# Basic Aerodynamics Question Total Drag from Glide Polar

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#### proppastie

##### Well-Known Member
Log Member
Picking through my second reading of Hiscock and trying to sub my numbers (no surprise) I have problems. His data for the Dox is RN=10,000,000 and my calculations show my bird has RN=1,000,000. (row 7) I am trying to calculate the total drag for HP required in climb at best L/D along with reading along with Hiscock. I do not have the airfoil data for my bird but do have the Glide polar. I would think one could calculate the total Drag with the glide polar. Looking at a spread sheet with the expressions, the questions arise. Maybe one could point to a thread on the site, surly it has been discussed. [FONT=&amp] [/FONT] 1: rows 8, 12, 13….if the glider is sinking ….. they can not all be correct? 2: row 13, 9 same question

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#### WonderousMountain

##### Well-Known Member
Sure,

Take (weight x sink)/V. You'll need forward velocity in FPS.

If you have a sink rate of 32 FPS, I think you have bigger problems than engine hp needs. Isn't that the rate of acceleration from gravity?

LuPi

Using 320# x 1.75 sink at 47fps, I got twelve

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#### Matt G.

##### Well-Known Member
If you have a sink rate of 32 FPS, I think you have bigger problems than engine hp needs. Isn't that the rate of acceleration from gravity?
No, because feet/second is velocity. Acceleration would be feet/second/second.

#### proppastie

##### Well-Known Member
Log Member
Sure,

Take (weight x sink)/V. You'll need forward velocity in FPS.

If you have a sink rate of 32 FPS, I think you have bigger problems than engine hp needs. Isn't that the rate of acceleration from gravity?

LuPi

Using 320# x 1.75 sink at 47fps, I got twelve

32 mph=47 ft/sec 320*1.75/47=12 lb (lb*ft/sec)/ft/sec=lb

yes thanks if it is that easy

so now I have the total drag

are my questions 1 & 2 valid questions?

I guess when I factor in the Drag it will all work out.

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#### Birdman100

##### Well-Known Member
proppastie,

with glide polar available you can extract a lot of info from it, you dont need airfoils characteristics at all. Actually airfoil itself is incorporated in polar by the person who draw it.
Glide polar (or speed) polar has same shape as Cl-Cd polar. Ok, as I can see from graph best L/D is 25. That means your drag at that point (speed) is 25 times less than lift, and cause lift is equal weight in horizontal flight you have:
320/25=12.8 lb of drag. Your engine (prop actually) needs to produce same amount of thrust (=12.8 pounds)to maintain that (horizontal) flight.

If you want to climb, you need to add thrust needed for climb and that is simple: =Weight*SIN (climb angle)

So, required overall thrust is = 12.8 + 320 *SIN (climb angle)

If for example climb angle is 10 deg you need 68.4 lb. The power of flight is simply speed times force (thrust) and for the example given it is 4350 watts. If you have 60% efficient prop. you need 4350/0.6=7.25 kW of engine power for that flight regime or that is 9.7 HP

#### Birdman100

##### Well-Known Member
excel seems correct. Why are you suspicious?

#### proppastie

##### Well-Known Member
Log Member
excel seems correct. Why are you suspicious?
Novice, not sure of anything. Thank you very much for another basic simple easy to understand addition to my quest.

#### proppastie

##### Well-Known Member
Log Member
 1hp=746 watts. 1hp=746 watts.=550ft-lb/sec 68.4lb*47fps=3215ft-lb/sec=4360watts=5.8hp

/.6=9.7hp

For completeness if you do not mind

I always like to see the units I think it is a good habit to get into.

Much of the literature makes way too many assumptions and often changes symbols from page to page, which is very frustrating to us novices.

Again thanks both of you very much you made it much clearer than anything I have seen.

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#### dino

##### Well-Known Member
HBA Supporter
I think the the hardest part is doing the calcs in Imperial units. Self inflicted injury. Occasionally the inability of society to accept change is staggering.

Dino

#### bmcj

##### Well-Known Member
HBA Supporter
I think the the hardest part is doing the calcs in Imperial units. Self inflicted injury. Occasionally the inability of society to accept change is staggering.

Dino
Don't worry, I'm sure the rest of the world will finally give in and accept the Imperial System. :gig:

#### BJC

##### Well-Known Member
HBA Supporter
I think the the hardest part is doing the calcs in Imperial units. Self inflicted injury. Occasionally the inability of society to accept change is staggering.

Dino
So use SI.

#### Highplains

##### Well-Known Member
When did we get 100 seconds to the minute or 100 minutes to the hour or 10 hour to the day.......

