Talking propellors

[bomber]

I hope that clears up the misunderstandings.

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.

I'd need proof of that because I've not seen anything that looks like this in my internet travels.

also and you really need to read this next sentance....

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

and I've highlighted the really important part of it...the axis of rotation is in the centre, ie a blade to the left and a blade to the right.... 2 blades... you need twice the mass therefore and the distance needs to be the total diameter... you can't shrink it by the hub diameter.

now you could have used this one...for a single blade and multiplied it by 2
Moments of Inertia for a slender rod with axis through end can be expressed as
I = 1/3 m L2
in which case you'd have to use the prop radius, you still can't shrink it by the hub radius.

...........................................

Mass moment of interia doesn't care how wide the blade is..it's all about having a mass rotating at a distance around an axis..The thing about the above equations however is that it's assumed that the mass of the object is evenly distibuted along it's length, so it doesn't matter if it 1inch or 2 inch wide... ie the mass isn't on the end of a 35inch length of string, but is spread evenly along it's length. (might have repeated myself)

[bomber]

Read this....

http://www.pilotfriend.com/training/fli ... /props.htm

If you take nothing from this document other than there being 3 types of propellor.... no not fixed pitch, constant speed, variable pitch....

but...... Climb, Cruise and Speed props....

The performance of aircraft fitted with fixed pitch or ground adjustable propellers is very much dependent on the chosen blade angle. Fixed pitch propellers limit the rpm developed by the engine at low forward velocity, such as occurs during the take-off ground roll and may also allow the engine rpm to exceed red-line maximum when the load on the engine is reduced, such as occurs in a shallow dive. Fixed pitch propellers operate at best efficiency at one combination of shaft power and airspeed. Blade angle is usually chosen to produce maximum performance at a particular flight condition, for example:
• Vy climb i.e. a climb propeller
• Vc cruise i.e. a cruise propeller
• High speed.

So when you're aiming for a plane that 'hits all the numbers'..... ask yourself how's that even possible if I can only use a single prop at a time ?

[bomber]

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?

Yes it does say at full throttle, but what this means is that as you're sat there with the brakes on and full throttle, the engine producing its maximum 100hp 75kW the pitch of the prop is such that it's creating such a load on the prop that the engine isn't powerfull enough to rotate at anything greater that 2400rpm.. you need a bigger engine with more grunt if you want to do that.

So you turn your brakes off trundling down the runway, picking up speed your prop will go through some very rough area's of it's work envelope.

Code: Select all

v          rpm      Power
[m/s]   [1/min]   [kW]
0.000   2394      75
3.480   2348      75
7.130   2405      75
10.64   2393      75
14.09   2378      75
17.54   2368      75
20.98   2361      75
24.45   2358      75
27.99   2361      75
31.63   2372      75
35.43   2391      75
39.51   2424      75
43.90   2469      75
48.69   2528      75
54.04   2605      75
60.19   2708      75
67.56   2850      75
76.89   3053      75
89.89   3370      75
111.5   3961      75
173.6   5860      75

bare in mind this plane can only fly at a max of 57m/s you can see that the propellor won't allow a full rpm but instead 2650rpm...

So this prop design is close but no cigar.

[jwocky]

Okay, some very basic things ...

ONE BLADE means ONE BLADE. It doesn't mean several blades. And since it is one blade, you can't use a simplified formula to get an approximation, you have actually to calculate it.

https://www.physicsforums.com/threads/propellers-moment-of-inertia.87204/
I suggest to read this. The same formulas are also in my handbook for engineers and the guys who use it there work as aerodynamic engineers in the industry, so, I am very hesitant to take a forum in which basically pilots and sim-enthusiasts discuss engineering at a higher value than handbooks for engineers who actually build planes.

The next thing is, in your "I just scavenge the internet and find three formulas" strategy is, you got well, three formulas. One for the calculation of the intertia for a rectangular piece of metal, one for a cylindrical one and one for a disk. So, not knowing what those formulas actually describe, you averaged the result. Which describes a thing that is 1/3 cylinder, 1/3 sheet and 1/3 disk ... and has not the faintest visual similarity with a propeller. THERE IS NO DISK INVOLVED! WHERE DO YOU SEE A DISK???????
So, given that the sheet formula works with 1/12, disk with 1/2, you are here already about factor 6/3 means about factor 2 too high.

