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Prop failures

I've been thinking about what happens.....

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John Cole03/08/2008 13:46:00
615 forum posts
24 photos

I've been thinking about prop failures.......

In looking at what happens when a prop fractures at speed you need to look at what the forces are (both at the instant of failure and immediately afterwards) and get a feel for the relative magnitudes of these.  It's raining today so I thought I would do a few sums.  I've based my sums on an old Master 10*6 prop.  I chose it as it is of relatively constant chord, which makes the sums easier.  I understand these props are made from nylon with chopped glass fibres added for strength.My prop is 25 cm in diameter and weighs 23 gms.  I guess each blade weighs about 10 gms and the hub 3 gms. 

My 40 engine will turn it at 12k rpm, or 200 revs per second.  It will just about hold a 2 Kgm model's weight so each blade is delivering about 1000 gms aerodynamic force forwards. The centripetal force is calculated from the acceleration of each element of the blade, converted to a force and summed. This is made arithmetically difficult by the way the blade section thins from root to tip, but again the Master prop is not too hard to approximate. 

The acceleration of each element is calculated as the square of the speed divided by the radius of the curved path, so for the tip (which is travelling at about 160 metres per second (mps) and pulled round a radius of 0.125 m) the acceleration is a little under 200,000 mpsps.  Divide that by 9.81 mpsps (the acceleration due to gravity) and you see that the acceleration at the tip is about 20,000 g.  So one gm at the tip would produce an outward pull of 20 kg.I calculate that the centripetal force at the blade hub totalling along the blade is about 84 kg (less than you might guess from the tip figure of 20 kg per gm as the inward pull decreases with the square of the radius as you move in towards the hub, and the mass is predominantly near the hub). 

This is a LOT more than the aerodynamic force and is the preponderant force in causing fracture, even allowing for the section-moment effect of leverage at the hub.  As well as being much greater than the aerodynamic force it is also much greater than the engine torque even allowing for the positive-negative variation in this during the engine's rotational cycle (which makes the peak instantaneous torque just after firing TDC much greater than the torque averaged round the engine's rotational cycle).But how does this force compare with the nominal strength of the prop? 

 I've seen the tensile strength of roving-reinforced GRP quoted as 1000 MPa, approximately 10,000 Kg / sq cm.  The material of a prop will be considerably weaker (chopped glass not roving, lower glass content, poor fibre alignment near the hub) but even if it's 10 times weaker this is much greater than the centripetally-caused stress: my Master prop has a cross-section near the hub of a bit under 1 sq cm and so at low estimate tensile strength of 1,000 Kg / sq cm would fail at over 36k revs, not 12k.  And we know, undamaged props don't fail at ...........(ctd next post)

John Cole03/08/2008 13:46:00
615 forum posts
24 photos

......"ordinary" revs.

So what's the damage that causes failure at "ordinary" revs?  I don't know but I would guess that both the backwards bending force of a dead-stick prop touching the ground on landing, and the rotational bending force of a fast-revving prop being stopped dead by ground contact, might fracture the glass filaments inside the prop (on the side put in tension by these actions).  T

his would not be visible but would weaken it in proportion to the amount of glass fractured.  Repeated impacts would cause additional damage and would leave it prone to failure at "ordinary" revs.  Note that the damage caused by the backwards bending force will weaken the FRONT of the blade-root - the part which the aerodynamic force will tend to put into compression, offsetting the centripetal tension.  This reduces the importance of the aerodynamic effect.So what happens when the prop fails? 

The forces that WERE acting on the blade (and caused it to fail) are now not relevant.  What will happen is that all parts of the blade will try to continue to move at unchanged speed along the straight line they were following at the instant of failure; these paths will be subsequently modified by the aerodynamic forces acting on them AFTER failure.  If the blade is upright at the instant of failure then it will fly off (initially) horizontally (NOT vertically outwards!) in the plane of the blade disc.  

The blade will fly off at the rotational speed of the C of G of the blade. For my Master prop I estimate the C of G to be at 5 cm from the hub, so rotating as 12k RPM the fractured blade will fly off at 65 metres per second horizontally, or about 145 mph.  But what's the effect of the aerodynamic forces?The blade as a whole will fly off at the speed of the C of G of the blade.  But every part of the blade will try to continue at its own speed.  So the tip will seek to continue horizontally at about 160 mps, while the fracture point will only move horizontally at about 10 mps. 

What this means may seem surprising, but the single blade will therefore initially continue to rotate at 12k rpm!  But now it's rotating about the blade's C of G rather than the prop hub.  This means that while the tip of the blade continues to generate thrust forward (that is, in the direction of the plane's axis) the part of the blade on the hub side of the blade's C of G is now generating BACKWARDS thrust!  Hold a prop at the blade's mid-point (with your fingertip on the blade front-face and your thumb on the blade rear-face) then rotate it and you'll see this is true).  So the blade instantly tumbles (tip forward and root backwards) and there is no significant net aerodynamic force to throw the blade off its course (in this example, horizontally in the prop-disc plane).Well, that's what I think!

