Quantcast

Basic Aerodynamics Question Total Drag from Glide Polar

HomeBuiltAirplanes.com

Help Support HomeBuiltAirplanes.com:

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
4,762
Location
NJ
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] TOTAL DRAG FROM GLIDE IG POLAR.jpg [/FONT] 1: rows 8, 12, 13….if the glider is sinking ….. they can not all be correct? 2: row 13, 9 same question carbon_dragon_polar_v_diagram_large.jpg
 

Attachments

Last edited:

WonderousMountain

Well-Known Member
Joined
Apr 10, 2010
Messages
2,124
Location
Clatsop, Or
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
 
Last edited:

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
4,762
Location
NJ
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.
 
Last edited:

Birdman100

Well-Known Member
Joined
Jun 12, 2013
Messages
807
Location
Novi Sad, Vojvodina
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
 

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
4,762
Location
NJ
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.
 
Last edited:

dino

Well-Known Member
HBA Supporter
Joined
Sep 18, 2007
Messages
681
Location
florida
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
Joined
Apr 10, 2007
Messages
13,541
Location
Fresno, California
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:
 

Highplains

Well-Known Member
Joined
Jun 2, 2014
Messages
356
Location
Over the Rainbow in Kansas, USA
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?
 

Aesquire

Well-Known Member
Joined
Jul 28, 2014
Messages
2,446
Location
Rochester, NY, USA
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
Joined
Jun 12, 2013
Messages
807
Location
Novi Sad, Vojvodina
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
Joined
Jan 11, 2009
Messages
1,860
Location
Sweden
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
 
Last edited:

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
4,762
Location
NJ
For those of us that like to check our trig with pictures. Thanks much Jan.


climb.jpg
 
Last edited:

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
4,762
Location
NJ
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

I found a good thread

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

Vigilant1

Well-Known Member
Lifetime Supporter
Joined
Jan 24, 2011
Messages
5,365
Location
US
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
 
Last edited:
Top