Daniel Burnam

Roger that.

So if I had a flat wall facing due south at high noon in the winter, the intensity of the sun at 5’ up the wall is no different than the intensity of the sun 10’ up the wall, assuming there’s no shading.

Window placement has to do with how far back into the space the sun will shine and so is about the relationship between the window and the floor, whether that floor is at ground level or 50’ off the ground. period...question mark...explanation point

To be sure I misunderstood that. Thank you for the clarity.

And for the links.

And yes, Glenn, I think I understand what you’re saying about the shortcomings of the sphere re: maximizing winter sun and minimizing summer sun. In the end, I will almost certainly wind up with an octagonal shape, but the sphere seemed like the easiest way to get my head around plotting out the sun path and also to play with my own ideas of having some part of the structure be above grade and also shaded by the larger diameter part of the building, for the sake of extra edge and potential microclimates.

YES thank you. That makes a lot of sense now and a good bit easier to wrap my head around. The parallel lines will be a huge help. The sun will be shining on everything not shaded, from it’s given angle, given the time of day and year. And yet there is a focal point on any given structure? And that focal point will coordinate to angle of the sun, measured from center of structure at ground level

For example, on the winter solstice, at 8:15 am, an hour after sunrise where I am, all of the structure facing 144 degrees SE will be irradiated by a sun shining at an angle of +8.5 degrees from the horizon. And the focal point of that irradiation will be concentrated on the structure as measured from the center at ground level at an angle of 8.5 degrees?

Did I get that right?

Sounds like I would be greatly aided by some computer modeling. I have so far remained ignorant of this wonderful modern tool, but I imagine it’s time for me to take the time to learn it. Any recommendations for easy CAD 3-D modeling type software that I could start to model without having to do a ton of self-education?

Thanks again for all the help

This is great I really appreciate that advice. I've been playing around with similar techniques using sun tracker apps and the like.

I live at roughly 40 degrees north latitude so my max winter sun angle is 26 degrees, 49 degrees on the equinox and 72 degrees at summer solstice.

What I understand you to mean by

Now you can just draw three lines on your graph that all pass through the center of your structure at GROUND level, one at 78degrees, one at 55degrees, and one at 32degrees.  Where they intersect the curved wall of the building will be were the sun will shine at noon.  You just substitute your latitude for the one I used in the example to get YOUR values.

is something like this first picture where every part of the sphere above grade is convex and that all seems to make sense where the winter sun would fall.

but if I were to raise the whole structure above grade and apply that same process it would look like the second picture, which looks funny. haha

my suspicion is that there would be an angle at which the sun would reach the equator of the structure and then start to track predictably. and that leaves a short portion of the day where the sun is tracking on the "underside" of the sphere and then it would begin tracking over the top and the underside would be shaded

any thoughts?

Hello Community! I'm designing my own off-grid home and I'm stuck on how to plot the travel of the sun across my building.

Since the structure is more or less spherical I could theorectically plot the sun's full arc across the sky across my home. The windows on the winter solstice could receive direct light from sun up to sun down. Where I am stuck is on how to find the actual height, on the structure, of where the sun will land, and how to change that calculation should the diameter of the home change, or the amount of the home above/below grade change.

the pic attached is an 18' diameter sphere, first 3' below grade, next 8' below grade. how would I find where the sun lands on the actual structure? how would I change my calculations if the sphere expanded to 24' diameter or sank below grade, or was elevated above grade on some type of foundation?

I appreciate all the genius minds at work here on this forum and I thank you for any advice you can give

What a wonderful project! This kind of tinkering is for sure the way the world is revolutionized. I'm excited to see this technology applied to something like water heating!

Excellent thank you.

I don’t have intentions on plugging into the grid. The wind on the coast of Maine is in my experience strong and consistent/predictable and the tidal swing is huge. My hope was to generate that sort of power using tidal and wind.

So that’s my next line of inquiry (probably for another forum) . Now that we have an idea of our basic power needs, how to determine the generator requirements to meet those needs

Thank you thank you for your attention here.

Yes I am imagining a direct battery to motor power transfer, with the batteries being charged from an on-shore battery bank while the boat is moored at night, rather than a diesel to electric conversion that S Benji was addressing.

So 35 us gallons of fuel = 5128 megajoules of energy

A Diesel engine at 20% efficiency is going to apply 1025.6 of those megajoules of energy to the motor

An electric motor at 90% efficiency drawing power from a battery pack at 85% efficiency is going to apply 3922.9 megajoules to that motor.
(5128 x .9)x .85 = 3922.9.  
Is that correct? Is that an oversimplification?

If the desired end product is about 1000 megajoules then is 285kwh enough?

https://www.rapidtables.com/convert/energy/Joule_to_kWh.html

That seems so much more doable. Thanks for helping me stumble through this.  Your insight and math skills are really valuable.

Hello to this wonderful forum and thanks to everyone!

I am working with a friend who is a commercial fisherman in Maine. We are talking about the possibility of running an electric motor on his boat rather than the diesel that everyone uses out there. His boat has a 425hp diesel caterpillar motor and he says he burns about 35 gallons of fuel/day on average. Just a quick search on the internet and some simple math says that he's using about 1 megawatt electrical equivalent in diesel fuel

Question 1: Can that be?

Question 2: Is the electric motor, in general, more efficient than a diesel which would then require less power input to do the same amount of work making the need for an actual megawatt of energy storage in the boat less relevant?

The boat wants ballast, so an enormous battery bank in the hull is not out of the question, I'm just trying to see if that's really the number we're aiming for in diesel-electrical conversion

thanks again

Hi! I'm a carpenter living in Brooklyn and our neighbor tosses out bags and bags of planer shavings. I was hoping to repurpose them (mostly ash and maple right now) to mulch our raised beds over the winter.

Does this work? i am new to the game and don't even want the take a stab at the myriad things that I should consider or may be overlooking. Jus hoping to send the wood some place positive and not a landfill (though we have donated some to a friend doing backyard mushrooms).

Also there is a big patch of yard that is mostly construction fill dirt. It is currently home to some hardworking pioneer plants, but I'd give this dirt a good coat of wood duff too if that's useful.

Thanks! I'm open to any and all info.