HAB Calculation HELP

[Shivaram S P]

  Hello,

I am new to HAB. I am planning to launch a weather balloon.

My balloon specification : 

My Parachute specification : 

My payload weight include weight of Payload + Parachute and Line etc. will be 2560g

I calculated by Burst altitude :  http://www.randomengineering.co.uk/Random_Aerospace/Balloons.html 

I predicted with : https://predict.sondehub.org/

And how accurate is this predict sondehub ?

I don't know how to find out my Descent Rate with my parachute (m/s): Can anyone let me know how to calculate.

How I have planned to launch the balloon : https://youtu.be/6_06Q_eWta8

calculate lift for a weather balloon : https://youtu.be/NxwlNnfKMHs

How to fill a weather balloon : https://youtu.be/5Z23L4QIgtQ

Also let me know If I am making any mistakes

[Steve]

Here are a few comments

    Steve G8KHW

On 22/07/2023 11:27, Shivaram S P wrote:

My payload weight include weight of Payload + Parachute and Line etc. will be 2560g

I calculated by Burst altitude :  http://www.randomengineering.co.uk/Random_Aerospace/Balloons.html 

I predicted with : https://predict.sondehub.org/

And how accurate is this predict sondehub ?

SR: Predication accuracy depends a lot on how accurately you measure free lift (and that depends a lot on how windy the balloon fill conditions are). Likewise Parachute descent rate. Another big factor is how long the flight path is - short flight paths give more accurate landing spot prediction, longer paths less absolute accuracy (accuracy more like a % of flight length).  wind changeability is also a factor (i.e. have the predictions have been changing a lot in the 48 hours ahead of the launch) .  Prediction is less accurate in the general direction of travel of the flight  more so at right angles to that.  As an example for a flight that travels 50Km from the launch site I'd generally expect to land within a 10Km x 5km oval centered on the prediction.

I don't know how to find out my Descent Rate with my parachute (m/s): Can anyone let me know how to calculate.

SR: to calculate descent rate you need to know the Cd (coefficient of drag) for your parachute - which will be dependent on its design (Parasheet/Hemispherical/Parabolic spill holes ... etc.).  The manufacturer should know this - I find it surprising it (or descent rate) is not listed in the data. A 3m chute seems a bit big for a 2.5Kg payload (as implied by the Payload Range data). Too big is no real problem if you don't mind traveling further to recover it.

There is a Parachute descent rate calculator here: http://randomaerospace.com/Random_Aerospace/Parachutes.html.

How I have planned to launch the balloon : https://youtu.be/6_06Q_eWta8

calculate lift for a weather balloon : https://youtu.be/NxwlNnfKMHs

How to fill a weather balloon : https://youtu.be/5Z23L4QIgtQ

SR: those generally seem OK (I didn't have time to watch them all through).  Personally I would not use the Free lift = 1.5x all up weight advice and calculate free lift using the Burst Calculator.   But what he says is golden:- overfill rather than underfill.  The desired 8m/sec ascent rate (In the burst estimator) is very high unless you have some specific reason. Such a high rate is generally a waste of gas and makes a lot of turbulence (which results in payload movement - bad for any video footage if that's an aim). I aim for 5.5m/sec these days.

Dave A's guides are a good starting point:

    http://www.daveakerman.com/?p=1732

Virus-free.www.avg.com

[Shivaram S P]

Hi,

Thanks for your valuable Info 

• Predication accuracy : I have taken a note on this
  
• calculate descent rate : I contacted my manufacturer and asked  I want to know the details for the field : Chute "Coefficent of Drag x Area" to calculate my descent speed / They replied :  This is closer to a hemispherical parachute and Cd will be around 1.5, Considering that the shape is hemispherical the area can be calculated using the canopy diameter given in the specifications. Can you help me on this if possible ?
• I have also changed my ascend speed to 5.5m/sec.

Thanks again :)

[John Laidler]

The drag formula is:

D = Cd * A * 0.5 * r * V * V

Where 

D = the drag in Newtons. You need to convert your payload from Kg to Newton's by multiplying by 9.81

A = area of the parachute in flight in metres. (pi * diameter /4)

r = air density,  Kg per cubic metres

V = the velocity of in your case the descent rate. 

So your descent rate (m/s) will be the square root of:

D/(Cd * A * 0.5 * r) 

I think the above is correct but the Earth was a lot younger when I learned all this stuff! 

