How much power and torque should we expect from Ford’s new 7.3L gasoline V8 engine? Let’s do some quick calculations and come up with an educated guess.
Pickup truck world is intrigued by the introduction of the new 7.3L gasoline V8 engine. The engine is referred to as a “big block” with an iron block, push-rod valve design, port injection, and more. There is just one gaping hole in the specs sheet. Ford did not announce any power specifications for the engine.
Let’s take Ford’s current 6.2L gas V8 as a baseline. It’s rated at 385 hp and 430 lb-ft of torque. Here is the power and torque per liter of displacement for the current engine.
- 385 hp / 6.2L = 62.1 hp/L
- 430 lb-ft / 6.2L = 69.4 lb-ft/L
This is just a rough estimate. Let’s see what the bigger 7.3-liter engine may be capable of.
- 62.1 hp/L * 7.3L = 453 hp
- 69.4 lb-ft/L * 7.3L = 506 lb-ft
453 horsepower and 506 lb-ft of torque is a very impressive estimate, considering that Ford’s own 6.8L gas V10 has a maximum torque rating of 460 lb-ft.
What if the numbers could be even better? What if Ford is able to get a 10% improvement in flow with the 7.3L? This will push the 7.3L V8 estimates to: 482 hp and 557 lb-ft of torque.
The 7.3L V8 will make its debut in the 2020 Ford Super Duty (F250 and F350) trucks later this year. The engine will be mated to Ford’s new 10-speed automatic transmission.
Here is a detailed interview with the chief engineer of the engine.