What is the optimum tilt angle for a solar panel in our area? What would the tilt?
That's actually a more complicated question than it might seem at first. The simple answer for anyone above 23 degrees latitude is to tilt at latitude, so if your house is at 35 degrees latitude, you would tilt at 35 degrees. However, the best financial payback may be to optimize for summer (latitude + 15 in the northern hemisphere), if you have air conditioning bills that kick up your electric rates into the expensive tier. On the other hand, if you live in an area that gets very hot, the panel efficiency will go down, so it might be better to go latitude + 5 or so. To optimize for Winter, the simple answer is latitude - 15, but again, not so simple, if there's a chance you'll get snow on the panels. I have our panels set up so that I could readjust the tilt twice a year if desired, but I have never done so, yet. Too much trouble for a rooftop installation. And I'm concerned about 100 mph wind gusts picking up the panels in winter if I were to tilt them at a high angle. As far as I know, the vast majority of residential installations just follow the slope of the south-facing roof and take what they can get. If you're talking about water heating panels, I think the general practice is to just match the slope of the roof, and point them southwest (not south) in our area, because that's the optimum balance between the sun and the daytime temperature.
At what angle should I set my launcher for the optimum range?
Here's the whole problem: How far away can that water balloon land? Galileo and others were confronted with the same problem -- at what angle should I set my launcher for the optimum range, by definition the furthest horizontal distance away from the launcher. a.) What equation is useful for describing the y-directed motion of the projectile? What are the initial and final conditions (positions, velocities) for the problem of projectile motion? Explain. I think this equation would be useful but I'm really not sure: y= y0 + v0y + 1/2at2 b.) What is the relationship between motion in the x- and y-directions and t, the time it takes fro the projectile to travel from the launcher to its landing point? From what I understand the relationship between y and t is a quadratic one, and between x and t it's a linear relationship. Again, I'm not sure on this one. c.) Derive an equation relating all the important physics variables to give R, the range. Can you use trigonometric relationships to relate these variables to the launch angle? I think the equation I could use is (v0)^2 * sin2theta / g Is this right? d.) From the equation derived in c, determine which angle gives the maximum range. Explain. considering I'm not sure if C is right, I don't think I can answer this. e.) So how sohould you launch that water balloon, anyway? Does its mass have any bearing on the problem? (If I launched two water balloons, each with different mass, would I choose a different launch angle for one? Any help is greatly appreciated!
What is the optimum angle of a projectile to achive maximun distance?
hi, Shrek is acceptable, you do no longer could desire to sparkling up for time. you're able to do away with the time by fixing simultaneous equations Time for ball to pass up and backpedal, utilising y-factor t=2*v*sin45/g Horizontal distance traveled by the ball in that ingredient X = v*cos45*t Now basically shove the 1st equation into the 2nd X = v*cos45*t X = v*cos45*(2*v*sin45/g) sparkling it up slightly X = (2 * v^2 * sin45 * cos45)/g X = (2 * (6 m/s)^2 * sin45 * cos45)/9.80 one m/s^2 X = 3.sixty seven m so as you will discover, you could sparkling up for time is you decide on for, yet you do no longer could desire to sparkling up for time. consistently attempt to maintain your equations with all the variables, in case you could, until the very final step. now and lower back they could basically fall away, and you have wasted "calculator time" for no longer something. i think of that shrek merits the wonderful answer, I basically stole his equations and elaborated slightly. good success!
This is a solved question, and you can get a quick-and-dirty answer in minutes. The following websites walk you through the procedure:Solar Panel Angle: how to calculate solar panel tilt angle? - Sinovoltaics - Your Solar Supply NetworkOptimum Tilt of Solar PanelsThis article gives a bit more depth about the issues involved:An Algorithm to Determine the Optimum Tilt Angle of a Solar Panel from Global Horizontal Solar Radiation…and remember that it’s important too to take the spacing between rows of panels into account, as well, and that such spacing is also dependent on latitude because of the shadow of the panels.
Why is a 45 degree angle the most optimal angle for distance?
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What is the optimal angle to throw a javelin?
Suppose the initial velocity is v at an angle θ to the ground. I'm assuming horizontal ground and no air resistance (throwing javelins in a vacuum is fun). Let h(t) denote the height at time t. h=vtsinθ-½gt² When the javelin lands, h=0 vtsinθ=½gt² t=2(v/g)sinθ In that time the horizontal distance travelled is 2(v/g)sinθ vcosθ. That's 2(v²/g)sinθcosθ/g or (v²/g)sin2θ. This is a maximum when 2θ=90°, ie. θ=45°. I hope this helps.
