Mathematical Models For Candle Balloon Design -- Overflite --

As the basis for theoretical design, different sized balloons are imagined, all heating to the same temperature.  As balloon volume increases, the bag, frame and engine weight increase too, but by less than the increase in volume and gross lift.  Hence, the relative net lift goes up, and the engine burntime can be increased, by adding on more wax.   See Design Synopsis

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Contents of Page: 

Geometric Mathematics General Heating Properties of Candle Powered Balloons
- Geometric Model for Equivilent Heat Rise
- Suggested Number of Candles for Different Sized Balloons
- Arithmetic and Geometric Models for Candle and Volume Changes
- Balloon Duration -- The Underlying Theoretical Basis for Hot Air Balloon Flight -

Geometric Mathematics

General Heating Properties of Candle Powered Balloons

The Maximum Sustainable Heat Rise depends on the wind conditions.  It is obviously less than the Peak Equilibrium Heat Rise.  Heat loss by conduction is less.  Heat loss by convection is more.  Heat rise can still be viewed as an equilibrium though, since long heating times are assumed.  Heat loss can still be viewed as an accurate proxy for the heating rate.

The Minimum Sustainable Heat Rise is more important.  When balloons are launched, they need to be able to recover quickly from wind blasts.  Hence, the assumed basis is how hot balloons get in a reasonably short heating time.  Here the heating rate is greater than the heat loss.  Technically this means that the heat loss is no longer a highly accurate proxy for the heating rate.  For modelling purposes though, it is still close enough to the actual reality to design balloons.

To compare balloons with different volumes and candlepowers, the mathematical model assumes Comparably Equivilent Heating Times.  Here, large balloons are "allowed" more time to heat than small balloons.  In a sense though, this works as a built-in self-fulfilling prophecy, since at some point in the heating time, the heat rise will match the prediction.

The model also assumes that different sized balloons will have similar ratios between the different measurements for heat rise.  But this is not strictly true. As balloons get larger the different measurements would seem to converge.  Hence, larger balloons probably have a higher Minimum Sustainable Heat Rise than predicted by the model.

With more candles, or less volume, the heat rise is more, and the heating rate is faster.  With fewer candles, or more volume, the heat rise is less, and the heating rate is slower.  Basically what happens is that balloons store up heat energy, over time, until they reach a maximum capacity, where the rate of heat loss becomes equal to the rate of heat input.

If the candles are increased, then the heating per candle goes down.  This demonstrates that the heating is less efficient.  Heat loss goes up, both through conduction and convection.  But the rate is less than the candle increase.  More heat energy is stored, and the gross lift is more.

Alternately, if the volume is increased, with the same number of candles, then the heat rise will go down, but by less than the volume increase.  Hence, the gross lift is more.  This suggests that the heating is more efficient.  But more so, what is really happening is that the balloon is storing up more heat energy, over a longer length of time.

In cold weather, the heat rise goes down, slightly. This is because there is more air to heat.  Equivilently, in hot weather, or at high elevations, the heat rise goes up, since there is less air to heat.  The effect is assumed to be around 1 % per 10 degree change in ambient temperature, and around 1 1/2 % per thousand feet of elevation.  The assumption here is that the heat rise changes by around the square root of the change in the mass of the ambient air displaced by the balloon.

Geometric Model for Equivilent Heat Rise

Change in Volume  =  Change in Dimensions ^3
Change in Surface Area  =  Change in Dimensions ^2

Change in Volume  =  ( Change in Surface Area ^1/2 ) ^3  =  Change in Surface Area ^3/2
Change in Surface Area =  ( Change in Volume ^1/3 ) ^2  =  Change in Volume ^2/3

Since the heated temperature is the same for all of the balloons, the heat loss through conduction changes by the difference in surface area.  This is equal to the change in volume^2/3.  This suggests that the difference in candles should probably also equal around the change in volume^2/3.  If so, then the airflow into and out of the balloon will change by this rate too, and the heat loss through convection will probably also change at around this rate.  All this suggests the following formula:

Change in Candles to get Equivilent Heat Rise  =  Change in Volume ^2/3

In reality though, larger balloons have less relative air turnover than small balloons.  This means that the air flowing out of the bag has more time to cool, so less heat is lost than predicted.  Hence, larger balloons should get roughly equivilent heating with somewhat fewer candles than predicted by this simplified mathematical model.  For simpler balloons this doesn't make much of a difference.  For advanced balloon designs the exact rate of roughly equililent heating has an effect on  Balloon Duration.
 

