Allied Gold-Match

Well, it is obvious, it wll slide to the right. But that does not mean it has no lift.

On the F4U there is an equal and opposite force created by the other wing, which is bent in the other direction.

The loss of lift from the wing angle is as I've shown. If you like, we can contact an expert to confirm it

Well, it is obvious, it wll slide to the right.

Click to expand...

It will also descend slightly.

But that does not mean it has no lift

Click to expand...

Not at all, but the lift will decrease along with the increase in angle.

The loss of lift from the wing angle is as I've shown. If you like, we can contact an expert to confirm it

Click to expand...

RG do we agree that at 90* the wing has no upwards lift nomore ?

If so, then by tilting the plane 22.5-25* in either direction the "upwards lift" will be reduced by 25-27.2%. (This is what I seem to remember from Flight-school, although it was a long time ago )

Ummm no.

Let me try to explain.

First I'm going to assert that the "lift" created by the wing passing through the air at a constant speed is perpendicular to the wing and does not change regaurdless of orientation. As long as the speed remains the same, the force in the direction of the top of the wing will remain the same. Furthermore, we will not consider the effect of changing angle of attack.

Furthermore, I'm going to define this force to be a unit vector with a strength of 1 Gravity. That is, we will assume that in level flight the total lift generated by the wing at the given constant speed is exactly sufficeint to counter the force of gravity. This is not necessary, but as you will see using 1 will make the math cleaner and easier to follow.

It is true that at 45 degrees the upward "lift" force equals the sideways "lift" force. However, it is not equal to 1/2 the wings lifting force!.

It is easier and more logical, at least for me, to work in radians. A radian is defined as the length of the radius of a circle meausured around its arc (i.e. in a curve). A circle is 2 x Pi radians in circumfrence. For a unit circle, a radian = 1 so that part drops out, and we can simply say a unit circle is 2 Pi in circumfrence.

So if a unit circle has a circumfrence of 2 Pi (radians). Half a circle has a of 1 Pi, a quarter circle (i.e. 90 degrees) is 1/2 Pi, an eight circle (45 degrees) is 1/4 Pi, and sixteenth circle (22.5 degrees) is 1/8 Pi.

Now, since we are working from the angle it is useful to use the sine and cosine (rather than derive them which would take a lot more text), which are defined as:

sine = In a right triangle, the ratio of the length of the side opposite an acute angle to the length of the hypotenuse. This is our sideways force vector.

cosine = In a right triangle, the ratio of the length of the side adjacent to an acute angle to the length of the hypotenuse. This is our upward force vector.

When the wing is angled at 45 degrees (1/4 Pi), down and to the left, the 1G lift vector will point up and to the left, and this unit vector forms the hypontenous of a right triangle, with the sides pointing horizontally to the left and vetically up.

sin(0.25 x Pi) = 0.707106781 {you can enter "sin (0.25 x Pi) =" into google and it will give you this result}

and (for this angle)

cos(0.25 x pi) = 0.707106781

So, as I hope you can see, at 45 degrees of bank the loss upward lift is only about 30%, not the 50% you thought.

You can confirm this by using the pythagorean theorim, which says that the squareroot of the sum of the squares of the sides of this triangle should equal 1.

Now, applying the same math to the angle of 22.5 degrees we can see that:

sin(0.125 x pi) = 0.382683432
cos(0.125 x pi) = 0.923879533

So for a bank angle of 22.5 degrees (1/8 Pi) the loss in upward lift is only about 8.6%.

=S=

Lunatic

RG_Lunatic said:

So for a bank angle of 22.5 degrees (1/8 Pi) the loss in upward lift is only about 8.6%.

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No no no !! Its "about 6.25%"

See your mistake ?

Sine and cosine? I thought they were restricted to math

Soren said:

RG_Lunatic said:

So for a bank angle of 22.5 degrees (1/8 Pi) the loss in upward lift is only about 8.6%.

Click to expand...

No no no !! Its "about 6.25%"

See your mistake ?

Click to expand...

LOL - I did it off a graphic and used scaling to estimate the loss. If the angle of the wing is 22.5 degrees down, then the loss would be 8.6%, I am not sure what the actual angle was.

Anyway, I hope you see that your %'s for lift are based upon a linearity that is incorrect?

