first question.

The main problem is that the magnitude of the centrifugal effect at the tips of the propeller would be much greater than by the center, requiring a stronger and longer chain, which would increase the weight of the airplane, offering no substantial advantage.

Please, see:
http://www.geversaircraft.com/ac/propdrive.htm

http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cfc

The tangential velocity of normal propellers easily reaches the speed of sound.
As you can see, the magnitude of the centrifugal effect depends on the square of that speed.

Another problem is the need to transfer the thrust induced by the propeller to the body of the airplane and wings.
Under the load of that thrust, each blade gets bent and its tip moves aft and forward, making the proper alignment with the chain/belt and engine's sprocket/pulley impossible.

It seems that the Wright brothers knew that.

 

[Sergio Silva]

Hi Lnewqban,I even speak about the angle of attack of the blades, and I have yet problems with the weight of the chain. I suppose the width of the chain could be smaller than length of the tip of the blades and independent blades angle. The length will be different, it must have the diameter of the propeller. One way to resolve the problem is to have two propellers with half length in each side of the engine. With more imagination you could have 4 propellers one after the other. (Rude example is Mars Helicopter, I don’t know the aerodynamic implications of that….)
Even more, the legth of the tips of the blades could be resolved with more blades but thinner.

What I thought is to have a cylinder with a propeller in one entry and a flap in middle or in the end. Sucking air on entry and flaps to sustain the plane.

I don’t want to construct a plane, but the first steps are always impossible, aren’t they? It’s only theoretical discussion.

By the way I didn’t know that the Wright Brothers have the knowledge to dream about speed of sound.

 

[Sergio Silva]

Hi guys, it’s me again!

Well back to basics. First, my apologies by my poor physics terms.

What I want to know is if like a windmill or animal traction mill when the force is applied in edge of the mechanism, we gain leverage.

If so, why so many machines use a central point of applying force to deliver motion.

I’m talking about wheels and axles, planes with propellers and drive shaft, boats equal and so on…

If this have some sense reply, please.

 

[Arjan82]

For propellers: rim drives exist. However the big disadvantage here is that the velocity at the edge is way bigger than at the root, this means more frictional resistance for the thruster. So it is often only done if driving from the center is somehow troublesome, like in a bow thruster.

Also note that leverage is increasing force, not power or efficiency. As long as torque is not an issue (which it often isn't, for propellers anyway) I don't see why it would be beneficial to drive from the edge.

 

[jbriggs444]

If so, why so many machines use a central point of applying force to deliver motion.

I’m talking about wheels and axles, planes with propellers and drive shaft, boats equal and so on…

One should be clear on the purpose of the mechanism. If you have a power source over here and a load over there, then transmission of power is the goal.

At one time, low speed belts were used to transfer power from a steam engine over here to a threshing machine over there. But there is only so much power you can transfer safely and effectively through a fabric belt. To say nothing of issues with safety and wear. With a solid steel shaft you can transmit hundreds of horsepower to a rear differential or to a PTO. Or hundreds of thousands of horsepower to a ship's propellors. Or tens of thousands of horsepower to an aircraft's ducted fans. Reliably. Safely. Efficiently. Cheaply. Compactly.

I shudder to think about a fabric belt running from an automobile engine to the rear wheels. Or needing to seal the rim of a twenty foot diameter hole below the water line for a twenty foot diameter hollow drive shaft. And then worrying about how to route water to the propeller so that the thing actually generates thrust. Or exposing an external chain drive to sea water for extended service. And then routing the chains through conduits to reach the engines without risking the crew. And replacing the conduits when they wear out. Shafts, bearings and seals are mature technologies.

 

[Sergio Silva]

Well, my knowledge of torque and power (horsepower) was much acquired when I bought a car. For instance, I had Alfa Romeo 147 that his maximum power was 118 hp at 6200 rpm. Now I have Peugeot 308 with slightly like more power (128 hp at 5500 rpm). it is useless to rev up or down that the maximum force is taken in that values.

The torque is how much the car is responsive to the throttle. My torque of 147 is 146nm (Newton meters) at 4200 rpm, on the other hand, the Peugeot, is 230 (Nm) at 1750 rpm. Compare the two cars my 147 was very sluggish at low revs but it was fun to drive in mountain roads because of his stiffer suspension. My Peugeot is more sensitive to throttle (230) and starting in low revs like 1750 rpm.

In conclusion, you might have a very powerful car, but you must wait a long time to achieve it.

About acceleration and friction, I have a link to a film in youtube about that:



What I have interest is about 2.59m until 3.21m when by this movie (and I want to believe that is right) the torque is calculated. It is basically the force of the axle multiply by the radius of the axle.