Even if you discard the disc part, averaging is not the way. Unless you have a propeller of which half the radius is hub, of course. Then your formula would be correct, but the propeller wouldn't pull much, would it? Lets see, do we have on this 70 inch propeller 35 inches hub? No, we don't!
The part of the radius covered by the blades is unequal bigger than the part covered by the hub. So, to be exact, you can't just average but have to look how much of the radius is blade and how much is hub. For example, if you have a 70in propeller with a hub diameter of 6 inches, you need to use
64in on the blades and 6in on the hub, not just 35in blade and 35in hub

Another basic ... if you have a 70in propeller, how much is the blade length? Well it is 35in-prop_diameter/2 ... because we talk radius here. Not diameter.


After we did the basics, lets go to the a little bit more complex things ...

Look at a propeller blade ... what do you see? Well, for sure no disk ... right? But you also don't see a rectangular sheet of metal. If you don't believe me, put your finger on the tip ... it's a tip, a point, not an edge ... that thing is not rectangular, it just isn't. It's more something between a triangle and some kind of leaf shape and twisted. Kind of like Tolkien's elven blades after a tank rolled over them.
So, if you want to be totally exact, you can only describe the inertia, such a blade produces by an integral over the whole length of the blade because every point on that blade had basically it's own distance from the center point, it's own run-back and it's own deviation from the center line of the blade. And I have no idea how to write such an integral in a text editor, so you may want to look it up.
However, this is for our purposes way too complex. If we use instead just a triangle, with the width at the widest point as base length, we are a little higher than the real result, but not so much that it would really matter. This is what I did. So my numbers, even they appear too low to you on your disk/rod/cylinder thing, are already on the high side.
Now, please look again at a propeller blade ... what else do you see? There is the end of the blade that connects to the hub, what shape does it have? Looks like a cylinder, right? So, there is a short piece looking like a cylinder that reaches into the hub. Well, actually that is the relevant piece of the hub, that is the majority of the mass that actually rotates in the hub. So, since this is an entirely different shape than the blade part, you need to calculate it on it's own and later add up.

So, I don't know where you travelled in the internet, I don't know what you found there. Honestly, I also lack the time and check on each site you may or may not have visited to look at the formulas there and check for what kind of application they are, but since you mixed already the disk formula into it, I assume, they didn't tell you for which application they are either. But I stand by my math and I think, backed up by a library full of engineering literature and the opinion of actual aircraft building engineers, I should. Although, I am aware, I took at least two short cuts that made me results a little bit too high, but still much lower than the unspeakable average formula. So, my opinion stands, the inertia you work with is too high ... which actually can be easily proven. If you take your pup on a runway, push full throttle and you need half the rudder way only to compensate for the torque, you built a plane that has a hard time to fly right-turns at all ... which is actually a disease, some planes and many helicopters in FG suffer from.

[bomber]

Your link wasn't conclusive in any way, just a conversation on a forum....kinda like this one.

Try this link

http://www.engineeringtoolbox.com/momen ... d_913.html

[bomber]

Here's the original link I was looking for

http://www.mh-aerotools.de/airfoils/pro ... of_Inertia

[bomber]

I've no problems you using this equation...

Moments of Inertia for a rectangular plane with axis through center can be expressed as

I = 1/12 m (a2 + b2) (5)

where

a, b = short and long sides

I've no problems with you adding to it with the hypothesis that the blade is a triangle.

I = 1/12 m ((a2 + b2)/2)

But I do have a problem with you not understanding that when it says short and long sides....it doesn't say half the long side... and no where does it say half the mass.

so

mass moment of inertia = 1/12 * 18.6kg * (1.778*1.778) + (0.142*0.142)/2)

mass moment of inertia = 4.9156173kgm2
mass moment of inertia = 3.6232slugft2

[jwocky]

If the whole propeller weights 41 lbs ...
can any SINGLE BLADE ever weight more than 20.5?

If the diameter of the propeller is 70 inches ...
can any SINGLE BLADE be longer than 35 inches?

I repeat again, I calculated ONE BLADE and multiplied only in the end by the number of blades

and about the link, you saw the point, where one of the guys linked to a lab setup for measuring?

[jwocky]

Thinking about it, I should have used your formula, that would have given me even 6.25in in the second term less ... not that it makes a difference, it's too small anyway, compared to the 4000something from the square of the long side.

[bomber]

Simple question....

what do you think this means

Moments of Inertia for a slender rod with axis through center can be expressed as

as opposed to this

Moments of Inertia for a slender rod with axis through end can be expressed as