A system based on the distance from the equator to north pole? Who said the Earth is not the center of the universe?

#### Birdman100

##### Well-Known Member
Don't worry, I'm sure the rest of the world will finally give in and accept the Imperial System.:gig:
Oh, yes... And then the absurdity to name that system "Imperial"...onder:

Oh, wait! I think I have more trouble with the word "System" there...

#### Aesquire

##### Well-Known Member
100 seconds to the minute was tried in post revolutionary France. Big fail, people are stubborn.

The U.S. officially went metric in the 19th century, 1895 iirc.

We just didn't pay attention. No one wanted to replace all the tools. So it goes.

Consider that the inch is based on a dead King's thumb.
The Meter is based on a platinum stick. Based on a measurement of the Earth's size.
Only later was a Meter redefined to a certain ( & seemingly random) number of wavelengths of a specific color of light.

If new thoughts on quantum physics are correct a Meter will be a different length elsewhere. ....... which just messes everything up. ( and probably proves Nicola Tesla right, again)

It's amazing how much data you can get out of a polar. ( and how much goes in )

#### Birdman100

##### Well-Known Member
If new thoughts on quantum physics are correct a Meter will be a different length elsewhere. ....... which just messes everything up. ( and probably proves Nicola Tesla right, again)
Thats why this definition of meter is obsolete, newer one is based upon speed of light in vacuum (which is constant) and time... That also means we are capable to measure time more precisely than length. And yes, AFAIK inch is 25,4 mm... :gig: Or one Kings thumb for those who prefer it more.

#### Jan Carlsson

##### Well-Known Member
You have the L/D at 32 MPH = 14,3 m/s

drag is W / L/D = 12,8 lb or 5,8 kg (57 N)

best climb speed is some 10% or so higher, (due to better efficiency and more rpm at higher speed) so 14,3*1,1 = 15,7 m/s

and L/D little lower say 22 or 20 so D= 7,25 kg (need to ad cooling drag and installation drag from engine)

you want a climb of 500 ft/min or 2,5 meter per second

climb rate is thrust minus drag / weight * speed

w 320 lb or 145 kg

2,5/V*145+D= thrust needed ~ 30 kg

thrust is HP x 76kg (or 75kg) * efficiency / V (the hard part is the efficiency)

to get 30 KG thrust at 15,7 m/s

Efficiency on a small prop might be 35-50% at climb

30*15,7/35%/76=17,7 HP imperial

30*15,7/50%/76=12,4 HP imperial

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#### BJC

##### Well-Known Member
HBA Supporter
It's amazing how much data you can get out of a polar. ( and how much goes in )

Or a post on HBA. The diversity of experience here is what keeps me logging on.

(Yes, I know that, just as with all internet forums, a filter can be useful.)

BJC

#### proppastie

##### Well-Known Member
Log Member
For those of us that like to check our trig with pictures. Thanks much Jan.

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#### proppastie

##### Well-Known Member
Log Member
30*15,7/35%/76=17,7 HP imperial

30*15,7/50%/76=12,4 HP imperial
Well I had not planned that this discussion would go this far.. but since it has, the current plan is (subject to change):

Desert Engine which claims 16 hp. at 8 lb. + exhaust + Prop + Gas and Tank, but perhaps all of that will be within the 320# gross.

Yes drag will be higher in climb.

Numbers work out to 300 ft. to over 50 ft obstacle, after I break ground, so I might be able to live with a little less climb, or who knows maybe the engine actually puts out 16 hp.

I am also thinking about curtains or hinged boards to cover the engine/prop after shut down, which should bring my performance back up. It will be a pusher installation, somewhat like the American Eaglet

There will be no provision for air re-start.

I am thinking I need good Spoilers (Frieze type?) in the wings.

Long as I have you here can you throw some numbers at me for "Profile Drag" of these devices.

Guess What I mean to say is what is considered safe aerodynamically as regards these devices., and or a good simple reference.

Edit

http://www.homebuiltairplanes.com/forums/aircraft-design-aerodynamics-new-technology/12028-desirable-l-d-approach.html

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#### Vigilant1

##### Well-Known Member
Efficiency on a small prop might be 35-50% at climb

30*15,7/35%/76=17,7 HP imperial

30*15,7/50%/76=12,4 HP imperial
Jan,
First, thanks very much for that post, it was really useful as we mulled over some issues on the "Beetlemaster" thread.
Question--why such a low efficiency for the prop? Would we expect just 35% - 50% for an appropriately-pitched 54" prop at 3200 RPM at climb (say 65 knots on a VW).

Mark

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