3 added 51 characters in body edited Aug 21 '17 at 12:53 Communisty 97644 silver badges1818 bronze badges This isn't answerable without running a climate model or some simplifying assumptions. I will try to answer the question with some simplifying assumptions that might then provide a coarse guess of the area needed. First lets assume that temperatures have risen $$1^{o}C$$ and that the cooling film acts as increasing surface albedo effectively reflecting that an additional fraction of sunlight (93/1366) from earth. By following the footsteps of the idealized greenhouse model we can derive the surface temperature as: $$T_{s} = \bigg(\frac{S_{0}(1 - \alpha_{p})}{4 \sigma (1 - \frac{\epsilon}{2})}\bigg)^{1/4}$$ Where $$T{_s}$$ is surface temperature, $$S_{0}$$ is the solar constant ($$1366W/m^ 2$$), $$\alpha_{p}$$ is the planetary albedo, $$\sigma$$ is the Stefan Boltzmann constant and $$\epsilon$$ is the emissivity. By approximating the planetary albedo as 0.3 and emissivity as 0.78 we get that the surface temperature of is 288K. We can now set that the surface temperature is one degree smaller and there is an addition to the planetary albedo due to the cooling films. Solving for albedo: $$\alpha_{p} = -\bigg(\frac{T_{s}^{4}4\sigma (1-\frac{\epsilon}{2})}{S_{0}} - 1\bigg)$$ By substituting the values we find that planetary albedo has to be about 0.31, so cooling of one degree has been reached by a 0.01 additional albedo. The ratio of the cooling film absorbency and solar constant $$\frac{93}{1366} \approx 0.07$$ can be seen as the additional planetary albedo if the cooling film is installed globally. This being seven times the needed change of albedo implies that a seventh of earths surface should be covered. This doesn't take into account any secondary effects, that all areas of earth aren't equally important, that earths albedo isn't constant, nor is the emissivity, angle of radiation matters, this equation doesn't really take into account greenhouse effect(it just uses values that provides in current state case reasonable numbers), et cetera. More accurate answer would come by running a climate model and tweaking albedo in fixed grid cells. This isn't answerable without running a climate model or some simplifying assumptions. I will try to answer the question with some assumptions that might then provide a coarse guess of the area needed. First lets assume that temperatures have risen $$1^{o}C$$ and that the cooling film acts as increasing surface albedo effectively reflecting that an additional fraction of sunlight (93/1366) from earth. By following the footsteps of the idealized greenhouse model we can derive the surface temperature as: $$T_{s} = \bigg(\frac{S_{0}(1 - \alpha_{p})}{4 \sigma (1 - \frac{\epsilon}{2})}\bigg)^{1/4}$$ Where $$T{_s}$$ is surface temperature, $$S_{0}$$ is the solar constant ($$1366W/m^ 2$$), $$\alpha_{p}$$ is the planetary albedo, $$\sigma$$ is the Stefan Boltzmann constant and $$\epsilon$$ is the emissivity. By approximating the planetary albedo as 0.3 and emissivity as 0.78 we get that the surface temperature of is 288K. We can now set that the surface temperature is one degree smaller and there is an addition to the planetary albedo due to the cooling films. Solving for albedo: $$\alpha_{p} = -\bigg(\frac{T_{s}^{4}4\sigma (1-\frac{\epsilon}{2})}{S_{0}} - 1\bigg)$$ By substituting the values we find that planetary albedo has to be about 0.31, so cooling of one degree has been reached by a 0.01 additional albedo. The ratio of the cooling film absorbency and solar constant $$\frac{93}{1366} \approx 0.07$$ can be seen as the additional planetary albedo if the cooling film is installed globally. This being seven times the needed change of albedo implies that a seventh of earths surface should be covered. This doesn't take into account any secondary effects, that all areas of earth aren't equally important, that earths albedo isn't constant, et cetera. I will answer the question with some simplifying assumptions that might then provide a coarse guess of the area needed. First lets assume that temperatures have risen $$1^{o}C$$ and that the cooling film acts as increasing surface albedo effectively reflecting that an additional fraction of sunlight (93/1366) from earth. By following the footsteps