At 45 degree's the loss of upward-lift is about 40% RG, not 30%, as it takes about 1.4G's to maintain constant height at a 45 degree bank angle(It takes 40% more lift to cancel out weight). At a 60 degree bank angle you need 2G's to maintain constant height. (Thats 100% more lift to cancel out weight)

This is how I was told it works atleast.

Soren said:

At 45 degree's the loss of upward-lift is about 40% RG, not 30%, as it takes about 1.4G's to maintain constant height at a 45 degree bank angle(It takes 40% more lift to cancel out weight). At a 60 degree bank angle you need 2G's to maintain constant height. (Thats 100% more lift to cancel out weight)

This is how I was told it works atleast.

Click to expand...

LOL... do the math Soren....

0.707106781 x 1.4 = 0.989949493 = about ONE!

And....

cos(60 degrees) = 0.5, so 0.5 x 2.0 = ONE!

So you see, there is no conflict between what they taught you in flight school and the math I've shown you at all!

I hope this all becomes clear to you. I've had jobs where I've had to do complex vector and matrix algebra all day long, so I have an advantage here. I can see what the outcome of the math is going to be almost instantly without even doing it.

I hope you will look over the math and try to understand it. If you have any questions feel free to ask. This kind of math has many applications and once you "get it" it is actually very easy. This is why I went into radians a bit above even though I could have avoided doing so. If you work things out a bit in radians you will get the big picture much faster than working in "degrees".

=S=

Lunatic

RG_Lunatic said:

Soren said:

At 45 degree's the loss of upward-lift is about 40% RG, not 30%, as it takes about 1.4G's to maintain constant height at a 45 degree bank angle(It takes 40% more lift to cancel out weight). At a 60 degree bank angle you need 2G's to maintain constant height. (Thats 100% more lift to cancel out weight)

This is how I was told it works atleast.

Click to expand...

LOL... do the math Soren....

0.707106781 x 1.4 = 0.989949493 = about ONE!

And....

cos(60 degrees) = 0.5, so 0.5 x 2.0 = ONE!

So you see, there is no conflict between what they taught you in flight school and the math I've shown you at all!

I hope this all becomes clear to you. I've had jobs where I've had to do complex vector and matrix algebra all day long, so I have an advantage here. I can see what the outcome of the math is going to be almost instantly without even doing it.

I hope you will look over the math and try to understand it. If you have any questions feel free to ask. This kind of math has many applications and once you "get it" it is actually very easy. This is why I went into radians a bit above even though I could have avoided doing so. If you work things out a bit in radians you will get the big picture much faster than working in "degrees".

=S=

Lunatic

Click to expand...

It is quite clear to me, however you said that at a 45 degree bank angle the loss of lift was about 30%, while it is actually 40%. I was just trying to make a simple correction, thats all.

Anyway the loss of lift is close to 9-10% on each wing, wich is more than what Wing-thickness can make up for, wich is the point.

Perhaps since you downloaded this "Airfoil program" you could calculate the exact difference of lift when thickness is increased by 1-5% ? I will get the program myself as soon as I find out how to make it work on XP, as it seems it will only run on Win95. And at the cost of 179$, I would like to find out about that first.

Soren said:

RG_Lunatic said:

Soren said:

At 45 degree's the loss of upward-lift is about 40% RG, not 30%, as it takes about 1.4G's to maintain constant height at a 45 degree bank angle(It takes 40% more lift to cancel out weight). At a 60 degree bank angle you need 2G's to maintain constant height. (Thats 100% more lift to cancel out weight)

This is how I was told it works atleast.

Click to expand...

LOL... do the math Soren....

0.707106781 x 1.4 = 0.989949493 = about ONE!

And....

cos(60 degrees) = 0.5, so 0.5 x 2.0 = ONE!

So you see, there is no conflict between what they taught you in flight school and the math I've shown you at all!

I hope this all becomes clear to you. I've had jobs where I've had to do complex vector and matrix algebra all day long, so I have an advantage here. I can see what the outcome of the math is going to be almost instantly without even doing it.

I hope you will look over the math and try to understand it. If you have any questions feel free to ask. This kind of math has many applications and once you "get it" it is actually very easy. This is why I went into radians a bit above even though I could have avoided doing so. If you work things out a bit in radians you will get the big picture much faster than working in "degrees".

=S=

Lunatic

Click to expand...

It is quite clear to me, however you said that at a 45 degree bank angle the loss of lift was about 30%, while it is actually 40%. I was just trying to make a simple correction, thats all.

