Day trip from Yingkou to Jinzhou
Jinzhou lies 104 km (65 mi) north-west of Yingkou, in the same time zone. What the numbers say about getting there.
At 104 km (65 mi) apart, Jinzhou is a realistic day trip from Yingkou: about 1 h 42 min each way by road.
| Yingkou | Jinzhou | |
|---|---|---|
| Country | China | China |
| Population | 591,159 | 1.5 million |
| Elevation | 4 m (13 ft) | 26 m (85 ft) |
| Time now | UTC+8 (Asia/Shanghai) | UTC+8 (Asia/Shanghai) |
| Postal codes | 115000 | 052200–121000 |
Where to book tours and tickets in Jinzhou
We take no bookings for these providers and hold no reservation with them. The provider you choose handles payment, confirmation, changes and refunds. How that works.
Staying in Jinzhou
Our booking partner lists 22 places to stay in Jinzhou. Of the 22 with a star rating, the largest group is 3-star (8), and 2 are five-star. Guests have left 189 reviews in total.
Book a hotel in Jinzhou
Live rooms and prices from our booking partner, booked and paid for here. Free cancellation is marked where it applies.
Common questions
How far is Jinzhou from Yingkou?
104 km (65 mi) in a straight line, north-west of Yingkou; by road roughly 127 km (79 mi).
How long does it take to get from Yingkou to Jinzhou by road?
About 1 h 42 min, estimated from distance and typical speeds. Timetables and traffic decide the real figure.
Is there a time difference between Yingkou and Jinzhou?
No, both are on Asia/Shanghai at the moment.
Where can I stay in Jinzhou?
Our booking partner lists 22 places to stay in Jinzhou, bookable on this site.
Where these facts come from
- GeoNames (CC BY 4.0): population, elevation, time zone, postal codes
- Wikidata (CC0): area, sister cities, visitor numbers, heritage listings, stations and ports
- OurAirports (public domain): airports, runways
- Natural Earth (public domain): map outlines
Data compiled 14 September 2026. Hotel and hostel figures are from our booking partner’s catalogue on that date. Journey times on this page are estimates from distance, not timetables; check the operator before you rely on one.