Edgtho
Miembro habitual
Subscribo, pero es que toooodo el interior es absolutamente horrible.Y el (en mi opinión) horrible interior en terciopelo azul.
Subscribo, pero es que toooodo el interior es absolutamente horrible.Y el (en mi opinión) horrible interior en terciopelo azul.
El ratio de autonomía/velocidad de recarga es prácticamente el único hándicap del eléctrico. Y es el mismo contratiempo que hace 100 años para el que de momento no hay solución satisfactoria para la mayoría.Es un problema gordo lo de la autonomía,
Un Porsche tipo 911 no es una berlina para viajar, es para ir a fuego y dudo que se recaliente como le pasa a Tesla, porque de ser así, no lo va a comprar nadie que sepa esa limitación.
Heat buildup is inevitable. After three or four laps at absolute tire-torturing full speed, the car begins to reduce power output. It's a balanced, gradual event. The motors and battery use cooling circuits that are independent but linked; as one component heats up, the system shifts cooling capacity where it's needed. It can even use the battery as a heat sink to shed excess thermal load from the motors.
This linked approach to thermal management means that, unlike previous Teslas, track driving won't lead to a total performance shutdown when one component's temperature spikes. Battery and motors heat up at the same rate; when the car starts approaching its thermal limits, it dials back power gradually, until the heat output can be managed by the car's cooling capability.
And you can keep lapping through it. The car's power output plateaus, the cooling system reaches a steady state. You're more than welcome to continue in this condition until you've drained the batteries. You'll miss some of the hard punch of acceleration coming out of the corners, but you end up driving it like a momentum car. It's still a ton of fun.
Lars Moravy (Tesla's Director of Chassis Engineering) told me that his team had run simulations to see how Track Mode would perform at this particular circuit. The data predicted that, after roughly three full-speed, perfect laps, the car would gradually start pulling power, hitting equilibrium at a pace about two to three seconds off the absolute quickest lap times the car is capable of. (...) Our experience matched his predictions.
Let's be honest. This isn't a "track car."
Viajando con tempomat a 125 km/h llevo un consumo más o menos en el diésel de entre 5.1 y 5.2 l/100. Te aseguro que a 60 no voy consumiendo cuatro veces menos, significaría 1.6 l/100, cosa que no he visto nunca en un coche familiar de 1450 kilos de peso.
En el gasolina por ejemplo, a 120 en quinta voy consumiendo más l/100 que a 140 también en quinta. Cuesta creer hasta que no lo pruebas, pero los kilómetros recorridos no mienten.
¿What? Yo te dije que es normal que si un coche tiene autonomía de 400 km a 100 km/h, a 200 km/h tendrá una autonomía de 100 km; Porque hace falta una fuerza cuatro veces mayor, joder, que es una fórmula matemática:Por eso he dicho que NO es cierto no que lo sea.
Tú decías que si consume 12 l/100 significaría que a 100 km/h consumiría 3 l/100.
Lo que yo dije es 5.2 l/100 a 120/130 km/h y sobre los 12 l/100 a 200 km/h en llano. El que no se lo cree eres tú
Drag dijo:Under the assumption that the fluid is not moving relative to the currently used reference system, the power required to overcome the aerodynamic drag is given by:
{\displaystyle P_{d}=\mathbf {F} _{d}\cdot \mathbf {v} ={\tfrac {1}{2}}\rho v^{3}AC_{d}}
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome aerodynamic drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW).[17] With a doubling of speed the drag (force) quadruples per the formula. Exerting 4 times the force over a fixed distance produces 4 times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, 4 times the work done in half the time requires 8 times the power.
When the fluid is moving relative to the reference system (e.g. a car driving into headwind) the power required to overcome the aerodynamic drag is given by:
{\displaystyle P_{d}=\mathbf {F} _{d}\cdot \mathbf {v_{o}} ={\tfrac {1}{2}}C_{d}A\rho (v_{w}+v_{o})^{2}v_{o}}
Where {\displaystyle v_{w}}is the wind speed and {\displaystyle v_{o}}it the object speed (both relative to ground).