Click to expand...

No Soren, it is not, it is about 30%. I just showed you this is the case.

At 1 G of lifting force off the wing, angled at 45 degrees the upward lifting force will be 0.707106781 G. That is about a 29.3% lost lift, which for purposes of discussion is "about 30%".

The fact that it takes ~40% more lift (via speed) off the wing to achieve 1 G of upward lifting force simply reflects the fact that about 30% of that lift will be lost as well. To get an additional 30% more upward lift you need an about an additional 43% lift at the 45 degree angle.

Soren said:

Anyway the loss of lift is close to 9-10% on each wing, wich is more than what Wing-thickness can make up for, wich is the point.

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No, the loss of lift is about 30% of the total lift for the angled part of the wing, which extends about 4 feet from root to elbow on each wing. The wingspan is about 41 feet, 4 feet of which is taken up by the fuselage. So figuring the total wing span of about 35 feet, only about 22% of that is bent down. So the loss in lift compared to a level wing is about 0.22 x 30% = ~6.6%. The higher wing thickness applies to the whole wing, so it will make up for this loss of lift, and probably exceed it substantially.

Soren said:

Perhaps since you downloaded this "Airfoil program" you could calculate the exact difference of lift when thickness is increased by 1-5% ? I will get the program myself as soon as I find out how to make it work on XP, as it seems it will only run on Win95. And at the cost of 179$, I would like to find out about that first.

Click to expand...

The airfoil program I downloaded is a 45 day evaluation version of "TraCFoil", which allows you to display the airfoils, it does not allow any simulations. It was sufficient to capture the airfoil shapes for this thread.

FoilSim from NASA may be sufficient to analyze the lift of the airfoil shapes. What I have not found yet is anything (for free) which will allow the whole wing to be analyzed. Use of something like FoilSim will probably be sufficient given 5 or more cross-sections - but the problem here is we don't know how fast the root airfoil transitions to the tip airfoil. Thus I'm still researching this (I have a few emails outstanding with ?'s about this).

=S=

Lunatic

Trying to keep up here. Wingspan of 41 feet. Fuselage takes up 4 feet. Each wing is 18.5 feet. 4 feet of each wing is ther downward root to elbow position that accounts for 22% of each wing suffering from the loss of lift of 30%.

Isn't that an overall lift loss of 9%?

No, the loss of lift is about 30% of the total lift for the angled part of the wing, which extends about 4 feet from root to elbow on each wing. The wingspan is about 41 feet, 4 feet of which is taken up by the fuselage. So figuring the total wing span of about 35 feet, only about 22% of that is bent down. So the loss in lift compared to a level wing is about 0.22 x 30% = ~6.6%. The higher wing thickness applies to the whole wing, so it will make up for this loss of lift, and probably exceed it substantially.

Click to expand...

RG stop making one statement and then another ! You said the loss of lift would be about 8.6% for a 22.5 degree angle, and now your changing it to 30% ! Make up your mind !!

Also the wing thickness will not increase lift by any big margin, only by a very small one ! The Tempest and Typhoon comparison test is a good example of this without going into the exact aerodynamic numbers. Tests between the Hurricane and Spitfire also revealed that they both turned very equally, and eventhough the Hurri had both lower wing-loading and a much thicker airfoil the Hurricane still turned only very slightly better.

The airfoil program I downloaded is a 45 day evaluation version of "TraCFoil", which allows you to display the airfoils, it does not allow any simulations. It was sufficient to capture the airfoil shapes for this thread.

FoilSim from NASA may be sufficient to analyze the lift of the airfoil shapes. What I have not found yet is anything (for free) which will allow the whole wing to be analyzed. Use of something like FoilSim will probably be sufficient given 5 or more cross-sections - but the problem here is we don't know how fast the root airfoil transitions to the tip airfoil. Thus I'm still researching this (I have a few emails outstanding with ?'s about this).

Click to expand...

I am prepared to pay money for the program, but not before finding out if it will at all work on a PC with XP installed on it. So what I need is a 'Demo' of some sort.

Jank said:

Trying to keep up here. Wingspan of 41 feet. Fuselage takes up 4 feet. Each wing is 18.5 feet. 4 feet of each wing is ther downward root to elbow position that accounts for 22% of each wing suffering from the loss of lift of 30%.

Isn't that an overall lift loss of 9%?

Click to expand...

