Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Does fire create air resistance/drag? So, for example, would it be harder to swing a flaming sword than a normal one?

share|improve this question
When I've been in practice juggling clubs I've wanted to juggle flaming torches at night. Just 'cause it looks cool, you understand. –  dmckee May 29 '13 at 14:47
add comment

2 Answers

No. Fire doesn't create resistance. Resistance is offered by drag force because we're pushing the medium (which pushes back on us) which obstructs our path. It depends on the density of the fluid, our velocity (for high Reynolds numbers), the area of the substance which obstructs the fluid flow. This can be seen in our theory $F=\frac{1}{2}\rho A v^2C_d$

A sudden definition for fire (as seen in everyday life) can be the burning of something by rapid oxidation. In other words, you're losing the mass of the material which catches fire because the it's slowly (when viewed in normal scales) stolen away by oxygen which leaves the molecules into some kinda excitation. You're lucky to see the fireworks which is actually due to the emission of blackbody radiation from the flying away molecules.

To be precise,, Fire actually helps reducing drag. After swinging for some time you'll notice that it'll be easier for you to swing the sword, because the it has been dissolving in air the whole time (i.e) the area of contact with the fluid flow $A$ is getting reduced (though a negligible effect). Also, the density of the fluid in front (shielding the whole object) is quite less than the density of fluid $(\rho_{air}>\rho_{fired})$, leading to easier motion of the sword...

share|improve this answer
Not sure actually...$F_D=\frac12 C_d \rho_{air} V^2 S$, and all variables remain the same for both situations, except the air density; that in fact decreases due to warming by the flame, therefore I'd say since $\rho_{air}^{normal}>\rho_{air}^{flame}$, a flaming sword actually has a slightly lower drag than a non-flaming one... –  Rody Oldenhuis May 29 '13 at 14:22
Hi @Rody: Agreed. But, I did say that it'll be easier to swing the burning sword. But still, this is true only for smaller velocities and the system is somewhat closed like a room or something even smaller (else, molecules will be replaced bringing new ones) ;-) –  Waffle's Crazy Peanut May 29 '13 at 14:28
you said "No. Both are not at all related to each other." and you gave a different reason than I for why it would be easier to swing the firesword. I think your reasoning is incomplete. –  Rody Oldenhuis May 29 '13 at 14:31
add comment

I think it is easier to swing the fire sword than the normal sword.

Take the (subsonic) equation for air drag $F_D$:

$$ F_D = \frac12 \rho_{\mathrm{air}} C_DV^2 S $$

where $C_D$ is the sword's drag coefficient, $S$ the frontal area, $V$ the speed at which you swing it, and $\rho_{\mathrm{air}}$ the air density. Take also the ideal gas law,

$$ \frac{P}{\rho}= RT $$

with $P$ pressure, $\rho$ the specific density, $R$ the universal gas constant and $T$ the temperature.

From the ideal gas law, it is obvious that an increase in temperature would result in a decrease in density and/or increase in pressure. So,

$$ \rho_{\mathrm{air}}^{\mathrm{hot}} < \rho_{\mathrm{air}}^{\mathrm{cold}} $$

and since the fire sword will locally heat up the airflow,

$$ F_D^{\mathrm{fire\ sword}} < F_D^{\mathrm{normal\ sword}} $$

share|improve this answer
How does the difference change when the tip speed increases (transonic and supersonic behavior)? I think the problem has been dealt with for helicopters. Also, does fire make the airflow more turbulent? –  Deer Hunter May 29 '13 at 15:39
The dynamic and kinematic viscosity both increase with temperature, so that would offset some (and potentially all?) of the decrease in density. That would have to be taken into account somehow. –  tpg2114 Sep 26 '13 at 17:39
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.