Change in Candles to Get Roughly Equivilent Heating -- Approximated at Change in Volume ^ 2/3

Volume	One	Two	Three	Four	Five	Six	Seven	Eight	Nine	Ten
Vol.^2/3	1	1.59	2.08	2.52	2.92	3.30	3.66	4	4.33	4.64

Suggested Number of Candles for Different Sized Balloons

At 28 Fahrenheit (488 Absolute), the air weighs 635/488, or 1.30 ounces per cubic foot.  The heat rise goes down by a few degrees as described, and is assumed to be 134 degrees.  Here, at 162 Fahrenheit (622 Absolute), the air weighs 635/622, or 1.02 ounces per cubic foot.  Hence the gross lift is around .28 ounces per cubic foot, or around 12% more.

As a practical matter, especially where the lift is significantly more than the weight, it does not matter all that much exactly how many candles are used.  Fewer candles will work out perfectly okay.  In hot weather though, it is not a good idea to use more than the suggested number of candles, since the balloons could heat up past the melting point of the plastic.
 

Estimated  # of Candles for Equivilent Heating of +130 to 140 degrees Farenheit = Around 7 * (Volume ^ 2/3)

CubicFeet	one	five	ten	fifteen	twenty	twenty-five	thirty	thirty-five	forty	forty-five	fifty
# candles	7	20	32	42	50	58	66	73	80	86	93
69 F Lift	.25 oz.	1.25 oz.	2.50 oz.	3.75 oz.	5.00 oz.	6.25 oz.	7.50 oz.	8.75 oz.	10.00	11.25	12.50
28 F Lift	.28 oz.	1.40 oz.	2.80 oz.	4.20 oz	5.60 oz.	7.00 oz.	8.40 oz.	9.80 oz.	11.20	12.60	14.

Arithmetic Model for Candle and Volume Changes

So now the double-volume balloon gets a double-candlepower engine.  This heat rise is estimated.  It appears to be around 170% of the heat rise from the single-engine, or 119% of the heat rise of the single-volume balloon, ie. 170% of 70%.

To make the double-volume balloon get the same heat rise as the single-volume balloon, the 70% heat rise needs to increase by around 43%.  This should happen with 60% more candles, since the heat rise "yield" is around 70% of the candle increase.

 So now the double-volume balloon gets around the same heat rise as the original balloon.  This process can be repeated, to imagine larger or smaller scale balloons, or it can be fractionalized. The results are reasonably close to what actually happens.  But mathematically, the arithmetic model only works precisely when the volume is doubled.

Geometric Model for Candle and Volume Changes

The estimated  heating for the double-volume balloon, ie. 70%  equates to 1 / 1.41,  ie.  1 / ( 2 ^1/2).
The estimated heating for the double-engine balloon, ie. 170% equates to 1.68, ie.  2 ^3/4.
The estimated increase in candles to equalize the heating, ie. 60% equates to 1.59, ie.  2 ^2/3.
Hence the following formulas:

Change in Heat Rise  =  Change in Candles ^3/4  /  Change in Volume ^1/2
Change in Candles for Equivilent Heat Rise  =  Change in Volume ^2/3

Candle Equalizer Power, ie. 2/3  =  Volume Power, ie. 1/2 /  Candle Power, ie. 3/4
Volume Power  =  Candle Equalizer Power * Candle Power
Candle Power =  Volume Power / Candle Equalizer Power

The Candle Power is more efficient if it is higher.  The Volume Power is more efficient if it is lower.  The Candle Equalizer Power is more efficient if it is lower.  As demonstrated by the Geometric Model for Equivilent Heat rise, it makes sense that it should be around 2/3.  As long as the ratio between the Volume and Candle Powers is 2/3, then the Candle Equalizer Power will also be 2/3.

As example, the Candle Power could become less efficient, while the Volume Power becomes more efficient.  With a small balloon, or with a balloon with a relatively high candle to volume ratio, more candles would seem to increase the heat rise by less than expected, while more volume would seem to reduce the heat rise by less than expected.  Hence the Candle and Volume Powers would both seem to drift down.

Alternately,  the Candle Power could become more efficient, while the Volume Power becomes less efficient.  With a large balloon, or with a balloon with a relatively low candle to volume ratio, more candles would seem to increase the heat rise by more than expected, while more volume would seem to reduce the heat rise by more than expected.  Hence, the Candle and Volume Powers would both seem to drift up.