Hmmm, the total wingspan is 41 feet. The fuselage eats up about 4 feet (at the level of the wing join) leaving 37 feet. The bent part is 8 feet of 37 feet,

8/37 = 0.216216216 = ~22%

0.22 x 0.30 = 0.066

I don't see where you come up with 9% Jank????

=S=

Lunatic

Soren said:

No, the loss of lift is about 30% of the total lift for the angled part of the wing, which extends about 4 feet from root to elbow on each wing. The wingspan is about 41 feet, 4 feet of which is taken up by the fuselage. So figuring the total wing span of about 35 feet, only about 22% of that is bent down. So the loss in lift compared to a level wing is about 0.22 x 30% = ~6.6%. The higher wing thickness applies to the whole wing, so it will make up for this loss of lift, and probably exceed it substantially.

Click to expand...

RG stop making one statement and then another ! You said the loss of lift would be about 8.6% for a 22.5 degree angle, and now your changing it to 30% ! Make up your mind !!

Click to expand...

Oops I goofed you are right! Thank you for catching my error! I applied the 30% loss figure, when in fact it should be only 8.6%.

So the lost lift should be 0.22 x 0.086 = ~1.9%! Barely any loss at all!

Soren said:

Also the wing thickness will not increase lift by any big margin, only by a very small one ! The Tempest and Typhoon comparison test is a good example of this without going into the exact aerodynamic numbers. Tests between the Hurricane and Spitfire also revealed that they both turned very equally, and eventhough the Hurri had both lower wing-loading and a much thicker airfoil the Hurricane still turned only very slightly better.

Click to expand...

The Tempest and Typhoon are not a good comparison because the Tempest is a little lighter which should make it turn better but has a thinner laminar flow type airfoil which should make it turn worse. Reports I've read indicate the Typhoon turned noticably better than the Tempest. As for the Hurc vs. the Spitfire, there are too many other variables involved to make much of a comparison.

Soren said:

I am prepared to pay money for the program, but not before finding out if it will at all work on a PC with XP installed on it. So what I need is a 'Demo' of some sort.

Click to expand...

Well, it would be interesting to see the results. However, I think it is foolish to spend the kind of money you will need to spend on a quality program unless you plan to build planes (or model planes). But of course that's your call 8)

=S=

Lunatic

Oops I goofed you are right! Thank you for catching my error! I applied the 30% loss figure, when in fact it should be only 8.6%.

So the lost lift should be 0.22 x 0.086 = ~1.9%! Barely any loss at all!

Click to expand...

No problem.

The loss of lift on the bent sections is 8.6%, but the outer wings are also bent slightly at approx.10 degree's leaving almost no loss there though.

The Tempest and Typhoon are not a good comparison because the Tempest is a little lighter which should make it turn better but has a thinner laminar flow type airfoil which should make it turn worse.

Click to expand...

RG the Tempest was only lighter by a mere 380lbs, not enough to make any difference at all. However the thing that did make a difference was the Laminar airfoil !

Reports I've read indicate the Typhoon turned noticably better than the Tempest.

Click to expand...

RG im going to have to need a sourcee on that, as all my available sources state the Typhoon and Tempest to be VERY equal in turning circles.

As for the Hurc vs. the Spitfire, there are too many other variables involved to make much of a comparison.

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Variables like what ?

Well, it would be interesting to see the results. However, I think it is foolish to spend the kind of money you will need to spend on a quality program unless you plan to build planes (or model planes). But of course that's your call 8)

Click to expand...

Im not low on finances , and the program will be of very good use to me as a matter of fact ! 8)

Just need to figure out how to get the damn thing to work on XP !

Soren said:

Oops I goofed you are right! Thank you for catching my error! I applied the 30% loss figure, when in fact it should be only 8.6%.

So the lost lift should be 0.22 x 0.086 = ~1.9%! Barely any loss at all!

Click to expand...

No problem.

The loss of lift on the bent sections is 8.6%, but the outer wings are also bent slightly at approx.10 degree's leaving almost no loss there though.

Click to expand...

Yes, 8.6% for the ~22% of the total wing that is bent down, for a total loss of lift across the whole wing of about 1.9%.

The remaining part of the wing is bent up at 6.5 degrees, causing a whopping 0.64% loss, which I think we can agree is insigificant.

Soren said:

The Tempest and Typhoon are not a good comparison because the Tempest is a little lighter which should make it turn better but has a thinner laminar flow type airfoil which should make it turn worse.

