Talking propellors

[jwocky]

That looks right to me

[bomber]

http://www.airpowerinc.com/productcart/ ... did=390848

here's the prop and it don't look like wood....they're normally a bit chunky.

[bomber]

so there are 3 way of calculating inertia...

Moments of Inertia for a slender rod with axis through center can be expressed as
I = 1/12 m L2

Moments of Inertia for a slender rod with axis through end can be expressed as
I = 1/3 m L2 

Moments of Inertia for a Circular Disk with axis at its center
I = 1/2 m r2

m = weight.

If you average the results you get..

5.716655233 kgm2 4.2136 slugft2

Code: Select all

  <ixx>2.31</ixx>

the ixx appears a little high

Well if you thought it was high before

[jwocky]

I was at the wrong propeller when I said wood, that was one from the guys who made the prop for the 160, but wrong line in the product table, so you're right, alu it is

Yeah, but then, it is no circular disk at all, so leave that one out. No need to bother with 1/2 m r2.
Go for slender rod and for that, we have two forumlas:

i=1/12 m L2 (shouldn't that be 1/12 m L2+B2)?
and i= 1/3 m L2

now, I am bit confused here, because both of those formulas are a bit contradictve and at the same time in the extreme. The 1/12 is valid for a perfectly rectangular piece of metal, not a propeller blade, those things are not flat, they are twisted blades. Now, normally, you can just measure the inertia of one blade out if you have one. You don't. So, lets guess here a bit ...
A single propeller blade is basically a twisted blade you can describe as an object surrounded by a cylinder. That would be somewhat about 1/4 m L2. Bot that is only entirely true for the base where the blade comes out of the propeller cap. Your second problem is, your blade goes narrower the nearer you come to the tip. So for a what 67 inch long blade sitting on a 3 inch long connection to the cap, you have actually a more or less triangular piece of metal that roughly is a triangle (actually a spherical triangle, but we don't want to make it too complex).
So, you have, per blade

1/3 * m_one * 9 + 1/12 * m_two * (4489 + b2)/2

in which
m_one is the socket weight,
m_two is the blade weight without socket
b2 the width of the blade at the widest point
and this is only the rough approximation for one blade

so, if that thing is already in total 41 lbs, this means, one rod, the cap, two blades (is that right?), we have to guess a bit how this is shared over all the parts. Lets just assume for the moment, roughly 34 go to the two blades, that leaves 6 for the two socket parts and 1 for the rest, which is already pretty much on the cautious side. And what would be the blade width at the biggest point? 3.5 inches?
that gives us for one blade

1/3 * 3lbs * 9in2 + 1/12 * 17lbs * (4489in2 + 12.25in2)/2
1lbs*9in2 + 1/12 * 17lbs* 2250in2
3196.5lbsin2
22.1979176lbsft2
10.06kgft2 (scratches head ... why did I calculate kgft2? Oh, yes, we wanted m2?)
0.9344604kgm2 for one blade
1.86921kgm2 for two blades

so, roughly 1.37866slugft2 for both together. I still think, 2.31slugft2 is a little high ...

[bomber]

if I could find it there's somewhere on the web where it explains having measured the inertia of a prop that the slender rod method provides to small an inertia value and the disk method to great a value...this is why I chose to use an average of the 3 methods.

[bomber]

upon review...

i=1/12 m L2 (shouldn't that be 1/12 m L2+B2)?

Not to my understanding...

you're also confusing your slender rod methods I think.

the first half of the question uses

Moments of Inertia for a slender rod with axis through center can be expressed as
I = 1/12 m L2

the second half uses
Moments of Inertia for a slender rod with axis through end can be expressed as
I = 1/3 m L2

don't have a problem with you doing that...it's just that you're neither using the correct mass or distance.

1/3 * 3lbs * 9in2 + 1/12 * 17lbs * (4489in2 + 12.25in2)/2

1/2 the hub + both prop blades.

1/3 * 3lbs * 9in2

9lbsin2.... the whole hub is 18lbsin2

1/12 * 17lbs * (4489in2 + 12.25in2)/2

this is only half the weight of the prop it's got to be 34lbs

1/12 * 17lbs * (4489in2 + 12.25in2)/2

I'm not sure you can use some pythagorous here it's more about distance from the rotating point.

so this value should be 70inch squared = 4900.

1/12 * 34lbs * 4900

13,883lbsin2

with the 18lbsin2 for the hub added total inertia is..

13,901 lbsin2
3.0 slugft2

but as I said in my previous post the basic slender rod method produces too small an inertia value compared to measured.

[bomber]

if anyone has an alli (i have a wooden) prop diagram with sections and distances etc I could model this in solidedge give it a property value of alli and let solid edge work out it's inertia...

We could then create an equation that gives a better estimation based on prop diameter.

[bomber]

I think, I miss here something
if you hang a propeller with a max rpm for 2400 on an engine with a max rpm of 2750, you don't have a gear ratio of 1.0
.

Because it's allready been touched on we might as well talk about this now

Now normally there are 2 limitations to the diameter size of a prop...