In reality, the Candle Equalizer Powers should tend to be less than 2/3.  As described, large balloons lose relatively less heat by convection than small balloons.  In addition, they will lose less heat when they get pummelled by the wind.  For comparison purposes, the Candle Equalizer Power can be idyllically imagined as drifting down towards around 3/5.  Another adjustment is to be liberal rather than conservative when making assumptions about the Minimum Sustainable Heat Rise.

Balloon Duration

As example, if a balloon volume is doubled, then 60% more candles should double the lift too.  Meanwhile the total weight increases by around 60%, if the engine burntime stays the same.  Hence, the total balloon weight can be increased by up to around 25%, by adding extra wax onto the engine, thereby increasing the burntime of the balloon.

Duration can be analyzed in terms of the Candle Equalizer Power.  For modelling purposes it is assumed to be 2/3.  As an idyllic comparison, it is also presented by what would happen if it were 3/5 instead:
 

Volume	1	2	4	6	8	16	24	32	48	64	128
CEP 2/3	1	1.59	2.52	3.30	4.00	6.35	8.32	10.08	13.20	16.00	25.40
CEP 3/5	1	1.52	2.29	2.93	3.48	5.27	6.73	8.00	10.20	12.12	18.37

As point of inquiry, the question is:  How much does the volume have to increase to double the Theoretical Minutes of Burntime?  Here, if we assume the Candle Equalizer Power to be 2/3, the answer is 8 times.  The balloon lift increases by 8 times.  The weight increases by 4 times.  Hence the balloon weight can be doubled.  This leads to the following formulas:

If the CEP = 2/3, then the Change in Balloon Duration =  Change in Volume ^1/3
If the CEP = 3/5, then the Change in Balloon Duration = Change in Volume ^2/5

Generically, through some calculus process, the equation can be written as follows:

Change in Balloon Duration =  1 / ( Change in Volume ^ ( Candle Equalizer Power - 1) )

Change in Balloon Duration =   Change in Volume ^ (  | Candle Equalizer Power - 1 | )

Calculations for Balloon Duration

The twenty candle five cubic foot dry cleaner bag balloon is assumed to lift approximately 1.25 ounces at 69 degrees Fahrenheit, and approximately 1.40 ounces at 28 degrees Fahrenheit.  Hence it would take 22.50 minutes for the candles to burn through 1.25 ounces of wax, and 25.20 minutes to burn through 1.40 ounces of wax.

These theoretical minutes of burntime are allocated between the different components of the balloon.  For the bag and frame, at 7/10 of an ounce, 12.60 minutes of burntime is accounted for.  The engine accounts for another 10.00 minutes.  In the case of the 28 degree ambient balloon, the net lift of .15 ounces accounts for another 2.70 minutes of burntime.

Theoretical Minutes of Burntime -- If Candle Equalizer Power = 2/3 -- At 69 Fahrenheit Ambient Sea Level

Volume	5	10	15	20	25	30	40	50	60	70	80
Units	1	2	3	4	5	6	8	10	12	14	16
Units^1/3	1	1.26	1.44	1.59	1.71	1.82	2.00	2.15	2.29	2.41	2.52
Minutes	22.50	28.34	32.45	35.71	38.47	40.88	45.00	48.47	51.51	54.22	56.70

How does the table get used?  Basically the weight of the bag and frame gets converted into theoretical burntime, then gets subtracted from the total theoretical burntime.  Also, for larger balloons a certain number of minutes should get subtracted for net lift, since the heating takes longer.  The number of minutes left is how long the candles can burn for.

If the bag size increases in scale, its theoretical burntime will stay the same.  In reality though it typically would get fatter, which would reduce its theoretical burntime.  But as a practical matter the difference isn't much.

With the 1/3 mil material, being lighter, 9 minutes of burntime is a reasonable subtraction.  For designing paper balloons, though, which are heavier, 20 to 30 minutes of theoretical burntime is probably called for.  Hence 30 to 40 cubic feet would apparently be needed to get a birthday candle powered paper balloon off the ground.  Stay tuned for more information here.

The table assumes the Candle Equalizer Power to be 2/3.  In reality though it is less.  Hence the numbers are conservative...

Stay Tuned For Advanced Calculations on Model hot Air Balloon  Design
By Thomas Taylor -- balloons@overflite.com
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