#### pictsidhe

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I'm going to go for 600fpm, if the powerplant can climb it at full throttle.

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I'm going to go for 600fpm, if the powerplant can climb it at full throttle.

Then I'm not understanding what this "stall behavior" you saw is. On my plane (and every other Velocity I've flown), when the canard stalls, the nose drops and I lose about 30-50' of altitude until the speed picks up and the nose comes back up. If that happened on takeoff at 8', that would be a bad thing. Which is why I always have the mains on the ground at no less than 5kts over the canard stall speed and the nose wheel on the ground before the canard stalls.Maybe 8 feet

Edit: swapped email with the pilot. Said he was fully loaded and had the trim too far back. Got about three small pitch bucks by the time he got the trim set. Apparently they were not full stalls on the canard.

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Hubris, just like the Capt. of the TitanicI’m trying to imagine just what the NTSB, with so many options, will identify as the root cause.

BJC

Ok, feel free to start a new thread and advise. ThanksNot on this thread.

Best,

Tom

To be fair to the captain of the Titanic, he didn't know the iceberg was there until he hit it, where as PM has known from the 'get go' that the Craptor has numerous issues that need attention, and failed in almost every area to address them satisfactorily.Hubris, just like the Capt. of the Titanic

To be fair to the captain of the Titanic, he didn't know the iceberg was there until he hit it, where as PM has known from the 'get go' that the Craptor has numerous issues that need attention, and failed in almost every area to address them satisfactorily.

Captain of the Titanic was an idiot...he took the fastest course ( through the north where the icebergs were ) as he tought Titanic is invincible.To be fair to the captain of the Titanic, he didn't know the iceberg was there until he hit it, where as PM has known from the 'get go' that the Craptor has numerous issues that need attention, and failed in almost every area to address them satisfactorily.

At least with all the excess heat the Craptor is producing icebergs won't be a problem with PM !!!Captain of the Titanic was an idiot...he took the fastest course ( through the north where the icebergs were ) as he tought Titanic is invincible.

Well, if PM test flies it then 'pilot error' would cover everything!I’m trying to imagine just what the NTSB, with so many options, will identify as the root cause.

BJC

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How do you figure that?...he needs about 6 additional thrust hp to climb 60 feet per minute or about 60 additional thrust hp to climb at 600 feet per minute. (3400 pounds)

Pv = m×g×vh

Vh = vertical speed.

Vh = vertical speed.

I just estimate in my head. I know a typical 550 pound ultralight needs one thrust hp per second for 60 feet per minute climb.How do you figure that?

So his weight of 3300-3400 is about 6 times more than an ultralight.

Ok. It didn't look small on the outside. I've never seen another canard takeoff and do that..

Edit: swapped email with the pilot. Said he was fully loaded and had the trim too far back. Got about three small pitch bucks by the time he got the trim set. Apparently they were not full stalls on the canard.

I've been told EZ's can bob up and down without loosing altitude if they have some power on. That's why I assumed he stalled the canard. My mistake.

Sometimes people do dumb things in front of a big crowd of airplane enthusiasts.

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Planes with the main gear placed well aft seem prone to PIO on departure. Even the MU-2 is like that and it has its front parts in the front and back parts in the back. Doesn’t have to be a canard to draw a pilot into PIO.

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Well, there's a little more to it than that...but your rule of thumb turns out to be fairly close, at least for this class of aircraft.I just estimate in my head. I know a typical 550 pound ultralight needs one thrust hp per second for 60 feet per minute climb.

So his weight of 3300-3400 is about 6 times more than an ultralight.

In order to realistically predict an airplane's climb performance we must first have a realistic idea of a couple of things: the airplane's drag characteristics, and its power.

I had previously detailed an analysis method for extracting the key drag parameter, parasite drag coefficient Cdo, from airplanes for which we have accurate data on engine power and best glide speed, since best glide speed is the aerodynamic condition in which zero-lift drag [parasite] exactly equals drag due to lift, Cdi.

From this, it is straightforward to extract the unknown Cdo as well as the Oswald efficiency factor and prop efficiency. This lets us extract the entire airplane drag polar as well as the thrust required curve, as shown in the drag calculator spreadsheet I worked up.

The power required curve comes from the thrust required curve. Power is physically force times velocity. We simply multiply our thrust force required [which is also our total drag] by the airplane velocity to get the power required to overcome drag in straight and level flight.

In order to climb, at any speed or altitude, we must have power available in

So if we look again at the SR22 example I have been using, we see that at SL ISA, best climb speed is given as 91 knots indicated, and a rate of climb of 1,304 ft/min. That climb speed is very close to the airplane's best glide speed of 88 KIAS.

So how much excess power does this airplane have at this speed? Well, that is fairly easy to compute once you have run the spreadsheet to find the thrust required curve. For instance, at 150 ft/s speed which is 89 KIAS we get a total drag of 222 lbf, which is also our thrust required. Our total flight power at this speed is then 222 lbf * 150 ft/s = ~33,200 ft*lbf/s. That's only about 60 hp [1hp = 550 ft*lbf/s].