Regards,  John   

Myron Beaumont03/08/2008 15:56:00
5797 forum posts
51 photos

John     What a brilliant piece of logic / maths / deduction .  I think you might have put an end to the  "chaos " theories ie everything can somehow be explained like your explanation . I like it very much .Keep looking after your brain cell   Post some more will you on any subject .Good stuff ,WE have another Timbo in our midst


Tim Mackey03/08/2008 18:01:00
20920 forum posts
304 photos
15 articles
Isnt there piece about this ( the exact same stuff I think ? ) in the mag this / last month in readers letters or summat??
John Cole03/08/2008 18:08:00
615 forum posts
24 photos
Yes there's a letter in All Write, but I was not convinced by what it said.  So I did some sums and came to different conclusions.
Brian Parker03/08/2008 20:29:00
538 forum posts

I also thought about this after reading ‘All right’ ..So without the maths..

Point ‘p’ on the tip of a prop will follow a circular path on a circumference of radius ½ prop diameter. The direction of velocity of 'p' will always be perpendicular to the radius, ie not straight, the direction changes in order to keep ‘p’ on a curved path. Note the direction changes not the magnitude. Force is accelerating ‘p’ inwards, this change in velocity is centripetal acceleration (inward acceleration in a circular path). The same rule applies at any point on the prop.

Centrifugal force is equal and opposite to the centripetal force. On prop. breakage centripetal force is released and centrifugal force will throw the prop outwards in a straight tangential line, i.e. Newton’s equal and opposite forces.

I think I am agreeing with John.

John Cole08/08/2008 17:25:00
615 forum posts
24 photos

Brian: No you're not.  What you say is exactly what I thought before I did the sums.  The prop is a bit of plastic; it has no brains - and in particular no memory! 

In my example the prop fails when in the vertical position.  Your argument (and my predjudice) is that the separated blade will go off vertically upwards (if it's pointing up).

My considered argument is different: that the instant after failure the blade "does not know" it was ever connected to the hub - so the blade (and all parts of it) will try to continue moving at the same speed and direction as before failure, subject only to the after-failure aerodynamic forces.

At the instant of failure, if the prop is in the vertical position then all parts of the (failed) blade will be travelling HORIZONTALLY.  The complication is that all the bits of the blade are travelling at different speeds.  In my example, the tip is travelling at 160 mps (about 360 Mph) and the base much slower.  The CofG is moving at 65 mps (145 Mph) and that's the speed at which the blade moves off (HORIZONTALLY).

However, the tip will try to move faster than this (and the base slower) and as a result the blade will TRY to continue to rotate STILL at 12k - but now round the CofG rather than the hub.

See my original note which shows why the blade then tumbles.

Brian Parker08/08/2008 19:52:00
538 forum posts


If you re-read my previous post, what I said was that the prop will be thrown at a tangent once the centripetal force is lost, ie If the prop is vertical then it will be thrown outwards horizontally in a straight line. See sketch.

The velocity at point ‘p’ at position ‘M’ is measured in feet/second (props being imperial) it also has an angular velocity in radians/second. M is the direction of velocity, always perpendicular to the radius, its path is constantly changing (angular) to keep it on its curved path unless centripetal force is lost (ie. when the prop breaks) then it will  take a tangential path. The same applies for any point along the radius of the prop the velocity appropriate to the radius. The now ‘free’ section of the prop will not generate thrust.

Myron Beaumont08/08/2008 20:58:00
5797 forum posts
51 photos

Must throw a wobbly in here chaps   --- LITERALLY ---.When the blade parts company I would imagine it's not a completely clean break .Although it happens in a fraction of a second there is probably a point where the blade is"Partly" attached ie twisting and changing its angle before letting go. This of course means that its departure angle cannot be worked out mathematically (&  who ever has a high speed camera on hand to discover what the source of chaos theory is  at that moment?

Brian Parker08/08/2008 22:39:00
538 forum posts


Once the blade detaches, by whatever means, then it must follow Newton’s Law and the most dangerous place to be will always be inline with the prop.

One other aspect to consider is the potential energy stored due to flexing and its effect on the point of release and also its release velocity. Then there's the radius of gyration to consider as opposed to the CofG, if anyone can do the maths

Getting bored with this, best to just arrange to stand safely behind the model

Eric Bray09/08/2008 10:59:00
6600 forum posts
2 photos
Gliders don't HAVE props! 
Brian Parker09/08/2008 11:24:00
538 forum posts

Lee Burke09/08/2008 11:34:00
203 forum posts
1 photos

 Brian Parker wrote (see)

Getting bored with this, best to just arrange to stand safely behind the model


No, no!  Don't lets leave this exciting thread just yet, we are now getting to the really interesting part.  The question is now at exactly which angle will the prop blade which has parted company from its hub and is now entering the solar plexus of its owner (and not, hopefully, a bystander) , at which angle will it cause the most devastating lesions?  Now, who is up for some testing here?

Not me, of course.

Eric Bray09/08/2008 12:15:00
6600 forum posts
2 photos
Then, of course, wehave the engine, turning half a prop at 30,000rpm. Which direction will THAT go in when the bearers let go?

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