Hot Air Balloons: how is the Lift related to environmental conditions? I'd like to build a small hot air balloon big enough to carry 0.8kg (a camera). The first thing I need to know is: once I have a big enough plastic bag and a big enough source of fire, how much will ambient environmental conditions effect the amount of lift (i.e. whether or not it can get off the ground)?
For example:


*

*How does atmospheric pressure contribute to lift?

*How does ambient air temperature contribute?

*How does hot sun contribute?

*Are there any simple formulae where I can estimate the percent difference in lift I will get given various values for the above variables?

 A: Punchline:  The lift is directly proportional to the ambient pressure, and inversely proportional to the ambient (Kelvin) temperature.  
Here's the derivation with some other related observations:
Let's say that the balloon has volume $V$.  The gas inside has some temperature $T_\mathrm{in}$ and pressure $P_\mathrm{in}$.  The gas outside has some temperature $T_\mathrm{out}$ and pressure $P_\mathrm{out}$.  The gas on the inside of the balloon satisfies the ideal gas law:
\begin{align}
  P_\mathrm{in}V = N_\mathrm{in}k T_\mathrm{in}
\end{align}
Where $N_\mathrm{in}$ is the number of gas molecules, and $k$ is Boltzmann's constant.  If we let $\mu$ denote the mass of each gas molecule, then $\mu N/V = \rho_\mathrm{in}$ is the density of the gas inside the balloon, so we can rewrite the ideal gas law inside as
\begin{align}
  P_\mathrm{in} = \frac{\rho_\mathrm{in}}{\mu}kT_\mathrm{in}
\end{align}
We can therefore write a similar expression for the gas outside the balloon:
\begin{align}
  P_\mathrm{out} = \frac{\rho_\mathrm{out}}{\mu}kT_\mathrm{out}
\end{align}
If the gas pressure inside and the gas pressure outside are unequal, gas will flow through the opening in the balloon, and the pressures will equalize, so the inside and outside pressures are taken to be equal $P_\mathrm{in} = P_\mathrm{out}$.  If we use this fact with the inside and outside ideal gas law statements, we obtain
\begin{align}
  \rho_\mathrm{in}T_\mathrm{in} = \rho_\mathrm{out} T_\mathrm{out}
\end{align}
This show that if the temperature on the inside is greater then the temperature on the outside, then the density on the inside will be lower.  This means that the balloon will float: here's why.
Archimedes' principle says that the bouyant force on an object in a fluid equals the weight of fluid displaced by the object.  In this case, the balloon will displace a mass of fluid given by $\rho_\mathrm{out} V$ where $\rho_\mathrm{out}$ is the density of the ambient air.  It follows that the buoyant force generated by the balloon is
\begin{align}
  F_\mathrm{buoyant} = \rho_\mathrm{out} Vg
\end{align}
On the other hand, the weight of the air in the balloon is
\begin{align}
  W_\mathrm{in} = \rho_\mathrm{in} Vg
\end{align}
So if the density inside is less than the density outside, then the buoyant force will be great than the weight, and the balloon will float.  Note that we are assuming here that the balloon's material is light enough that its weight can be neglected.
What about if you want to support something else of mass $M$ with the balloon (like a camera)?  Well in this case we want the buoyant force to counteract both the weight of the air, and the weight of the object.  So we get the condition
\begin{align}
  \rho_\mathrm{out} Vg \ge \rho_\mathrm{in} V g + Mg = (\rho_\mathrm{in} V  + M)g
\end{align}
we can rewrite this as
\begin{align}
  \rho_\mathrm{out} - \rho_\mathrm{in} \ge \frac{M}{V}.
\end{align}
Using the ideal gas law for the air inside and outside, we can write this as
\begin{align}
  \frac{1}{T_\mathrm{out}} - \frac{1}{T_\mathrm{in}} \ge \frac{kM}{\mu P_\mathrm{out}V}
\end{align}
which gives
\begin{align}
  \boxed{T_\mathrm{in} \ge \left(\frac{1}{T_\mathrm{out}}-\frac{kM}{\mu P_\mathrm{out}V}\right)^{-1}}
\end{align}
This allows you to compute the minimum temperature you need to keep the inside of the balloon at so that you can support the mass $M$ given the volume $V$ of the balloon, the molecular mass $\mu$ of the ambient air (which you can look up), and the outside temperature and pressure.  You can also look up the value of Boltzmann's constant.
As for the differences in percent lift values you get when you change the ambient temperature and pressure, take the expression for the buoyant force written above, and note that using the ideal gas law for the outside air gives
\begin{align}
  F_\mathrm{buoyant} = \frac{\mu Vg}{k}\frac{P_\mathrm{out}}{T_\mathrm{out}}
\end{align}
In other words, the buoyant force is directly proportional to the ambient pressure, and inversely proportional to the ambient temperature.  Note that I have written the ideal gas law here in the Kelvin temperature scale.  So if the pressure of the ambient air increases by a factor of two, so will the lift, while if the Kelvin temperature increases by a factor of two, then the lift will be halved.
A: Lift is proportional to the density of the material and the hot to cold temperature ratio.
The density of the air is proportional to pressure and inversely proportional to temperature.  The air is densest therefore on a cold day in a high pressure zone.  Pressure, and therefore density, also goes down with altitude.
I wouldn't agonize too much over the details.  Heat loss thru the surface of the balloon will be the biggest determinant of how much heat you have to pump in at the bottom.  Pick a reasonable representative figure for air density and make sure you have a reasonable margin.
You can do a few back of the envelope calculations quite easily.  Air is mostly nitrogen and oxygen.  Nitrogen has atomic weight 14 and oxygen 16.  Both are diatomic, so their molecular weights are 28 and 32, so figure 29 for "air".  Again, there is no point getting too exact.  There will be much larger errors from various sources in practise.
We have decided a mole of air has a mass of 29 grams.  We know that one mole of gas at STP (0°C = 273°K, 1 amosphere pressure) occupies 22.4 l.  Let's say we are targeting 300°K (81°F), so that would be about 25 l.  Let's say we can heat the air in the baloon to a average of 80°C, which is 353°K.  At that temperature it will occupy 29 l.  That displaces 29/25 = 1.16 moles of atmosphere, which has a mass of 33.6 g, for a total mass-hole in the atmosphere of 4.6 g per 29 l of hot air.  You say you want to lift 800 g, so that requires about a 5000 l volume of hot air, or about 5 cubic meters.
You also have to consider the weight of the balloon and the heater, not just the payload.  A bare minimum - again for this quick and dirty back of the envelope calculation - is therefore at least 10 cubic meters.  You probably need more, depending on how light the balloon is and how light a heater you can manage.  A small backpacking stove might be good for this.
To get a feeling of size, let's see how big a 10 cubic meter sphere is.  The volume of a sphere is (4/3) π r2, so about 2.7 meters diameter gets you 10 cubic meters volume.