With an engine of about 320 hp we see that we have plenty of power in reserve for climbing. But we also must take prop efficiency into account. We found in the drag and takeoff analysis that during the takeoff run our prop efficiency is hovering only around 50 percent.

At this climb speed our prop efficiency turns out to be about 61 percent. [We find power available by rearranging our above relation: Pa = W * Rc + Pr = 3,400 lbm * 22 ft/s + 33,200 ft*lbf/s = ~107,000 ft*lbf/s = ~ 195 THP.]

Our prop thrust horsepower of 195 is only about 0.61 of the full engine power of 320.

So we see that we have about 135 THP in reserve, if we subtract the 60 THP required to fly at that speed. That works out to about 10 ft/min per each THP, which is in fact similar to your shorthand observation of 600 ft/min for 60 THP.

So where does this leave us with Raptor? Can we apply the same shorthand and expect to get a realistic result?

Let's see. We know that he needs to reach about 90 knots just to rotate, so let's say he can climb at 110 KIAS [185 ft/s]. To first get this airplane's power required in straight and level at this speed, we have to have some realistic idea about its drag.

Unfortunately we can only try to make a reasonable estimate here. Let's look at the SR22: at the same speed of 110 KIAS it is going to have 243 lbf of drag in straight and level. Its drag area is then 243 / dynamic pressure of ~41 lbf/ft^2 = 6 ft^2.

Notice that this is a lot higher than the SR22's drag area in cruise of 4.5 ft^2. The reason is that we are flying at this low speed with a lot more alpha. Our CL here is about 0.6, which means an AOA of about 6 degrees. This means the airplane is presenting a lot more cross section area to the wind, as the wing tilts up and the aft part of the wing [and fuselage] that is normally tucked neatly behind the LE at low alpha is now in the wind stream.

Now obviously the Raptor with its much bigger fuselage and 30 percent bigger wing is going to have a bigger drag area, perhaps quite a bit [and that's not even taking cooling drag into consideration, which is a mess on Raptor]. Assuming conservatively that Raptor's drag area at the same 110 knots indicated is only 25 percent bigger than SR22, then that still give us 7.5 ft^2.

At the same 41 lbf/ft^2 dynamic pressure that is a total drag and thrust required of just over 300 lbf. That is power required of 300 lbf * 185 ft/s = 55,500 ft*lbf/s = 100 THP. Considering prop efficiency is only about 0.61 at this speed* that is already 165 engine hp just to go straight and level.

We previously found engine power from his takeoff acceleration to be about 230 hp. That gives power available of 230 * 550 * 0.61 = ~77,000 ft*lbf/s. Subtracting the ~55,000 power required leaves only about 22,000 ft*lbf/s, which is only about 40 hp.

Rate of climb is then Rc = Pa - Pr / W = 77,000 - 55,000 / 3,500 = 6 ft/s = 360 ft/min.

We see also that your approximate rule of thumb is still decently close, 40 hp * 10 ft/min/hp = ~400 ft/min. Not a bad rule of thumb.

* Why is prop efficiency so low at lower speeds? It has to do with the work required to draw air into the prop disk at low speeds. This is called the induced velocity in momentum theory of propulsion. At high speeds the air is already rushing into the disk so very little work is required to be expended on the induced velocity. At low speeds a lot of the work going into the prop is just bringing the air in to begin with. At the static condition, the overall prop efficiency of a piston airplane will typically be only 1 or 2 percent.

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Thanks for that confirmation.We see also that your approximate rule of thumb is still decently close, 40 hp * 10 ft/min/hp = ~400 ft/min. Not a bad rule of thumb.

I mentioned in post 6611 that I think he could do a careful 5-10' agl level run of 2000 feet to record the fuel flow and rpm at 90kts and estimate hp.

From there he might be able to estimate if the extra 40 hp at a higher rpm and fuel flow would be available. The actual hp isn't reliably measured but he could at least estimate a reasonable percentage needed. Then he might know if flight should or should not be attempted until more power is found.

If the test run calculations were at 90kts (somewhat below best climb) then there would be even more climb reserve for his actual takeoff using best climb speed.

Comparing Orion (or most other canards) to Raptor would to tantamount to comparing a slick newThe Orion canard is very similar to Raptor. It would be interesting to compare the weight and performance of the Orion.

to a

Yes, both are mid/aft-engined, but only one draws the ladies, and the other was mired in controversy!

Never needed any mechanical assistance ....Yes, both are mid-engined, but only one draws the ladies!

BJC

Likewise. I was on foot in Riga, Latvia when I met my lass some years ago!Never needed any mechanical assistance ....

BJC

These two airframes were largely designed and built by the same person, as far as I can tell from research on the web.Comparing Orion (or most other canards) to Raptor would to tantamount to comparing a slick new...

The engine and pressurization is of course different. That's why I am curious of the Orion weight.

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