I now officially need reading glasses.
Remember my last blog posting in which I provided a handy equation for determining the bases of cumulus clouds--H=ST-DP x 125 (height = the difference between surface temperature and dew point multiplied by 125? Well, there was a tiny six-letter word after that H in my meteorology book that I didn't see--meters. To figure out the height in feet, you have to multiply by 228.
Perhaps this textbook publisher should consider Large Type editions....
Funny, last weekend, my husband and I were out hiking and looking at clouds and trying to estimate their height. There was a perfect batch of cumulus clouds over Olympia on Sunday and once we logged onto the National Weather Service website to find the current temperature and dew point, we did our multiplication and declared the clouds over us to be 825 feet high. It seemed low, but I had just posted the equation on my blog so I knew it was correct. Well, it wasn't. The clouds were 875
high. Recalculating this in feet, we have 1,916 feet. That's more like it, footwise.
....or I should consider wearing reading glasses.
We did see lower clouds that day--stratus clouds--clinging to tops of the Black Hills west of Olympia. But the equation I have provided here (and in my earlier posting) does not work for stratus (layer) clouds. It works only for cumulus clouds--ones that form when warm air rises and cools--and it works best on a sunny day in the afternoon. We guessed that the stratus clouds were about 500 feet high. We had no idea really how high the Black Hills were, though we look at them pretty much every day. Our USGS topo sheets showed the ridge of the Black Hills where we had seen the clouds to be 600 feet high. Now we know.
Because you are reading my blog and not meteorologist Cliff Mass's, I know you prefer relatively simple, Accidental Naturalist-style explanations. Here is mine: Rising air cools 10°C for every 1000 meters of altitude (fine: 50° F for every 3,300 feet). The temperature at which water vapor in the air condenses into liquid water (dew point temperature) drops 2°C for every 1000 meters of altitude (36°F for every 3,300 feet). So, for every 1000 meters of rise, air temp and dewpoint temp drop steadily, but get 8°C closer.
This is beginning to sound like a word problem involving two people driving at different speeds on the same road--who arrives at grandma's house first?
And because word problems are math, I am having a hard time understanding how to get to the magic number 125 or 288. Here is the explanation offered in my meteorology book:
"Rising surface air with an air temperature and dew point spread of 8°C would produce saturation and a cloud at an elevation of 1000 m. Put another way, a 1°C difference between the surface air temperature and the dew point produces a cloud base at 125 m."
To give you an example of how I handle word problems, I once "helped" my son solve one in which he was asked to provide the dimensions of a chain-link dog enclosure using 70 feet of fencing. I knew how to calculate area--length x width--so I told him to get a dacschund and build a 1' x 70' enclosure.
So, while I figure out the math and why we can calculate the height of cumulus clouds, enjoy trying it yourself, keeping your and meters separate. Maybe one day we won't be so afraid of meters--a meter is really just a yard and so much easier to visualize than feet or miles when it comes to clouds.