John

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[Shivaram S P]

Hello 

Thanks for the valuable info: 
I made this calculation. But I don't think its right. Can you correct me where I am making a mistake 

Payload: 2100 grams

Coefficient of Drag (Cd): 1.5 (as provided by the manufacturer)

Canopy Diameter: 300 cm

Air Density (r): Approximately 1.225 kg/m³ (average at sea level)

Step 1: Convert Payload to Newtons

Payload in Newtons (N) = (2100 grams / 1000) * 9.81 m/s² ≈ 20.58 N

Step 2: Calculate Area of the Parachute in Flight

Area of Parachute (A) = π * (Canopy Diameter / 2)^2

A = π * (300 / 2)^2 ≈ 706.86 cm² ≈ 0.0707 m²

Step 3: Calculate Descent Rate (m/s)

Descent Rate (V) = √(D / (Cd * A * 0.5 * r))

Substitute the values:

V = √(20.58 N / (1.5 * 0.0707 m² * 0.5 * 1.225 kg/m³))

V ≈ √(20.58 N / 0.0530 kg*m²/s²)

V ≈ √(388.68 m²/s²)

V ≈ 19.72 m/s (approximately)

 

[David Akerman]

A = π * (300 / 2)^2 ≈ 706.86 cm² ≈ 0.0707 m²

You're out by a factor of 100 here.  Just do it in metres:  

A = π * (3 / 2)^2 ≈ 7.065 m²

Which, btw, is huge for your payload weight.

Dave

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[Steve]

You will find a lot of guff about parachute areas and Cd on the internet.  Do not use Cd as a way of comparing parachutes effectiveness unless the manufacturer tells you the way Area is being calculated and how that relates to their "size" measurement.  What the parachute reference area A is seems to vary from manufacturer to manufacturer:

e.g. one manufacturer might measure area as cross sectional area (CSA) in flight and have a Cd of X - but a different manufacturer might measure the Surface Area (SA) of  the same parachute design and have a Cd of Y.  By choosing to use CSA rather than Surface Area (SA) you can get a higher Cd number and make the parachutes look better on paper.  For a hemispherical chute the SA will be double the CSA (and hence SA measured Cd will be half of that CSA).

Which method is correct? - I believe the professionals use Surface Area.

Also - what is the dimension the parachute "size" is measured and how is this turned into a reference area (that they are measuring Cd against).

As John says correct drag force D = Cd * A * 0.5 * rho * V * V 

In terminal descent that force is in balance with the weight W of the payload + parachute.  The force (in N) is weight x acceleration due to gravity.

so (W x 9.81) = Cd * A * 0.5 * rho * V * V

and hence V * V = (W * 9.81) / (Cd * A * 0.5 * rho)

hence velocity V = SQRT( (W * 9.81) / (Cd * A * 0.5 * rho))

the usual value of air density "rho" is taken to be approximately 1.225 at sea level (15C 101.325 kPa)

A is the reference Area.

    Steve

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[Steve]

If the parachute is one of these (T-PF01-290):

    http://nwzimg.wezhan.cn/contents/sitefiles2045/10226828/images/20013375.jpg

Then its probably a para-sheet (a single circular sheet of fabric with lines attached) - the diameter is probably that of the sheet laid flat.

Para-sheets are said to have a Cd of about 0.75 using the sheet surface area as a reference.  (The claimed 1.5 must be relative to in flight CSA).

In general para-sheets perform poorly relative to their size and weight - as Dave says generally a 3m chute would be considered huge for 2.1Kg of payload.

So surface area A = Pi * D / 4 = 3.142 * 3.0 / 4  = 2.3565 sq m

The payload + parachute weight is 2.1Kg + 0.37Kg = 2.47Kg       * 9.81 =  24.231 Newtons of force

using rho of 1.225Kg/cu m

V = SQRT( (W * 9.81) / (Cd * A * 0.5 * rho))

V = SQRT (24.231 / (0.75 * 2.3565 * 0.5 * 1.225)   =  4.7m/sec

    Steve

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[John Laidler]

I get a slightly different figure as 1.5 * 0.0707 * 0.5 * 1.225 = 0.06496 not 0.0530

This gives a velocity of 17.8 m/s which is still very high but it seems reasonable for a 2.1Kg payload under a 30cm diameter parachute.  It is the sort of speed which is likely to cause damage to the payload when it lands and would certainly hurt someone or damage property.

A typical payload is usually just few hundred grams say in the range 200g - 300g.  I suggest you need to look at a bigger parachute and aim for a descent rate of around 5m/s perhaps?  Alternatively, try and get the weight of the payload down.

John   

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[Steve]

The chute is 3.0m (300cm) in diameter:

     Steve

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[David Akerman]

It's a 300cm chute John, not 30cm.

Unless it's a rather poor chute (which it sounds like it is - see Steve's post) then it's too large, not too small.

Dave

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[John Laidler]

Dave,

Thank you, I didn't spot that - but should have guessed from the payload it can carry!

John

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