Another interesting film is about planetary gears set to automatic gears. To this thread I have no interest in automatic gears.



So, for me is only have interest until 38s and remove the carrier. (I’m not joking but who are reading could laugh).

If the sun gear will be the axle, attached to multiple planet gears, which, will be attached to interior of wheel (in this case is the ring gear) what could be the result?

The principal gear will rotate in the one direction, the planet gears in the opposite and, the wheel in the direction of the axle (In the film it doesn’t rotate. It was the closest model that I could grab).

Viewing this you will see a lot of friction and weight. But what be the formula to achieve acceleration? You have the force and radius of the sun gear (axle) and the force divided in this case by three (each per planetary gear) multiply by the radius.

I think this is much speculation but for now I want to see your reactions.

 

[jbriggs444]

What I have interest is about 2.59m until 3.21m when by this movie (and I want to believe that is right) the torque is calculated. It is basically the force of the axle multiply by the radius of the axle.

If you consider that the "force of the axle" is the shear stress that is present along the axle and pretend that this stress is all concentrated at the rim then yes. The torque carried by the axle is equal to the "force of the axle" multiplied by the radius of the axle.

The power transmitted by this force is then given by the torque of the axle multiplied by its rotation rate.

[Alternately, one can take a surface integral of the stress tensor times a directed incremental area element times the rotation rate times the radius vector over a cross-section of the shaft.

The stress tensor times a directed area gives you the force vector that is transmitted across the area element. Multiply that (dot product) by velocity to get power. But velocity of an area element is given by the cross product of the rotation pseudo-vector and the radius vector for that area element. Add that up for all of the incremental area in a cross section of the shaft and you have the total transmitted power.

The stress tensor or Cauchy stress tensor is a 3x3 matrix that tots up the three dimensional force per unit area in each of three directions. 3 directions times 3 components of force per direction = 3x3 matrix. This completely describes the stress at a point in a material.]

But what be the formula to achieve acceleration?

As long as you are not traction limited, power divided by velocity gives you force. Divide that by mass to get acceleration. If you are traction limited, you may be spinning the wheels and dissipating power into smoke.

 

[Sergio Silva]

If you have a planetary gear system with a sun gear and to 4 planetary gears with one-to-one relationship and fitted on the interior of rear wheel of a bicycle, and sun the connected to the cassete (plates) of the bicycle, you will not gain force advantage?

 

[Arjan82]

force, sure. But force is not energy. It is ultimately energy that you care about.

 

[sophiecentaur]

human help, who pull the propeller by the edge of a blade.

The reason for pulling the tip of the blade is that you need too be as far away as possible when the engine fires so you can't stand in front and pull near the shaft. Many ground crew used to get arms chopped off when they didn't get out in time. I believe there have been numerous injuries from model aircraft engines due to bad 'flicking' technique.

But the same maths applies to propellors, wheels and levers. The principle of moments is the thing.

If you have a planetary gear system with a sun gear and to 4 planetary gears

There are two reasons for having four planetary wheels. Firstly, they share the load so they can be made lighter / cheaper and secondly it means that the system is balanced so (again) is can save on material. The actual gear ratio is the same with any number of planet wheels; it just depends on the ratio of numbers of gear teeth.

 

[sophiecentaur]

You have the force and radius of the sun gear (axle) and the force divided in this case by three (each per planetary gear) multiply by the radius.

I just read this. It is not relevant to the mechanical advantage of the system - just the physical strength of those gears. How ever many there are, the resulting torque is the same.

 

[Sergio Silva]

About leverage and propellers, I think i’m repeating myself, but I think that these two images are somehow related:

 

[Arjan82]

They aren't...

(what can be stated without proof can be dismissed without proof...)

 

[Drakkith]

About leverage and propellers, I think i’m repeating myself, but I think that these two images are somehow related:

Sure, in that a propeller has certain properties in common with a lever. If one wished to turn the engine by exerting a force on the propeller, a longer propeller would result in a lower required force, and vice-versa. Broken down, a propeller is just a double-handled lever with a twist applied to turn it into a screw.

That being said, you don't care very much about the lever properties of a propeller. You care about how well it turns the power output from the engine into a force applied to the air, which is more about prop speed, shape, and engine power.

Note that, as has been said already, mechanical advantage is about force multiplication, not about energy or power multiplication. You will never do more work or provide more power than the engine can provide no matter how you drive the prop, and aircraft are, ultimately, limited by the power of their engine more than anything else.

 

[berkeman]

About leverage and propellers, I think i’m repeating myself

You are, and this thread is done.