of the idealized greenhouse model we can derive the surface temperature as: $$T_{s} = \bigg(\frac{S_{0}(1 - \alpha_{p})}{4 \sigma (1 - \frac{\epsilon}{2})}\bigg)^{1/4}$$ Where $$T{_s}$$ is surface temperature, $$S_{0}$$ is the solar constant ($$1366W/m^ 2$$), $$\alpha_{p}$$ is the planetary albedo, $$\sigma$$ is the Stefan Boltzmann constant and $$\epsilon$$ is the emissivity. By approximating the planetary albedo as 0.3 and emissivity as 0.78 we get that the surface temperature of is 288K. We can now set that the surface temperature is one degree smaller and there is an addition to the planetary albedo due to the cooling films. Solving for albedo: $$\alpha_{p} = -\bigg(\frac{T_{s}^{4}4\sigma (1-\frac{\epsilon}{2})}{S_{0}} - 1\bigg)$$ By substituting the values we find that planetary albedo has to be about 0.31, so cooling of one degree has been reached by a 0.01 additional albedo. The ratio of the cooling film absorbency and solar constant $$\frac{93}{1366} \approx 0.07$$ can be seen as the additional planetary albedo if the cooling film is installed globally. This being seven times the needed change of albedo implies that a seventh of earths surface should be covered. This doesn't take into account any secondary effects, that all areas of earth aren't equally important, that earths albedo isn't constant, nor is the emissivity, angle of radiation matters, this equation doesn't really take into account greenhouse effect(it just uses values that provides in current state case reasonable numbers), et cetera. More accurate answer would come by running a climate model and tweaking albedo in fixed grid cells. 2 added 117 characters in body edited Aug 21 '17 at 12:36 Communisty 97644 silver badges1818 bronze badges This isn't answerable without running a climate model or some simplifying assumptions. I will try to answer the question with some assumptions that might then provide a coarse guess of the area needed. First lets assume that temperatures have risen $$1^{o}C$$ and that the cooling film acts as increasing surface albedo effectively reflecting that an additional fraction of sunlight (93/1366) from earth. By following the footsteps of the idealized greenhouse model we can derive the surface temperature as: $$T_{s} = \bigg(\frac{S_{0}(1 - \alpha_{p})}{4 \sigma (1 - \frac{\epsilon}{2})}\bigg)^{1/4}$$ Where $$T{_s}$$ is surface temperature, $$S_{0}$$ is the solar constant ($$1366W/m^ 2$$), $$\alpha_{p}$$ is the planetary albedo, $$\sigma$$ is the Stefan Boltzmann constant and $$\epsilon$$ is the emissivity. By approximating the planetary albedo as 0.3 and emissivity as 0.78 we get that the surface temperature of is 288K. We can now set that the surface temperature is one degree smaller and there is an addition to the planetary albedo due to the cooling films. Solving for albedo: $$\alpha_{p} = -\bigg(\frac{T_{s}^{4}4\sigma (1-\frac{\epsilon}{2})}{S_{0}} - 1\bigg)$$ By substituting the values we find that planetary albedo has to be about 0.31, so cooling of one degree has been reached by a 0.01 additional albedo. The ratio of the cooling film absorbency and solar constant $$\frac{93}{1366} \approx 0.07$$ can be seen as the additional planetary albedo if the cooling film is installed globally. This being seven times the needed change of albedo implies that a seventh of earths surface should be covered.   This doesn't take into account any secondary effects, that all areas of earth aren't equally important, that earths albedo isn't constant, et cetera. This isn't answerable without running a climate model or some simplifying assumptions. I will try to answer the question with some assumptions that might then provide a coarse guess of the area needed. First lets assume that temperatures have risen $$1^{o}C$$ and that the cooling film acts as increasing surface albedo effectively reflecting that an additional fraction of sunlight (93/1366) from earth. By following the footsteps of the idealized greenhouse model we can derive the surface temperature as: $$T_{s} = \bigg(\frac{S_{0}(1 - \alpha_{p})}{4 \sigma (1 - \frac{\epsilon}{2})}\bigg)^{1/4}$$ Where $$T{_s}$$ is surface temperature, $$S_{0}$$ is the solar constant ($$1366W/m^ 2$$), $$\alpha_{p}$$ is the planetary albedo, $$\sigma$$ is the Stefan Boltzmann constant and $$\epsilon$$ is the emissivity. By approximating the planetary albedo as 0.3 and emissivity as 0.78 we get that the surface temperature of is 288K. We can now set that the surface temperature is one degree smaller and there is an addition to the planetary albedo due to the cooling films. Solving for albedo: $$\alpha_{p} = -\bigg(\frac{T_{s}^{4}4\sigma (1-\frac{\epsilon}{2})}{S_{0}} - 1\bigg)$$ By substituting the values we find that planetary albedo has to be about 0.31, so cooling of one degree has been reached by a 0.01 additional albedo. The ratio of the cooling film absorbency and solar constant $$\frac{93}{1366} \approx 0.07$$ can be seen as the additional planetary albedo if the cooling film is installed globally. This being seven times the needed change of albedo implies that a seventh of earths surface should be covered.   This isn't answerable without running a climate model or some simplifying assumptions. I will try to answer the question with some assumptions that might then provide a coarse guess of the area needed. First lets assume that temperatures have risen $$1^{o}C$$ and that the cooling film acts as increasing surface albedo effectively reflecting that an additional fraction of sunlight (93/1366) from earth. By following the footsteps of the idealized greenhouse model we can derive the surface temperature as: $$T_{s} = \bigg(\frac{S_{0}(1 - \alpha_{p})}{4 \sigma (1 - \frac{\epsilon}{2})}\bigg)^{1/4}$$ Where $$T{_s}$$ is surface temperature, $$S_{0}$$ is the solar constant ($$1366W/m^ 2$$), $$\alpha_{p}$$ is the planetary albedo, $$\sigma$$ is the Stefan Boltzmann constant and $$\epsilon$$ is the emissivity. By approximating the planetary albedo as 0.3 and emissivity as 0.78 we get that the surface temperature of is 288K. We can now set that the surface temperature is one degree smaller and there is an addition to the planetary albedo due to the cooling films. Solving for albedo: $$\alpha_{p} = -\bigg(\frac{T_{s}^{4}4\sigma (1-\frac{\epsilon}{2})}{S_{0}} - 1\bigg)$$ By substituting the values we find that planetary albedo has to be about 0.31, so cooling of one degree has been reached by a 0.01 additional albedo. The ratio of the cooling film absorbency and solar constant $$\frac{93}{1366} \approx 0.07$$ can be seen as the additional planetary albedo if the cooling film is installed globally. This being seven times the needed change of albedo implies that a seventh of earths surface should be covered. This doesn't take into account any secondary effects, that all areas of earth aren't equally important, that earths albedo isn't constant, et cetera. 1 answered Aug 21 '17 at 11:38 Communisty 97644 silver badges1818 bronze badges This isn't answerable without running a climate model or some simplifying assumptions. I will try to answer the question with some assumptions that might then provide a coarse guess of the area needed. First lets assume that temperatures have risen $$1^{o}C$$ and that the cooling film acts as increasing surface albedo effectively reflecting that an additional fraction of sunlight (93/1366) from earth. By following the footsteps of the idealized greenhouse model we can derive the surface temperature as: $$T_{s} = \bigg(\frac{S_{0}(1 - \alpha_{p})}{4 \sigma (1 - \frac{\epsilon}{2})}\bigg)^{1/4}$$ Where $$T{_s}$$ is surface temperature, $$S_{0}$$ is the solar constant ($$1366W/m^ 2$$), $$\alpha_{p}$$ is the planetary albedo, $$\sigma$$ is the Stefan Boltzmann constant and $$\epsilon$$ is the emissivity. By approximating the planetary albedo as 0.3 and emissivity as 0.78 we get that the surface temperature of is 288K. We can now set that the surface temperature is one degree smaller and there is an addition to the planetary albedo due to the cooling films. Solving for albedo: $$\alpha_{p} = -\bigg(\frac{T_{s}^{4}4\sigma (1-\frac{\epsilon}{2})}{S_{0}} - 1\bigg)$$ By substituting the values we find that planetary albedo has to be about 0.31, so cooling of one degree has been reached by a 0.01 additional albedo. The ratio of the cooling film absorbency and solar constant $$\frac{93}{1366} \approx 0.07$$ can be seen as the additional planetary albedo if the cooling film is installed globally. This being seven times the needed change of albedo implies that a seventh of earths surface should be covered.