First of all, figure 60deg F, zero humidity, and a 29.92" barometer as 100% air density - AKA the "Standard Day”
Now that we have the Standard Day figures, we can get to how each of the three measurements (temperature, humidity and barometer) affect air density. Actually, there is a moderately involved formula for calculating air density, but I'll skip that and get to how you can closely estimate pluses and minuses.
1. Barometer readings have a linear effect on air density. That is, if the barometer goes up or down 1% from the standard reading, so will air density. Therefore, a barometer of 30.22" gets you to 101% air density, 29.62" nets to 99%, and so on.
2. For each 5o change in temperature (away from 60o), you get about a 1% change in air density. Therefore, an 80o day means 96% air density, and a 40o day means 104% air density. Note: This is *not* linear, since the calculation has a base of absolute zero, but it's close enough so we don't have to worry about it.
3. At 60o, jumping from zero to 50% humidity will cost you roughly 1% in air density. However, that same 50% humidity at 90o will cost you more than 2% in air density. The reason for this is that it requires a *lot* more water vapor to get you to 50% humidity at 90o than it does at 60o, since air can hold more water as it's heated - like sugar in your coffee or tea. Hot tea can hold a *bunch* of sugar, but iced tea can't hold too much. "Relative humidity" is expressed as a percentage of water in the air compared to how much it *could* hold at a given temperature.
All 3 figures need to be juggled. As an example, a 50o day with 50% humidity and a 29.62" barometer nets you to just about 100% air density, since the 2% you pick up in temperature is offset by the losses you incur for the barometer drop of 1% and the humidity loss of 1%.
Now, how does air density affect power?
This is also something that is moderately involved, but figure around 75-80% power delta compared to the air density delta. That is, an 8% drop in air density will likely cost you somewhere around 6% in power.
Actually, barometer changes give you something right around a 1.2-1% swing in power, temperature changes give you something right around a .7-1% swing, and humidity is near 1-1%. In practice, temperature is likely to be the biggest variable, hence the 75-80% "rule of thumb".
In general, temperature has twice the effect on the car than humidity does. If the humidity goes up 2 points, it affects the car the same as if the temperature goes up one degree. And vice-versa if the air temp goes down. This is a rule of thumb that holds fairly well for all cars.
Also, as air pressure increases, the Cd (co-efficient of drag) increases. In other words, denser air means more air resistance and this slows the car down some. Total aerodynamic resistance is determined by multiplying Cd times the frontal area of the vehicle. Dragsters have a small frontal area compared to a door car although their Cd might be higher then a swoopy door car. So, in general, a dragsters aerodynamic drag is less affected by changes in air pressure and temperature than a door car.
If both cars have identical motors, both will loose or gain the same amount of horsepower due to weather changes, however, the door car will have a wider variation in E.T. due to weather changes because of the change in aerodynamic drag affects the door car to a greater degree. The door car will fall off faster or pick up more than the dragster. Thus, the dragster will appear to be more consistent than a door car.
Now, how does power affect quarter-mile speeds?
Use the formula:
Horsepower = car weight (with driver) times the quantity [quarter-mile-speed over 232.3] cubed.
HP = car weight x [1/4 mile speed/232.3]3
So, assuming your car (with you aboard) goes around 3450 pounds, we get:
96/232.3 = .4132587, times .4132587 = .1707827, times .4132587 = .0705774, times 3450 = 243.49 HP.
On the other hand, 93.5 MPH gives you "only" 224.96 HP, so you might have lost a computed 7.6% in power from outing to outing. This implies a 10% loss of air density, which sounds like a bit much.
As a BTW, an ET number that one might aspire to is:
ET = 1353/mph
This assumes good street rubber and decent traction.
So, 1353/96 = 14.09, and 1353/93.5 = 14.47.
Back at the beginning I mentioned "other environmental factors".
A key one is wind. A headwind will drop MPH significantly, and the opposite
for a tailwind. Such winds will likely have less effect on ET than they
will on MPH. HP needs rise as the *square* of the speed, so the first
330 feet (where you make or break ET) will hardly be affected, while the
last 330 feet (where ET is hardly affected at all) will be dramatically
affected by wind.