Is the heat from a flame mainly radiation or convection? I'm learning about the three types of heat transfer conduction, convection, and radiation.
An example the teacher gives us is a campfire which involves all three. Now I know that the hot air rising is convection and I also know that the fire gives off electromagnetic waves in the form of visible light and infrared.
My question is where the majority of the heat is coming from if it let's say heats an object overhead the fire? Since the radiant waves are directed everywhere(including upwards) and the hot air also goes up, is there a way of determining which is more dominant?
You could say the fire emits thermal radiation without needing a medium but in this case, it's not in a vacuum and it could also be convective heating since there is a fluid. I hope my explanation of my question wasn't too convoluted, thanks!
 A: 
My question is where the majority of the heat is coming from if it
let's say heats an object overhead the fire?

This is very hard to call, without some seriously complicated math.
But we can make some general observations.
Heat radiation (infrared) is emitted by a hot object in roughly randomly all directions, so only a small part of the total radiation will hit the object. And some of it will simply be reflected by it, especially if the object is shiny.
Convective heat tends to rise vertically, so most of it reaches the overhead object. But heat transfer takes time, so not all the convected heat energy will be absorbed by the object.
The bottom line is that this kind of heating is not very efficient!
A: This source addresses your question directly. It states that conduction, due to the fact that air is a good thermal insulator, is of least importance in the heat from a campfire. The heat transferrered via convection shoots up in the air, hotter air rises, and probably won't warm the people around it. Hence radiation is the way the people around the campfire feel the warmth.
A: Here's a fun experiment you can do. Take a marshmallow and hold it a fixed distance r, horizontally from the fire, and measure the time $t_h$ it takes to cook it to your preference. Take a second marshmallow and do the same at a distance r above the fire to get a cook time $t_v$. We expect that
$t_h = c/(P_{rad})$
$t_v = c/(P_{rad} + P_{conv})$
where c is some constant, $P_{rad}$ is the radiation energy output of the fire experienced by the marshmallow at distance r, and $P_{conv}$ is the convective output of the fire experienced by the marshmallow at distance r. Thus with both these quantities we can measure the relative strength of convective and radiative heating
$P_{conv}/P_{rad} = t_h/t_v - 1$
Caveats: I assumed crucially that the radiant power output of the fire is the same in all directions (so that $P_{rad}$ is the same quantity in both $t_h$ and $t_v$). That would also be a nice thing to test. $P_{conv}$ and $P_{rad}$ also depend on details of how well the marshmallow absorbs heat. For instance if we somehow had a black marshmallow, all other properties held the same, it would have a higher $P_{rad}$. We could maybe use painted marshmallows to test whether $P_{rad}$ is bigger in the horizontal or vertical direction, or to figure out the ratio for objects of emissivity 1. Unfortunately we wouldn't be able to eat the marshmallows afterwards (and it might be hard to tell when they're "done"), so future research is needed :)
A: This depends a lot on the relative position of the fire and what you want to heat up. For instance, you typically put food above the fire, where convection is dominant.
On the other side, if you're in open space, the convention component can decrease more rapidly than the radiation one while getting further away (horizontally) and that actually also depends on other factors like wind. In a more controlled environment like a fireplace, you should see that convection is the leading contribute. I didn't do the math on that, but as a personal experience I have a fireplace and once the fan granting convection stopped working for days and you could that the heat generated was only a very small percentage.
