It’s nice to see someone sharing the design process in little snippets like this, thanks for that.

- Thread starter Tiger Tim
- Start date

It’s nice to see someone sharing the design process in little snippets like this, thanks for that.

Glad you like it, I’m just hoping someone speaks up if they see me making a mistake.It’s nice to see someone sharing the design process in little snippets like this, thanks for that.

Oh, they will!

Glad you like it, I’m just hoping someone speaks up if they see me making a mistake.

"Horsepower" is a slippery subject. One wants to consider "Thrust", more than calculated horsepower.

An example: the Curtiss OX-5 was rated at 90hp. The Continental C-90 was rated the same. If One mounted a C-90 in a Curtiss Jenny, it would not fly.

More closely related to your project, let's compare the larger Mosler 40hp, and a Rotax 447 with a 2:1 reduction drive. Both are "rated" at 40hp. The Mosler turns a 54x24 prop at about 3200rpm. The 447/2:1 combination turns a 54x30 at 3250rpm. There will be a significant difference in thrust generated. You begin to see why "horsepower" is a less than perfect means of calculating speed/load-carrying capabilities.

Hope this is of assistance.

A quick re-calc to see if I could make a more sensible HP calculation:

“The fastest way to get the right answer online is to first post the wrong answer”Oh, they will!

“The fastest way to get the right answer online is to first post the wrong answer”

Even then it's still uncertain as big slower-moving prop giving X amount of thrust will do better for takeoff than a smaller faster prop giving the same thrust.On the subject of horsepower:

"Horsepower" is a slippery subject. One wants to consider "Thrust", more than calculated horsepower.

An example: the Curtiss OX-5 was rated at 90hp. The Continental C-90 was rated the same. If One mounted a C-90 in a Curtiss Jenny, it would not fly.

More closely related to your project, let's compare the larger Mosler 40hp, and a Rotax 447 with a 2:1 reduction drive. Both are "rated" at 40hp. The Mosler turns a 54x24 prop at about 3200rpm. The 447/2:1 combination turns a 54x30 at 3250rpm. There will be a significant difference in thrust generated. You begin to see why "horsepower" is a less than perfect means of calculating speed/load-carrying capabilities.

Hope this is of assistance.

Are you saying that two airplanes with equal thrust will accelerate differently just because of a difference in prop diameter and RPM? Gee whiz, Mr. Wizzard! I thought that thrust was thrust.Even then it's still uncertain as big slower-moving prop giving X amount of thrust will do better for takeoff than a smaller faster prop giving the same thrust.

BJC

Looks like a five foot root chord is about right, later I’ll see where that leaves me for a wing planform but for now I’ve laid out where the main fuselage tube would go (and limited it to 12’). I’ve also put the seat about nine inches ahead of where it was on the Piper Skycyle since a UL’s pilot makes up such a large fraction of its takeoff weight. Putting the pilot right on the CG should allow a wide range of pilot weights.

From here I should be able to sort out the tail surface sizes next, working backwards from required tail volume.

That’s a vertical tail area of 12.8% of the wing area and a horizontal area of 17.8%. Compared to my model experience the stabilizer and elevator are on the small side while the fin and rudder seem quite large. I don’t think either of those numbers is too far out though but would like a second opinion.

I do appreciate you taking the time to publicly document your process. I hope you don't mind and tag along without comment and learn.

Sure, as long as you understand I’m only doing it for my own education. And by all means, feel free to comment.I hope you don't mind and tag along without comment and learn.

At this point I’m working out of Beaujon’s book, which is more or less an extremely simplified approach to UL design. Maybe too simple. I’ll have to look for a proper tail volume formula and see how it compares.What is your reference?

This hiccup is a good thing though, it’s a chance to either learn why the recommended fin/rudder area is enormous or it will help prove or disprove the book.

Just looking at your tails sizes as you list them so far, calculated your tail volumes per How big the tail.

For your sizes so far as I understand them:

Sw=130

Sv=16.7

Sh=23.2

L=12

MAC=6.5

From this, calculate a span of 20 (Sw/MAC)

Can then calculate the volume coefficients.

Vh=23.2*12/(130*6.5)=0.33 - if you look at the formula you posted it is just running this calculation the other way to get an area based on a Vh of 0.33. 0.33 strikes me as kinda low, might lead to a limited CG range.

Vv=16.7*12/(130*20)=0.077 Which is above average for Vv. The calculation you posted before is odd, using the gross weight (presumably in lbs). The volume coefficients will work out in any measurement system (the lengths divide out giving a dimensionless number), but the calculation you have is lbs*ft/ft^2 = lbs/ft or a wing loading.

Vertical tail area is based on wing span, not weight.

I didn’t recall using weight for this either. I wonder what the intention was when it was put in Beaujon’s formula, maybe it originally had wing loading (for some reason) but after a bunch of simplification gross weight was left?Vertical tail area is based on wing span, not weight.

Either way, I’m going to go in another direction with it.

Last edited:

I couldn’t agree more with your whole post and thanks for bringing it up.On the subject of horsepower...

At this early stage of design I chose to go with the FAA’s guidance on maximum horsepower, which may or may not even be accurate, as a sort of go/no-go reality check. At some later point I plan to determine an approximate minimum thrust for level flight then actual engine selection can have some power level between the two, depending on desired climb performance... I think.