1) if it's too big the bottom might touch the ground (this is a no brainer)
2) The tip speed could become supersonic (prop produces less thrust)

to be honest having done the calcs neither of these 2 conditions come into play..

at the top speed of the plane ie 204km/h and a prop rpm of 2400rpm then the helical tip speed is only 223m/s out of a 335m/s max possible or other words 66% of max.

with a gear ratio of 1:1 this prop at 2750rpm will only ever get to 255m/s or 75% of max

McCauley 1A105 SCM7053
Diameter range 70” max. -68.5” min.
Static r.p.m. at maximum possible throttle setting (cross wind) :
Not over 2400, not under 2300 r.p.m.
No additional tolerance permitted

But what it's NOT saying is that the prop has a max rpm of 2400.... it's saying static rpm

it's saying that if you open the throttle to it's max, whilst static the prop will only rotate at between 2400 and 2300rpm.

it's digging in... and this is another clue so if you build an engine/prop configuration and whilst static on the runway you can rev it up above 2400rpm's then regardless of the fdm hitting all the numbers, or how long you spent on it, or how much everyone says it's a great flight model.... the things wrong.

simon

[bomber]

Ok last challenge I think before it gets real exiciting..

The last four digits in the designation of a McCauley propeller give the
diameter in inches and the geometric pitch of the propeller.
The geometric pitch is the circumference of the propeller disc at the blade
station multiplied by the tangent of the blade angle at that point. The blade
station is the point at 0.75 of the radius.
The blade pitch angle can be calculated using this Python snippet:
GP = 53
D = 70
(180 / math.pi) * math.atan((GP) / (D * 0.75 * math.pi))
The pitch angle has been rounded to the nearest degree.

<maxpitch>18.0</maxpitch>

The text is correct just that I think the pitch angle works out at 18.74

Just to say I've a spreadsheet for all the calcs done so far if anyones interested.

[jwocky]

Hi Bomber,

I didn't mess up, I calculated ONE blade. The 3 inches with 3lbs weight is only the almost cylindrical part of ONE blade where it goes into the hub. The socket. I later multiply by two for two blades.

http://eaavideo.org/video.aspx?v=2010764613001 Here is a little video how to measure, but it works only if you have the propeller.

The disk method is, as far as I know, only in cheap simulation used because it's nice and easy, one formula, but well, since there is no disk only blades, it simply doesn't work for a propeller. You have a different situation though in jet engines in which the wheel full of blades behaves more like a disk because well, there are basically blades everywhere.
For a single propeller blade, you have a socket (the part in connected into the hub) and the blade part itself. And again FOR A SINGLE BLADE. You calculate ONE blade and multiply later with the number of blades.

I = 1/12 m L2 This formula is, as far as I know, an incomplete form of the inertia if you rotate a rectangular sheet of metal. It is based on the assumption that you sheet is 1 unit wide. Since our unit is inches in this case and the blade is more than 1 inch wide (I went with three and a half, but that is guess), you need to add the square of the width for the actual propeller part, but then divide by two because your blade part is a triangle not a rectangle.
Ergo, I use actually the correct mass and distance FOR ONE BLADE.

1/3 * 3lbs * 9in2 + 1/12 * 17lbs * (4489in2 + 12.25in2)/2

This is the cylindrical part at the inside end of ONE PROPELLERBLADE:
3lbs=weight of the socket
9in^2=(3in)^2

and now the blade part
17lbs=the weight of the blade part and only the blade part of ONE PROPELLERBLADE
4489in^2=(67in)^2
12.25in2=(3.5in)^2 for the blade width (the sum is divided by 2 because the area rotating here is actually more of a triangle than a rectangle)

So for ONE BLADE nothing is messed up, the masses and distances are all okay (in fact all a bit on the high side estimated)
3 inches+67 inches = 70 inches
3 lbs+17 lbs=20 lbs per blade gives us for two blades 40lbs for the blades alone and 1 lb for the hub (very optimistic).

so again:
1/3 * 3lbs * 9in2 + 1/12 * 17lbs * (4489in2 + 12.25in2)/2 IS NOT half a hub+both blades, it is just one blade, JUST ONE BLADE! JUST ONE BLADE! JUST ONE BLADE!

See, you can't measure, you don't have this propeller. You can't use just one formula or even worse the average over three different formulas of which two are usable for different parts of the blades and one is not usable at all in this case (the disk). So you have to go back to ONE BLADE. The blade consists of a socket shaped part (which connect to the hub) and a blade part which is basically for purposes of inertia a TRIANGULAR sheet of metal.
Only then, after you calculated ONE BLADE, you multiply with the number of blades. Sorry, you fell for this misunderstanding.

About the propeller limitations: There are actually three
1.) if it's too big it hits the ground
2.) tip speed can become supersonic
3.) The mass of the propeller including the torque are too big for the power of the engine or reach the structural limit for the axle in the hub or the gear.


About the gear problem:

McCauley 1A105 SCM7053
Diameter range 70” max. -68.5” min.
Static r.p.m. at maximum possible throttle setting (cross wind) :
Not over 2400, not under 2300 r.p.m.
No additional tolerance permitted

But what it's NOT saying is that the prop has a max rpm of 2400.... it's saying static rpm

it's saying that if you open the throttle to it's max, whilst static the prop will only rotate at between 2400 and 2300rpm.

it's digging in... and this is another clue so if you build an engine/prop configuration and whilst static on the runway you can rev it up above 2400rpm's then regardless of the fdm hitting all the numbers, or how long you spent on it, or how much everyone says it's a great flight model.... the things wrong.

Well yeah, it also said at maximum possible throttle setting ... so how do you want to go over the maximum possible throttle setting? Second lever? Hidden turbo?

The last challenge: Well I came out at 18.73, but I was a bit sloppy rounding. So, aboutish the same. Since we have a float for pitch angle, we can use the actual calculated one and don't need to round back to exact 18 degrees.

I hope that clears up the misunderstandings.