Click to expand...

RG the Tempest was only lighter by a mere 380lbs, not enough to make any difference at all. However the thing that did make a difference was the Laminar airfoil !

Reports I've read indicate the Typhoon turned noticably better than the Tempest.

Click to expand...

RG im going to have to need a sourcee on that, as all my available sources state the Typhoon and Tempest to be VERY equal in turning circles.

Click to expand...

Generally the Typhoon suffered from "winding up" in a turn and very heavy elevator response, which diminished its turn quality, so it makes this comparision against the Tempest kind of mute. It was not the wings that made the two planes almost identical in evaluated turn quality.

Soren said:

As for the Hurc vs. the Spitfire, there are too many other variables involved to make much of a comparison.

Click to expand...

Variables like what ?

Click to expand...

Control surface composition, size, and actuation. Engine power (depends on model), etc... In the early models with fabric coverings on both, the Hurc was noticably more nimble.

Soren said:

Well, it would be interesting to see the results. However, I think it is foolish to spend the kind of money you will need to spend on a quality program unless you plan to build planes (or model planes). But of course that's your call 8)

Click to expand...

Im not low on finances , and the program will be of very good use to me as a matter of fact ! 8)

Just need to figure out how to get the damn thing to work on XP !

Click to expand...

Well, as long as you have a use for it beyond these discussions

If you can tell me exactly what you want out of such a program I might be able to help you find one. I've run through a lot of them over the last few weeks. Do you plan to build wings? If so, would these be for scale model aircraft with wood ribs/struts/spars? There is software specifically for models which differs from that for real aircraft.

=S=

Lunatic

Yes, 8.6% for the ~22% of the total wing that is bent down, for a total loss of lift across the whole wing of about 1.9%.

The remaining part of the wing is bent up at 6.5 degrees, causing a whopping 0.64% loss, which I think we can agree is insigificant.

Click to expand...

Agreed.

Generally the Typhoon suffered from "winding up" in a turn and very heavy elevator response, which diminished its turn quality, so it makes this comparision against the Tempest kind of mute. It was not the wings that made the two planes almost identical in evaluated turn quality.

Click to expand...

Well IIRC the Tempest had a 4mph lower stall speed I guess the Aspect ratio of the Tempest's wing was what made it equal to a Typhoon in a turn.

Control surface composition, size, and actuation. Engine power (depends on model), etc... In the early models with fabric coverings on both, the Hurc was noticably more nimble.

Click to expand...

The Hurri was never more nimble than the Spit, however it turned slightly better in the early stages, but as soon as the fabric coverings got extinct this all changed.

Well, as long as you have a use for it beyond these discussions

Click to expand...

Indeed I have

If you can tell me exactly what you want out of such a program I might be able to help you find one. I've run through a lot of them over the last few weeks. Do you plan to build wings? If so, would these be for scale model aircraft with wood ribs/struts/spars? There is software specifically for models which differs from that for real aircraft.

Click to expand...

Well I was planning to maby use it for 1:20-25 scale "flying" aircraft models, but I like the Idea of having all the info on full scale airfoils aswell.

I tried to download this Foilsim program but it doesnt seem to work, even after downloading all the updates required. So im suspecting it doesnt work on XP.

FoilSim works, but you may need to update your Java to enable it, and it may be blocked by your browser security. However, it's kinda primative and only useful for very general stuff. It's intended for school kids to use to get some feel for this topic.

The thing is the model airplane airfoil programs help you to put wooden ribs/struts/spars where needed, the real plane programs use a different dynamic for this. So I'm not sure you are going to find one program that does both. I'll look for some of both kinds so you can choose.

=S=

Lunatic

If 22% of each wing has got got inly 70% lift, then the totasl loss over the entire wing is 9%.

each wing is 18.5 feet. 11.43 feet of which are developing 100% lift. 4.07 feet of which are developing only 70% lift. (30% less)

With 11.43 feet developing 100% lift and 4.07 feet developing 70% lift therre is 9% less lift than if the entire 18.5 feet developd 100% lift.

The thing is the model airplane airfoil programs help you to put wooden ribs/struts/spars where needed, the real plane programs use a different dynamic for this. So I'm not sure you are going to find one program that does both. I'll look for some of both kinds so you can choose.

Click to expand...

Thanks RG, I appriciate it.