First of all, I’m assuming that air friction and drag is not considered. So here is a little illustration:What we’re now trying to maximize is [math]s_x.[/math]The first well-known formula we need is the following:[math]s(t)=v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2[/math]In our case, [math]s(t)[/math] will be the vertical height, [math]v_0[/math] the vertical initial speed and [math]a[/math] will be the acceleration due to gravity. Rewriting:[math]s_y(t)=v_y \cdot t - \frac{1}{2} \cdot g \cdot t^2[/math]The acceleration is negative, because the positive y direction is upwards in this case.We want to know, when [math]s_y(t)=0, so:[/math][math]0=v_y \cdot t - \frac{1}{2} \cdot g \cdot t^2[/math]Solving for [math]t[/math] we get two solutions: [math]t_1=0[/math] & [math]t_2=\frac{2 \cdot v_y}{g}[/math]The first solution is at the start of this scenario, so we’re only interested in the second one.Now, that we have the time until the projectile hits the ground, we can calculate the horizontal distance:[math]s_x=v_x \cdot t_2=v_x \cdot \frac{2 \cdot v_y}{g}[/math]Using basic trigonometry, we can see that:[math]v_x=v_0 \cdot cos(\alpha)[/math] and [math]v_y=v_0 \cdot sin(\alpha)[/math]Inserting that we get:[math]s_x=v_0 \cdot cos(\alpha) \cdot \frac{2 \cdot v_0 \cdot sin(\alpha)}{g}=\frac{2 \cdot v_0^2}{g} \cdot cos(\alpha) \cdot sin(\alpha)[/math]Now we need to find the maximum of [math]s_x[/math], which we can get by using simple calculus. First, we need the derivative in respect to [math]\alpha[/math]:[math]s_x'(\alpha)=2 \cdot v_0^2 \cdot (\frac{2 \cdot cos(\alpha)^2 - 1}{g})[/math]And then we need to find at which [math]\alpha[/math] [math]s_x'[/math] is equal to [math]0[/math]:[math]s_x'(\alpha)=0[/math]-> [math]\alpha = \frac{\pi}{4}=45°[/math]Edit: [math]\alpha[/math] should be [math]x[/math]
Put em at 15 degrees, or thereabouts, to promote self-cleaning when it rains. There will be negligible loss of output from a theoretically optimum horizontal array. The direction of orientation is therefore not crucial. However, in an offgrid system where one wishes to maximise solar output during the season of lowest irradiance, ie the monsoon period, it may help to tilt the array to the part of the sky the sun inhabits at that time of the year. In northern australia and indonesia, for example, that would be 15 degrees to the south for the december to march monsoon season.
The angle that the sun will be at in its peak position will be equal to your latitude, plus or minus up to 20 degrees according to season. That 15 degrees is a fudge-factor attempt to make sure it's producing more on the non-exact-peak part of the season, at the expense of sacrificing a little bit on the exact peak of the season.Image source: NOAA / NASA / Jet Propulsion LabSorry about the font, for those who hate informal fonts. It's not my figure.The goal of doing this is to minimize the angle of the sun off the solar panels -- the more directly it faces the sun, the more power it produces.You can play around with this here:Solar Power Effect On Solar PanelsOr, you can do what my wife's dad did and point some panels at the sun at peak, point a bunch more at the sun's position at 4:00 with an adjusting bar to switch the angle between winter and summer, and put up one more set on a pole with some actuators to track the sun all day.
What is the angle of tilt of the earth with the vertical axis?
this question is basically responded by technique of the adjustments contained in the seasons. Our planet tilts 23 ranges 27 minutes. this is shown as I reported by technique of the replace contained in the seasons. could our planet be upright we could have only one season 365 days round. depending on the position you stay in the international could count upon your variety north or south, it may be gentle or spring and fall like temperatures could prevail above and reduce than the tropics.and not using a tilt, there could be no adjustments in seasons in any respect. because of the tilt, wintry climate contained in the norther hemisphere, the position you stay, begins on or about December 21-23. this is the wintry climate solstice. this is even as the solar dips decrease than the equator for basically about six months. At that precise second the southern hemisphere begins summer season. There are 2 equinoxes. Equinox potential equivalent evening. this is as close to as upright because the planet receives. Twelve hours of day and twelve hours of evening. those ensue contained in the spring and fall. there is very a lot extra, although for the reply for your question, its no longer only a conception this is a medical actuality. all of us recognize the plane of the ecliptic and for this reason we are able to gauge absolutely the posture of each and each and every of the planets on there axis.